How to convert a 4-channel image to 3-channel image in an elegant way?

I have a 4-channel image (RGBA). I want to reduce its channel count from 4 to 3.

The method I am using currently is to Right click and Save as a .png file. After that when I Import it in Mathematica, the channel count is reduced to 3. Is there a more elegant way to do that?

Note: RemoveAlphaChannel function won't work. Because that will change the appearance of the original image (see the example below).

This is a 4-channel picture from ref / AlphaChannel:

4-channel picture

 img = Import @ "https://i.stack.imgur.com/XvzDc.png"

This is the 3-channel picture which is what I want:

3-channel picture

If I use RemoveAlphaChannel, I will get this picture instead:

Picture after RemoveAlphaChannel

edited Jan 24 at 8:13

Andreas Rejbrand

10314

asked Jan 24 at 6:59

Voldikss

709

3

$begingroup$
does RemoveAlphaChannel[img, White] not work for you? it does give 3-channel image ( ImageChannels@RemoveAlphaChannel[img, White] is 3).
$endgroup$
– kglr
Jan 24 at 7:26

$begingroup$
@kglr Yes it works for that situation. It's amazing, thank you.
$endgroup$
– Voldikss
Jan 24 at 7:57

add a comment |

I have a 4-channel image (RGBA). I want to reduce its channel count from 4 to 3.

Note: RemoveAlphaChannel function won't work. Because that will change the appearance of the original image (see the example below).

This is a 4-channel picture from ref / AlphaChannel:

4-channel picture

 img = Import @ "https://i.stack.imgur.com/XvzDc.png"

This is the 3-channel picture which is what I want:

3-channel picture

If I use RemoveAlphaChannel, I will get this picture instead:

Picture after RemoveAlphaChannel

edited Jan 24 at 8:13

Andreas Rejbrand

10314

asked Jan 24 at 6:59

Voldikss

709

3

$begingroup$
does RemoveAlphaChannel[img, White] not work for you? it does give 3-channel image ( ImageChannels@RemoveAlphaChannel[img, White] is 3).
$endgroup$
– kglr
Jan 24 at 7:26

$begingroup$
@kglr Yes it works for that situation. It's amazing, thank you.
$endgroup$
– Voldikss
Jan 24 at 7:57

add a comment |

I have a 4-channel image (RGBA). I want to reduce its channel count from 4 to 3.

Note: RemoveAlphaChannel function won't work. Because that will change the appearance of the original image (see the example below).

This is a 4-channel picture from ref / AlphaChannel:

4-channel picture

 img = Import @ "https://i.stack.imgur.com/XvzDc.png"

This is the 3-channel picture which is what I want:

3-channel picture

If I use RemoveAlphaChannel, I will get this picture instead:

Picture after RemoveAlphaChannel

edited Jan 24 at 8:13

Andreas Rejbrand

10314

asked Jan 24 at 6:59

Voldikss

709

I have a 4-channel image (RGBA). I want to reduce its channel count from 4 to 3.

Note: RemoveAlphaChannel function won't work. Because that will change the appearance of the original image (see the example below).

This is a 4-channel picture from ref / AlphaChannel:

4-channel picture

 img = Import @ "https://i.stack.imgur.com/XvzDc.png"

This is the 3-channel picture which is what I want:

3-channel picture

If I use RemoveAlphaChannel, I will get this picture instead:

Picture after RemoveAlphaChannel

image-processing image

edited Jan 24 at 8:13

Andreas Rejbrand

10314

asked Jan 24 at 6:59

Voldikss

709

edited Jan 24 at 8:13

Andreas Rejbrand

10314

asked Jan 24 at 6:59

Voldikss

709

edited Jan 24 at 8:13

Andreas Rejbrand

10314

edited Jan 24 at 8:13

Andreas Rejbrand

10314

edited Jan 24 at 8:13

Andreas Rejbrand

10314

asked Jan 24 at 6:59

Voldikss

709

asked Jan 24 at 6:59

Voldikss

709

asked Jan 24 at 6:59

Voldikss

709

3

$begingroup$
does RemoveAlphaChannel[img, White] not work for you? it does give 3-channel image ( ImageChannels@RemoveAlphaChannel[img, White] is 3).
$endgroup$
– kglr
Jan 24 at 7:26

$begingroup$
@kglr Yes it works for that situation. It's amazing, thank you.
$endgroup$
– Voldikss
Jan 24 at 7:57

add a comment |

3

$begingroup$
does RemoveAlphaChannel[img, White] not work for you? it does give 3-channel image ( ImageChannels@RemoveAlphaChannel[img, White] is 3).
$endgroup$
– kglr
Jan 24 at 7:26

$begingroup$
@kglr Yes it works for that situation. It's amazing, thank you.
$endgroup$
– Voldikss
Jan 24 at 7:57

does RemoveAlphaChannel[img, White] not work for you? it does give 3-channel image ( ImageChannels@RemoveAlphaChannel[img, White] is 3).

