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#721
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Is RGB to Lab lossy? - was( Lenses and sharpening)
On 10/5/2014 10:37 PM, Alan Browne wrote:
On 2014.10.05, 20:55 , PeterN wrote: On 10/5/2014 6:57 PM, Alan Browne wrote: On 2014.10.05, 14:42 , PeterN wrote: We went through all this some many months ago. I demonstrated clearly that the amount of 'loss' was negligible in practical terms. I would use the terem "color change." anstead of loss. Any change is a quality loss. Whether that is colour difference, tone, brightness, sharpness ... whatever, it's a loss. Then you are using a different definition of quality. Not at all. A non lossy process would have: RGB-A -- X-format -- RGB-B with RGB-A identical to RGB-B But - the fact is that with Lab RGB-A -- Lab -- RGB-B RGB-A =/= RGB-B, therefore there was quality loss. It seems to me that the assumption in that logic is: the quality of RGB-A quality of RGB-B. LAB has a larger color gamut than RGB. If there is no processing in LAB I would think that there would be no need for interpolation on the return trip. Do the round trip x + 10 times without processing and one might see a difference. It is doubtful that there will be a noticable difference from 10 round trips. Meanwhile there are color modification processes that are easier to perform in LAB than RGB. I would think that if the changes made in LAB created color outside the RGB gamut there would have to be some interpolation. The interpolation coud mae a better image, or it could make the changed image horrific. In another area, I have found images to be fine with a color cast, but when I remove the cast, to my eye the image looks horrific. -- PeterN |
#722
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Is RGB to Lab lossy? - was( Lenses and sharpening)
On 10/6/2014 4:31 AM, Eric Stevens wrote:
On Sun, 05 Oct 2014 21:05:19 -0400, nospam wrote: In article , PeterN wrote: I gaae him some common uses. He typically uses "edge case' to give him wriggle room. wrong again. what i call an edge case is an edge case and what you're calling common can be done *without* lab more easily and with better quality results. in other words, you're blaming others for your own lack of knowledge and unwillingness to learn. Well then explain with facts and detail. Warning. I have Dan's book and will use it as a reference. that's your problem. dan is wrong and reading his books has led you astray. if you read other books, you'll see that they consistently prove just how much of an idiot dan really is. i've mentioned two such books in this thread and other books in other threads. You seem to have a thing about Lab color. Your comments are quite unbalanced. You don't seem to have a good comment about *any* aspect of it. What happened to you? He just likes to argue. Even, when as pointed out earlier, it's against his own argument. -- PeterN |
#723
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Is RGB to Lab lossy? - was( Lenses and sharpening)
On 2014.10.06, 09:19 , PeterN wrote:
On 10/5/2014 10:37 PM, Alan Browne wrote: On 2014.10.05, 20:55 , PeterN wrote: On 10/5/2014 6:57 PM, Alan Browne wrote: On 2014.10.05, 14:42 , PeterN wrote: We went through all this some many months ago. I demonstrated clearly that the amount of 'loss' was negligible in practical terms. I would use the terem "color change." anstead of loss. Any change is a quality loss. Whether that is colour difference, tone, brightness, sharpness ... whatever, it's a loss. Then you are using a different definition of quality. Not at all. A non lossy process would have: RGB-A -- X-format -- RGB-B with RGB-A identical to RGB-B But - the fact is that with Lab RGB-A -- Lab -- RGB-B RGB-A =/= RGB-B, therefore there was quality loss. It seems to me that the assumption in that logic is: the quality of RGB-A quality of RGB-B. LAB has a larger color gamut than RGB. If there is no processing in LAB I would think that there would be no need for interpolation on the return trip. It's not about colour gamut since the "larger gamut" of LAB can't inherit anything better from the lesser gamut in the RGB image. If it was missing in the RGB it won't magically appear in the LAB. You'll just have more room to maneuver in the LAB version when you edit there. It's not about "need for interpolation" - it's about what changed. Any change is loss. Whether it occurred in step RGB- Lab or Lab-RGB is not relevant to the end product. If the end product changed then there was loss. Any deviation from the original, whether more or less colour, lighter or darker tone, etc. is loss of information (quality) from the original. Period. There is no interpretation to do. It's a strictly technical thing. From a quality standpoint it is loss. *to re-iterate: this is all quite pedantic as the actual amount of loss (change) is not visible in practical terms, edge cases aside. -- Among Broad Outlines, conception is far more pleasurable than “carrying [the children] to fruition.” Sadly, “there’s a high infant mortality rate among Broad Outlines—they often fall prey to Nonstarters.” "Bestiary of Intelligence Writing" - CIA |
#724
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Is RGB to Lab lossy? - was( Lenses and sharpening)
On 06/10/2014 14:19, PeterN wrote:
On 10/5/2014 10:37 PM, Alan Browne wrote: On 2014.10.05, 20:55 , PeterN wrote: On 10/5/2014 6:57 PM, Alan Browne wrote: On 2014.10.05, 14:42 , PeterN wrote: We went through all this some many months ago. I demonstrated clearly that the amount of 'loss' was negligible in practical terms. I would use the terem "color change." anstead of loss. Any change is a quality loss. Whether that is colour difference, tone, brightness, sharpness ... whatever, it's a loss. Then you are using a different definition of quality. Not at all. A non lossy process would have: RGB-A -- X-format -- RGB-B with RGB-A identical to RGB-B But - the fact is that with Lab RGB-A -- Lab -- RGB-B RGB-A =/= RGB-B, therefore there was quality loss. It seems to me that the assumption in that logic is: the quality of RGB-A quality of RGB-B. LAB has a larger color gamut than RGB. If there is no processing in LAB I would think that there would be no need for interpolation on the return trip. The problem is not one of interpolation but that there are unavoidable minor rounding errors in the nonlinear transform from RGB to CIELAB and also on the way back due to the finite representation of the results. See: http://en.wikipedia.org/wiki/Lab_color_space There is no assumption here beyond the very definition of a lossless transform. Applying a lossless transform and then its inverse to the result you must be able to get to back your original image *EXACTLY*. Do the round trip x + 10 times without processing and one might see a difference. It is doubtful that there will be a noticable difference from 10 round trips. Meanwhile there are color modification processes that are easier to perform in LAB than RGB. I would think that if the changes made in LAB created color outside the RGB gamut there would have to be some interpolation. The interpolation coud mae a better image, or it could make the changed image horrific. It isn't interpolation - if anything it is a quantisation effect with very different steps in CIELAB than in RGB. See for example: http://en.wikipedia.org/wiki/Lab_color_space Some pixel values may converge to an attractor on the first or second pass around the loop and after that do not change at all. A few unlucky ones may drift away from the original colour (but probably not by much). Out of gamut values pretty much end up clipped against the nearest RGB boundary on a minimise the absolute perceived colour error basis. In another area, I have found images to be fine with a color cast, but when I remove the cast, to my eye the image looks horrific. CIELAB does a better job of managing just noticeable visual differences and allows better adjustment of visual saturation and lightness. -- Regards, Martin Brown |
#725
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Is RGB to Lab lossy? - was( Lenses and sharpening)
On 2014.10.06, 11:33 , Martin Brown wrote:
On 06/10/2014 14:19, PeterN wrote: On 10/5/2014 10:37 PM, Alan Browne wrote: On 2014.10.05, 20:55 , PeterN wrote: On 10/5/2014 6:57 PM, Alan Browne wrote: On 2014.10.05, 14:42 , PeterN wrote: We went through all this some many months ago. I demonstrated clearly that the amount of 'loss' was negligible in practical terms. I would use the terem "color change." anstead of loss. Any change is a quality loss. Whether that is colour difference, tone, brightness, sharpness ... whatever, it's a loss. Then you are using a different definition of quality. Not at all. A non lossy process would have: RGB-A -- X-format -- RGB-B with RGB-A identical to RGB-B But - the fact is that with Lab RGB-A -- Lab -- RGB-B RGB-A =/= RGB-B, therefore there was quality loss. It seems to me that the assumption in that logic is: the quality of RGB-A quality of RGB-B. LAB has a larger color gamut than RGB. If there is no processing in LAB I would think that there would be no need for interpolation on the return trip. The problem is not one of interpolation but that there are unavoidable minor rounding errors in the nonlinear transform from RGB to CIELAB and also on the way back due to the finite representation of the results. See: Excellent point. -- Among Broad Outlines, conception is far more pleasurable than “carrying [the children] to fruition.” Sadly, “there’s a high infant mortality rate among Broad Outlines—they often fall prey to Nonstarters.” "Bestiary of Intelligence Writing" - CIA |