– kglr
Jan 24 at 7:26

@kglr Yes it works for that situation. It's amazing, thank you.

– Voldikss
Jan 24 at 7:57

add a comment |

2 Answers
2

active

oldest

votes

The second example from RemoveAlphaChannel:

Remove opacity by blending with a white background:

img2 = RemoveAlphaChannel[img, White]

enter image description here

ImageChannels /@ {img, img2}

{4, 3}

ColorSeparate /@ {img, img2} // Grid

enter image description here

Row[{Labeled[Framed@AlphaChannel@img, "img", Top], 

     Labeled[Framed@AlphaChannel@img2, "img2", Top]}]

enter image description here

edited Jan 24 at 7:45

answered Jan 24 at 7:36

kglr

189k10205422

$begingroup$
Thank you very much @kglr. I wonder if RemoveAlphaChannel[#, White]& is universal. If it also works for various alpha channels. (I've tried some pictures with different alpha channels and it works, but I am still not sure whether it works for all the cases...)
$endgroup$
– Voldikss
Jan 24 at 7:54

$begingroup$
@VoldikSS, I am not sure if it works for all cases.
$endgroup$
– kglr
Jan 24 at 8:04

add a comment |

@kglr 's solution is very convenient and powerful. However RemoveAlphaChannel[#, colorname]& does not work for all the cases. So I spent some time searching in the web about image channels and found the principle of the alpha channel.

Based on that, I get the following solution which can used for other cases(different images with different alpha channels).

reduceImageChannels[img_] :=

 Map[Most,

    Map[(#*#[[4]] + 1*(1 - #[[4]])) &,

     (Flatten[ImageData@img, 1])]] // Partition[#, First@ImageDimensions[img]] & // Image

To check if it works

img = Import @ "https://i.stack.imgur.com/XvzDc.png"

enter image description here

img2 = reduceImageChannels@img

enter image description here

ImageChannels@img2

3

It given me the image exactly what I want. Exciting!

BTW, I am a mathematica newbie, so be free to correct my code and English grammer.

edited Jan 26 at 13:46

answered Jan 26 at 13:37

Voldikss

709

$begingroup$
I haven't ran the code, but Most gives all but the last element of a list. And what is 120 doing there, is that something to do with the ImageDimensions?
$endgroup$
– KraZug
Jan 26 at 13:42

$begingroup$
Thank you for the suggestion! I will replace Drop[#, {4}] & with Most. You are right, ImageDimensions@img returns {120,120}
$endgroup$
– Voldikss
Jan 26 at 13:45

$begingroup$
you can also use reduceImageChannels2[img_] := Map[Most[#*#[[4]] + 1*(1 - #[[4]])] &, ImageData@img, {2}]
$endgroup$
– kglr
Jan 26 at 13:56

$begingroup$
@kglr Thanks, it's much more elegant! I even forgot the Map's levelspec option...
$endgroup$
– Voldikss
Jan 26 at 13:59

1

$begingroup$
for an arbitrary color in the second argument,say Orange, RemoveAlphaChannel[img, Orange] and ImageApply[Most[#]*#[[4]] + {1, 0.5, 0}*(1 - #[[4]]) &, img] gives the same image as ImageApply[Most[#]*#[[4]] + {1, 0.5, 0}*(1 - #[[4]]) &, img] (that is, Max@(ImageData@ ImageApply[Most[#]*#[[4]] + {1, 0.5, 0}*(1 - #[[4]]) &, img] - ImageData[RemoveAlphaChannel[img, Orange]]) .) Note that Orange is RGBColor[1,.5,0].
$endgroup$
– kglr
Jan 26 at 14:26

|
show 6 more comments

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

The second example from RemoveAlphaChannel:

Remove opacity by blending with a white background:

img2 = RemoveAlphaChannel[img, White]

enter image description here

ImageChannels /@ {img, img2}

{4, 3}

ColorSeparate /@ {img, img2} // Grid

enter image description here

Row[{Labeled[Framed@AlphaChannel@img, "img", Top], 

     Labeled[Framed@AlphaChannel@img2, "img2", Top]}]

enter image description here

edited Jan 24 at 7:45

answered Jan 24 at 7:36

kglr

189k10205422

$begingroup$
Thank you very much @kglr. I wonder if RemoveAlphaChannel[#, White]& is universal. If it also works for various alpha channels. (I've tried some pictures with different alpha channels and it works, but I am still not sure whether it works for all the cases...)
$endgroup$
– Voldikss
Jan 24 at 7:54