#726
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Is RGB to Lab lossy? - was( Lenses and sharpening)
On Sun, 05 Oct 2014 22:37:46 -0400, Alan Browne
wrote: On 2014.10.05, 20:55 , PeterN wrote: On 10/5/2014 6:57 PM, Alan Browne wrote: On 2014.10.05, 14:42 , PeterN wrote: We went through all this some many months ago. I demonstrated clearly that the amount of 'loss' was negligible in practical terms. I would use the terem "color change." anstead of loss. Any change is a quality loss. Whether that is colour difference, tone, brightness, sharpness ... whatever, it's a loss. Then you are using a different definition of quality. Not at all. A non lossy process would have: RGB-A -- X-format -- RGB-B with RGB-A identical to RGB-B But - the fact is that with Lab RGB-A -- Lab -- RGB-B RGB-A =/= RGB-B, therefore there was quality loss. But hang on: we do accept a certain degree of quality loss as part of the normal process of editing. It doesn't take much manipulation to turn a smooth histogram into something like http://pe-images.s3.amazonaws.com/ba.../fix-white.gif Push things a bit harder and you can get http://www.snoopy.me.uk/misc/365proj...gram/comb3.jpg or even https://aperture64.files.wordpress.c...09/combing.gif The production of histograms like the first one is common and generally acceptable. The second histogram is worse but even then may be acceptable. Only the last one is so bad that it will nearly always be unacceptable. The point of all this is that some degree of quality loss is virtually inevitable as soon as you start to manipulate an image. In the context of the present discussion, the question is, does the conversion to Lab colour incur any more damage than one can expect in the course of ordinary editing? My understanding of nospam's claim is that it does. My (admittedly limited) experience with it suggests that conversion to Lab causes no significant damage; certainly less than I am going to inflict on the image by the changes I want to make. As to the extent of the damage, I can only refer to my original experiment described in Message-ID: ------------------------------------------ This one continues to bother me. I am still inclined to agree with Dan Margulis. I'm not quite sure what procedure Andrew Rodney is proposing to prove his point so, using Photoshop CC, I have carried out my own test as follows: 1. Find a JPG with a suitable range of colors. This one came from my wife's collection: https://dl.dropboxusercontent.com/u/...20IMG_2154.jpg I saved a copy as a PSD (see below for the reason). 2. Copy and convert to Lab. I couldn't save to JPG from Lab so I saved to PSD. See https://dl.dropboxusercontent.com/u/...54-via-Lab.jpg 3. I then loaded the two PSD files into a new file as separate layers. (1) above was the background layer and (2) was the next. I subtracted the 2nd layer from the first with the result shown in https://dl.dropboxusercontent.com/u/...Difference.jpg That's right: solid black. 4. To confirm the point I took a screen shot. See https://dl.dropboxusercontent.com/u/...t%20Screen.jpg Note the histogram. All of the pixels appear to be down at the zero end of the scale: that is, jet black. The only conclusion I can reach is that there is no difference between a PSD created from a RGB file and a PSD created from the same image when it has first been converted from RGB to Lab. I'm not wedded to the perfection of the method I have used and I would be interested to hear from anyone who has a meaningful criticism. -------------------------------- and my later comments in Message-ID: ======================== I've just compared the original JPG with a copy -- Lab -- JPG again. JPGs are RGB are they not? Anyway I still got an apparently all-black screen and here is the screen shot showing the histogram: https://dl.dropboxusercontent.com/u/...screen%202.jpg An even tighter all-black bar than previously. ============================= The histogram of the JPG(RBG) --- Lab --- PSD when compared with JPG(RBG) --- PSD shows only very slight evidence of differences between the two. However JPG(RBG) --- Lab --- JPG(RBG) appears to be identical to the original JPG(RBG). So far I don't think I have found any evidence of damage worth worrying about. -- Regards, Eric Stevens |
#727
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Is RGB to Lab lossy? - was( Lenses and sharpening)
On 2014.10.06, 17:27 , Eric Stevens wrote:
On Sun, 05 Oct 2014 22:37:46 -0400, Alan Browne wrote: On 2014.10.05, 20:55 , PeterN wrote: On 10/5/2014 6:57 PM, Alan Browne wrote: On 2014.10.05, 14:42 , PeterN wrote: We went through all this some many months ago. I demonstrated clearly that the amount of 'loss' was negligible in practical terms. I would use the terem "color change." anstead of loss. Any change is a quality loss. Whether that is colour difference, tone, brightness, sharpness ... whatever, it's a loss. Then you are using a different definition of quality. Not at all. A non lossy process would have: RGB-A -- X-format -- RGB-B with RGB-A identical to RGB-B But - the fact is that with Lab RGB-A -- Lab -- RGB-B RGB-A =/= RGB-B, therefore there was quality loss. But hang on: we do accept a certain degree of quality loss as part of the normal process of editing. It doesn't take much manipulation to turn a smooth histogram into something like http://pe-images.s3.amazonaws.com/ba.../fix-white.gif Push things a bit harder and you can get http://www.snoopy.me.uk/misc/365proj...gram/comb3.jpg or even https://aperture64.files.wordpress.c...09/combing.gif The 'issue' refers to the questions: "If I take my JPG and throw it into LAB ('cause I want to do something easier done there) and then throw it back, is there a loss? Is it important?" 1. Yes. 2. Negligible. So in a "normal process of editing" where one goes from a high quality image (raw) to the Adobe "editing space" format and then save as a: PSD: no loss (other than editing effects) TIFF: no loss (other than editing effects) JPG: lossy But if one went to Lab space and back along the way, then it will always be lossy even if nothing was done in Lab space. The production of histograms like the first one is common and generally acceptable. The second histogram is worse but even then may be acceptable. Only the last one is so bad that it will nearly always be unacceptable. The point of all this is that some degree of quality loss is virtually inevitable as soon as you start to manipulate an image. In the context of the present discussion, the question is, does the conversion to Lab colour incur any more damage than one can expect in the course of ordinary editing? My understanding of nospam's claim is that it does. My (admittedly limited) experience with it suggests that conversion to Lab causes no significant damage; certainly less than I am going to inflict on the image by the changes I want to make. As to the extent of the damage, I can only refer to my original experiment described in Message-ID: ------------------------------------------ This one continues to bother me. I am still inclined to agree with Dan Margulis. I'm not quite sure what procedure Andrew Rodney is proposing to prove his point so, using Photoshop CC, I have carried out my own test as follows: 1. Find a JPG with a suitable range of colors. This one came from my wife's collection: https://dl.dropboxusercontent.com/u/...20IMG_2154.jpg I saved a copy as a PSD (see below for the reason). 2. Copy and convert to Lab. I couldn't save to JPG from Lab so I saved to PSD. See https://dl.dropboxusercontent.com/u/...54-via-Lab.jpg 3. I then loaded the two PSD files into a new file as separate layers. (1) above was the background layer and (2) was the next. I subtracted the 2nd layer from the first with the result shown in https://dl.dropboxusercontent.com/u/...Difference.jpg That's right: solid black. 4. To confirm the point I took a screen shot. See https://dl.dropboxusercontent.com/u/...t%20Screen.jpg Note the histogram. All of the pixels appear to be down at the zero end of the scale: that is, jet black. The only conclusion I can reach is that there is no difference between a PSD created from a RGB file and a PSD created from the same image when it has first been converted from RGB to Lab. I'm not wedded to the perfection of the method I have used and I would be interested to hear from anyone who has a meaningful criticism. I'll try your method above when I have a chance. I've gone through this exercise in the past and the difference (by subtraction) was visible (faint, but unmistakable). https://www.dropbox.com/s/3uwyuwun56...02-SD.jpg?dl=0 The only issues a 1. Lossy formats. Repeatedly opening and then saving a JPG at a lower quality level will increase loss. This will be visible at some point. 2. Format conversion. For all the mentioned reasons, including the quantization error the Martin Brown pointed out, there is a change in the image and therefore it is quality loss. There is no other term. ============================= The histogram of the JPG(RBG) --- Lab --- PSD when compared with JPG(RBG) --- PSD shows only very slight evidence of differences between the two. The better way to see is to do image subtraction. While a histo may bear witness to change, an imaage subtraction will always bear witness. However JPG(RBG) --- Lab --- JPG(RBG) appears to be identical to the original JPG(RBG). So far I don't think I have found any evidence of damage worth worrying about. Never said different. -- Among Broad Outlines, conception is far more pleasurable than “carrying [the children] to fruition.” Sadly, “there’s a high infant mortality rate among Broad Outlines—they often fall prey to Nonstarters.” "Bestiary of Intelligence Writing" - CIA |
#728
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Is RGB to Lab lossy? - was( Lenses and sharpening)
On Mon, 06 Oct 2014 17:42:50 -0400, Alan Browne
wrote: On 2014.10.06, 17:27 , Eric Stevens wrote: On Sun, 05 Oct 2014 22:37:46 -0400, Alan Browne wrote: On 2014.10.05, 20:55 , PeterN wrote: On 10/5/2014 6:57 PM, Alan Browne wrote: On 2014.10.05, 14:42 , PeterN wrote: We went through all this some many months ago. I demonstrated clearly that the amount of 'loss' was negligible in practical terms. I would use the terem "color change." anstead of loss. Any change is a quality loss. Whether that is colour difference, tone, brightness, sharpness ... whatever, it's a loss. Then you are using a different definition of quality. Not at all. A non lossy process would have: RGB-A -- X-format -- RGB-B with RGB-A identical to RGB-B But - the fact is that with Lab RGB-A -- Lab -- RGB-B RGB-A =/= RGB-B, therefore there was quality loss. But hang on: we do accept a certain degree of quality loss as part of the normal process of editing. It doesn't take much manipulation to turn a smooth histogram into something like http://pe-images.s3.amazonaws.com/ba.../fix-white.gif Push things a bit harder and you can get http://www.snoopy.me.uk/misc/365proj...gram/comb3.jpg or even https://aperture64.files.wordpress.c...09/combing.gif The 'issue' refers to the questions: "If I take my JPG and throw it into LAB ('cause I want to do something easier done there) and then throw it back, is there a loss? Is it important?" 1. Yes. 2. Negligible. So in a "normal process of editing" where one goes from a high quality image (raw) to the Adobe "editing space" format and then save as a: PSD: no loss (other than editing effects) TIFF: no loss (other than editing effects) JPG: lossy But if one went to Lab space and back along the way, then it will always be lossy even if nothing was done in Lab space. True, but as I found in my experiments (as described again, below) the loss on conversion is close to zero. The argument is not whether or not there is any loss in going through Lab space but whether or not the loss is significant. nospam seems to equate even the smallest loss arising from Lab conversion as significant but he forgets that the fact that he has loaded the image into an editor is going to wreak considerably more damage to the original image. That's why I think he is talking nondense when he advocates not using Lab so as to avoid damage. The production of histograms like the first one is common and generally acceptable. The second histogram is worse but even then may be acceptable. Only the last one is so bad that it will nearly always be unacceptable. The point of all this is that some degree of quality loss is virtually inevitable as soon as you start to manipulate an image. In the context of the present discussion, the question is, does the conversion to Lab colour incur any more damage than one can expect in the course of ordinary editing? My understanding of nospam's claim is that it does. My (admittedly limited) experience with it suggests that conversion to Lab causes no significant damage; certainly less than I am going to inflict on the image by the changes I want to make. As to the extent of the damage, I can only refer to my original experiment described in Message-ID: ------------------------------------------ This one continues to bother me. I am still inclined to agree with Dan Margulis. I'm not quite sure what procedure Andrew Rodney is proposing to prove his point so, using Photoshop CC, I have carried out my own test as follows: 1. Find a JPG with a suitable range of colors. This one came from my wife's collection: https://dl.dropboxusercontent.com/u/...20IMG_2154.jpg I saved a copy as a PSD (see below for the reason). 2. Copy and convert to Lab. I couldn't save to JPG from Lab so I saved to PSD. See https://dl.dropboxusercontent.com/u/...54-via-Lab.jpg 3. I then loaded the two PSD files into a new file as separate layers. (1) above was the background layer and (2) was the next. I subtracted the 2nd layer from the first with the result shown in https://dl.dropboxusercontent.com/u/...Difference.jpg That's right: solid black. 4. To confirm the point I took a screen shot. See https://dl.dropboxusercontent.com/u/...t%20Screen.jpg Note the histogram. All of the pixels appear to be down at the zero end of the scale: that is, jet black. The only conclusion I can reach is that there is no difference between a PSD created from a RGB file and a PSD created from the same image when it has first been converted from RGB to Lab. I'm not wedded to the perfection of the method I have used and I would be interested to hear from anyone who has a meaningful criticism. I'll try your method above when I have a chance. I've gone through this exercise in the past and the difference (by subtraction) was visible (faint, but unmistakable). https://www.dropbox.com/s/3uwyuwun56...02-SD.jpg?dl=0 The only issues a 1. Lossy formats. Repeatedly opening and then saving a JPG at a lower quality level will increase loss. This will be visible at some point. 