$begingroup$
@VoldikSS, I am not sure if it works for all cases.
$endgroup$
– kglr
Jan 24 at 8:04

add a comment |

The second example from RemoveAlphaChannel:

Remove opacity by blending with a white background:

img2 = RemoveAlphaChannel[img, White]

enter image description here

ImageChannels /@ {img, img2}

{4, 3}

ColorSeparate /@ {img, img2} // Grid

enter image description here

Row[{Labeled[Framed@AlphaChannel@img, "img", Top], 

     Labeled[Framed@AlphaChannel@img2, "img2", Top]}]

enter image description here

edited Jan 24 at 7:45

answered Jan 24 at 7:36

kglr

189k10205422

$begingroup$
Thank you very much @kglr. I wonder if RemoveAlphaChannel[#, White]& is universal. If it also works for various alpha channels. (I've tried some pictures with different alpha channels and it works, but I am still not sure whether it works for all the cases...)
$endgroup$
– Voldikss
Jan 24 at 7:54

$begingroup$
@VoldikSS, I am not sure if it works for all cases.
$endgroup$
– kglr
Jan 24 at 8:04

add a comment |

The second example from RemoveAlphaChannel:

Remove opacity by blending with a white background:

img2 = RemoveAlphaChannel[img, White]

enter image description here

ImageChannels /@ {img, img2}

{4, 3}

ColorSeparate /@ {img, img2} // Grid

enter image description here

Row[{Labeled[Framed@AlphaChannel@img, "img", Top], 

     Labeled[Framed@AlphaChannel@img2, "img2", Top]}]

enter image description here

edited Jan 24 at 7:45

answered Jan 24 at 7:36

kglr

189k10205422

The second example from RemoveAlphaChannel:

Remove opacity by blending with a white background:

img2 = RemoveAlphaChannel[img, White]

enter image description here

ImageChannels /@ {img, img2}

{4, 3}

ColorSeparate /@ {img, img2} // Grid

enter image description here

Row[{Labeled[Framed@AlphaChannel@img, "img", Top], 

     Labeled[Framed@AlphaChannel@img2, "img2", Top]}]

enter image description here

edited Jan 24 at 7:45

answered Jan 24 at 7:36

kglr

189k10205422

edited Jan 24 at 7:45

answered Jan 24 at 7:36

kglr

189k10205422

answered Jan 24 at 7:36

kglr

189k10205422

answered Jan 24 at 7:36

kglr

189k10205422

$begingroup$
Thank you very much @kglr. I wonder if RemoveAlphaChannel[#, White]& is universal. If it also works for various alpha channels. (I've tried some pictures with different alpha channels and it works, but I am still not sure whether it works for all the cases...)
$endgroup$
– Voldikss
Jan 24 at 7:54

$begingroup$
@VoldikSS, I am not sure if it works for all cases.
$endgroup$
– kglr
Jan 24 at 8:04

add a comment |

$begingroup$
Thank you very much @kglr. I wonder if RemoveAlphaChannel[#, White]& is universal. If it also works for various alpha channels. (I've tried some pictures with different alpha channels and it works, but I am still not sure whether it works for all the cases...)
$endgroup$
– Voldikss
Jan 24 at 7:54

$begingroup$
@VoldikSS, I am not sure if it works for all cases.
$endgroup$
– kglr
Jan 24 at 8:04

Thank you very much @kglr. I wonder if RemoveAlphaChannel[#, White]& is universal. If it also works for various alpha channels. (I've tried some pictures with different alpha channels and it works, but I am still not sure whether it works for all the cases...)

– Voldikss
Jan 24 at 7:54

@VoldikSS, I am not sure if it works for all cases.

– kglr
Jan 24 at 8:04

add a comment |

Based on that, I get the following solution which can used for other cases(different images with different alpha channels).

reduceImageChannels[img_] :=

 Map[Most,

    Map[(#*#[[4]] + 1*(1 - #[[4]])) &,

     (Flatten[ImageData@img, 1])]] // Partition[#, First@ImageDimensions[img]] & // Image

To check if it works

img = Import @ "https://i.stack.imgur.com/XvzDc.png"

enter image description here

img2 = reduceImageChannels@img

enter image description here

ImageChannels@img2

3

It given me the image exactly what I want. Exciting!