2. Format conversion. For all the mentioned reasons, including the quantization error the Martin Brown pointed out, there is a change in the image and therefore it is quality loss. There is no other term. ============================= The histogram of the JPG(RBG) --- Lab --- PSD when compared with JPG(RBG) --- PSD shows only very slight evidence of differences between the two. The better way to see is to do image subtraction. While a histo may bear witness to change, an imaage subtraction will always bear witness. However JPG(RBG) --- Lab --- JPG(RBG) appears to be identical to the original JPG(RBG). So far I don't think I have found any evidence of damage worth worrying about. Never said different. -- Regards, Eric Stevens |
#729
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Is RGB to Lab lossy? - was( Lenses and sharpening)
On 06/10/2014 22:27, Eric Stevens wrote:
On Sun, 05 Oct 2014 22:37:46 -0400, Alan Browne wrote: On 2014.10.05, 20:55 , PeterN wrote: On 10/5/2014 6:57 PM, Alan Browne wrote: On 2014.10.05, 14:42 , PeterN wrote: We went through all this some many months ago. I demonstrated clearly that the amount of 'loss' was negligible in practical terms. I would use the terem "color change." anstead of loss. Any change is a quality loss. Whether that is colour difference, tone, brightness, sharpness ... whatever, it's a loss. Then you are using a different definition of quality. Not at all. A non lossy process would have: RGB-A -- X-format -- RGB-B with RGB-A identical to RGB-B But - the fact is that with Lab RGB-A -- Lab -- RGB-B RGB-A =/= RGB-B, therefore there was quality loss. But hang on: we do accept a certain degree of quality loss as part of the normal process of editing. It doesn't take much manipulation to turn a smooth histogram into something like http://pe-images.s3.amazonaws.com/ba.../fix-white.gif Push things a bit harder and you can get http://www.snoopy.me.uk/misc/365proj...gram/comb3.jpg or even https://aperture64.files.wordpress.c...09/combing.gif The production of histograms like the first one is common and generally acceptable. The second histogram is worse but even then may be acceptable. Only the last one is so bad that it will nearly always be unacceptable. The point of all this is that some degree of quality loss is virtually inevitable as soon as you start to manipulate an image. In the context of the present discussion, the question is, does the conversion to Lab colour incur any more damage than one can expect in the course of ordinary editing? My understanding of nospam's claim is that it does. My (admittedly limited) experience with it suggests that conversion to Lab causes no significant damage; certainly less than I am going to inflict on the image by the changes I want to make. Since CIELAB is a colour space intended to manage just noticeably colour differences more optimally than the naive RGB colour space it isn't too surprising that you cannot *see* a difference in the final JPG taken from RGB or via CIELAB. But they are very slightly different. As to the extent of the damage, I can only refer to my original experiment described in Message-ID: ------------------------------------------ This one continues to bother me. I am still inclined to agree with Dan Margulis. I'm not quite sure what procedure Andrew Rodney is proposing to prove his point so, using Photoshop CC, I have carried out my own test as follows: 1. Find a JPG with a suitable range of colors. This one came from my wife's collection: https://dl.dropboxusercontent.com/u/...20IMG_2154.jpg I saved a copy as a PSD (see below for the reason). 2. Copy and convert to Lab. I couldn't save to JPG from Lab so I saved to PSD. See https://dl.dropboxusercontent.com/u/...54-via-Lab.jpg 3. I then loaded the two PSD files into a new file as separate layers. (1) above was the background layer and (2) was the next. I subtracted the 2nd layer from the first with the result shown in https://dl.dropboxusercontent.com/u/...Difference.jpg That's right: solid black. It is only solid black so long as you don't use the histogram tool to look in detail at the noise floor (also I am not sure how you did the differencing - you may be missing half the differences if you did a simple subtraction which clipped to zero as opposed to an absolute difference where any discrepancy is rendered as a positive difference). Use the histogram tool and you will see that the images do differ in the luminance least significant bit. This would be undetectable in practice but it is non-the-less a difference (ie not lossless). 4. To confirm the point I took a screen shot. See https://dl.dropboxusercontent.com/u/...t%20Screen.jpg Note the histogram. All of the pixels appear to be down at the zero end of the scale: that is, jet black. The only conclusion I can reach is that there is no difference between a PSD created from a RGB file and a PSD created from the same image when it has first been converted from RGB to Lab. In practice you will not be able to see the difference and without pixel peeping you can't see the difference on a simple difference image but it is still there - just below your visual threshold. I'm not wedded to the perfection of the method I have used and I would be interested to hear from anyone who has a meaningful criticism. The only thing that did surprise me was that the resulting errors are entirely in luminance there is no chroma noise introduced at all. (this might be an artefact of how you did the differencing) -- Regards, Martin Brown |
#730
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Is RGB to Lab lossy? - was( Lenses and sharpening)
On Tue, 07 Oct 2014 08:44:45 +0100, Martin Brown
wrote: --- snip --- In the context of the present discussion, the question is, does the conversion to Lab colour incur any more damage than one can expect in the course of ordinary editing? My understanding of nospam's claim is that it does. My (admittedly limited) experience with it suggests that conversion to Lab causes no significant damage; certainly less than I am going to inflict on the image by the changes I want to make. Since CIELAB is a colour space intended to manage just noticeably colour differences more optimally than the naive RGB colour space it isn't too surprising that you cannot *see* a difference in the final JPG taken from RGB or via CIELAB. But they are very slightly different. Agreed, but the question is, does the difference matter? In particular does it matter enough to earn the reputation that nospam is trying to assign to it? I would generally answer 'no' to both of thos questions. As to the extent of the damage, I can only refer to my original experiment described in Message-ID: ------------------------------------------ This one continues to bother me. I am still inclined to agree with Dan Margulis. I'm not quite sure what procedure Andrew Rodney is proposing to prove his point so, using Photoshop CC, I have carried out my own test as follows: 1. Find a JPG with a suitable range of colors. This one came from my wife's collection: https://dl.dropboxusercontent.com/u/...20IMG_2154.jpg I saved a copy as a PSD (see below for the reason). 2. Copy and convert to Lab. I couldn't save to JPG from Lab so I saved to PSD. See https://dl.dropboxusercontent.com/u/...54-via-Lab.jpg 3. I then loaded the two PSD files into a new file as separate layers. (1) above was the background layer and (2) was the next. I subtracted the 2nd layer from the first with the result shown in https://dl.dropboxusercontent.com/u/...Difference.jpg That's right: solid black. It is only solid black so long as you don't use the histogram tool to look in detail at the noise floor (also I am not sure how you did the differencing - you may be missing half the differences if you did a simple subtraction which clipped to zero as opposed to an absolute difference where any discrepancy is rendered as a positive difference). Use the histogram tool and you will see that the images do differ in the luminance least significant bit. This would be undetectable in practice but it is non-the-less a difference (ie not lossless). 4. To confirm the point I took a screen shot. See https://dl.dropboxusercontent.com/u/...t%20Screen.jpg Note the histogram. All of the pixels appear to be down at the zero end of the scale: that is, jet black. The only conclusion I can reach is that there is no difference between a PSD created from a RGB file and a PSD created from the same image when it has first been converted from RGB to Lab. In practice you will not be able to see the difference and without pixel peeping you can't see the difference on a simple difference image but it is still there - just below your visual threshold. The histogram of a subtraction will give you the answer. I'm not wedded to the perfection of the method I have used and I would be interested to hear from anyone who has a meaningful criticism. The only thing that did surprise me was that the resulting errors are entirely in luminance there is no chroma noise introduced at all. (this might be an artefact of how you did the differencing) I did nothing fancy at all. -- Regards, Eric Stevens |
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