BTW, I am a mathematica newbie, so be free to correct my code and English grammer.

edited Jan 26 at 13:46

answered Jan 26 at 13:37

Voldikss

709

$begingroup$
I haven't ran the code, but Most gives all but the last element of a list. And what is 120 doing there, is that something to do with the ImageDimensions?
$endgroup$
– KraZug
Jan 26 at 13:42

$begingroup$
Thank you for the suggestion! I will replace Drop[#, {4}] & with Most. You are right, ImageDimensions@img returns {120,120}
$endgroup$
– Voldikss
Jan 26 at 13:45

$begingroup$
you can also use reduceImageChannels2[img_] := Map[Most[#*#[[4]] + 1*(1 - #[[4]])] &, ImageData@img, {2}]
$endgroup$
– kglr
Jan 26 at 13:56

$begingroup$
@kglr Thanks, it's much more elegant! I even forgot the Map's levelspec option...
$endgroup$
– Voldikss
Jan 26 at 13:59

1

$begingroup$
for an arbitrary color in the second argument,say Orange, RemoveAlphaChannel[img, Orange] and ImageApply[Most[#]*#[[4]] + {1, 0.5, 0}*(1 - #[[4]]) &, img] gives the same image as ImageApply[Most[#]*#[[4]] + {1, 0.5, 0}*(1 - #[[4]]) &, img] (that is, Max@(ImageData@ ImageApply[Most[#]*#[[4]] + {1, 0.5, 0}*(1 - #[[4]]) &, img] - ImageData[RemoveAlphaChannel[img, Orange]]) .) Note that Orange is RGBColor[1,.5,0].
$endgroup$
– kglr
Jan 26 at 14:26

|
show 6 more comments

Based on that, I get the following solution which can used for other cases(different images with different alpha channels).

reduceImageChannels[img_] :=

 Map[Most,

    Map[(#*#[[4]] + 1*(1 - #[[4]])) &,

     (Flatten[ImageData@img, 1])]] // Partition[#, First@ImageDimensions[img]] & // Image

To check if it works

img = Import @ "https://i.stack.imgur.com/XvzDc.png"

enter image description here

img2 = reduceImageChannels@img

enter image description here

ImageChannels@img2

3

It given me the image exactly what I want. Exciting!

BTW, I am a mathematica newbie, so be free to correct my code and English grammer.

edited Jan 26 at 13:46

answered Jan 26 at 13:37

Voldikss

709

$begingroup$
I haven't ran the code, but Most gives all but the last element of a list. And what is 120 doing there, is that something to do with the ImageDimensions?
$endgroup$
– KraZug
Jan 26 at 13:42

$begingroup$
Thank you for the suggestion! I will replace Drop[#, {4}] & with Most. You are right, ImageDimensions@img returns {120,120}
$endgroup$
– Voldikss
Jan 26 at 13:45

$begingroup$
you can also use reduceImageChannels2[img_] := Map[Most[#*#[[4]] + 1*(1 - #[[4]])] &, ImageData@img, {2}]
$endgroup$
– kglr
Jan 26 at 13:56

$begingroup$
@kglr Thanks, it's much more elegant! I even forgot the Map's levelspec option...
$endgroup$
– Voldikss
Jan 26 at 13:59

1

$begingroup$
for an arbitrary color in the second argument,say Orange, RemoveAlphaChannel[img, Orange] and ImageApply[Most[#]*#[[4]] + {1, 0.5, 0}*(1 - #[[4]]) &, img] gives the same image as ImageApply[Most[#]*#[[4]] + {1, 0.5, 0}*(1 - #[[4]]) &, img] (that is, Max@(ImageData@ ImageApply[Most[#]*#[[4]] + {1, 0.5, 0}*(1 - #[[4]]) &, img] - ImageData[RemoveAlphaChannel[img, Orange]]) .) Note that Orange is RGBColor[1,.5,0].
$endgroup$
– kglr
Jan 26 at 14:26

|
show 6 more comments

Based on that, I get the following solution which can used for other cases(different images with different alpha channels).

reduceImageChannels[img_] :=

 Map[Most,

    Map[(#*#[[4]] + 1*(1 - #[[4]])) &,

     (Flatten[ImageData@img, 1])]] // Partition[#, First@ImageDimensions[img]] & // Image

To check if it works

img = Import @ "https://i.stack.imgur.com/XvzDc.png"

enter image description here

img2 = reduceImageChannels@img

enter image description here

ImageChannels@img2

3

It given me the image exactly what I want. Exciting!

BTW, I am a mathematica newbie, so be free to correct my code and English grammer.

edited Jan 26 at 13:46

answered Jan 26 at 13:37

Voldikss

709

Based on that, I get the following solution which can used for other cases(different images with different alpha channels).

reduceImageChannels[img_] :=

 Map[Most,

    Map[(#*#[[4]] + 1*(1 - #[[4]])) &,

     (Flatten[ImageData@img, 1])]] // Partition[#, First@ImageDimensions[img]] & // Image

To check if it works

img = Import @ "https://i.stack.imgur.com/XvzDc.png"

enter image description here

img2 = reduceImageChannels@img

enter image description here

ImageChannels@img2

3

It given me the image exactly what I want. Exciting!

BTW, I am a mathematica newbie, so be free to correct my code and English grammer.

edited Jan 26 at 13:46

answered Jan 26 at 13:37

Voldikss

709

edited Jan 26 at 13:46

answered Jan 26 at 13:37

Voldikss

709

answered Jan 26 at 13:37

Voldikss

709

answered Jan 26 at 13:37

Voldikss

709

$begingroup$
I haven't ran the code, but Most gives all but the last element of a list. And what is 120 doing there, is that something to do with the ImageDimensions?
$endgroup$
– KraZug
Jan 26 at 13:42

$begingroup$
Thank you for the suggestion! I will replace Drop[#, {4}] & with Most. You are right, ImageDimensions@img returns {120,120}
$endgroup$
– Voldikss
Jan 26 at 13:45

$begingroup$
you can also use reduceImageChannels2[img_] := Map[Most[#*#[[4]] + 1*(1 - #[[4]])] &, ImageData@img, {2}]
$endgroup$
– kglr
Jan 26 at 13:56

$begingroup$
@kglr Thanks, it's much more elegant! I even forgot the Map's levelspec option...
$endgroup$
– Voldikss
Jan 26 at 13:59

1

$begingroup$
for an arbitrary color in the second argument,say Orange, RemoveAlphaChannel[img, Orange] and ImageApply[Most[#]*#[[4]] + {1, 0.5, 0}*(1 - #[[4]]) &, img] gives the same image as ImageApply[Most[#]*#[[4]] + {1, 0.5, 0}*(1 - #[[4]]) &, img] (that is, Max@(ImageData@ ImageApply[Most[#]*#[[4]] + {1, 0.5, 0}*(1 - #[[4]]) &, img] - ImageData[RemoveAlphaChannel[img, Orange]]) .) Note that Orange is RGBColor[1,.5,0].
$endgroup$
– kglr
Jan 26 at 14:26

|
show 6 more comments

$begingroup$
I haven't ran the code, but Most gives all but the last element of a list. And what is 120 doing there, is that something to do with the ImageDimensions?
$endgroup$
– KraZug
Jan 26 at 13:42

$begingroup$
Thank you for the suggestion! I will replace Drop[#, {4}] & with Most. You are right, ImageDimensions@img returns {120,120}
$endgroup$
– Voldikss
Jan 26 at 13:45

$begingroup$
you can also use reduceImageChannels2[img_] := Map[Most[#*#[[4]] + 1*(1 - #[[4]])] &, ImageData@img, {2}]
$endgroup$
– kglr
Jan 26 at 13:56

$begingroup$
@kglr Thanks, it's much more elegant! I even forgot the Map's levelspec option...
$endgroup$
– Voldikss
Jan 26 at 13:59

1

$begingroup$
for an arbitrary color in the second argument,say Orange, RemoveAlphaChannel[img, Orange] and ImageApply[Most[#]*#[[4]] + {1, 0.5, 0}*(1 - #[[4]]) &, img] gives the same image as ImageApply[Most[#]*#[[4]] + {1, 0.5, 0}*(1 - #[[4]]) &, img] (that is, Max@(ImageData@ ImageApply[Most[#]*#[[4]] + {1, 0.5, 0}*(1 - #[[4]]) &, img] - ImageData[RemoveAlphaChannel[img, Orange]]) .) Note that Orange is RGBColor[1,.5,0].
$endgroup$
– kglr
Jan 26 at 14:26

I haven't ran the code, but Most gives all but the last element of a list. And what is 120 doing there, is that something to do with the ImageDimensions?

– KraZug
Jan 26 at 13:42

Thank you for the suggestion! I will replace Drop[#, {4}] & with Most. You are right, ImageDimensions@img returns {120,120}

– Voldikss
Jan 26 at 13:45

you can also use reduceImageChannels2[img_] := Map[Most[#*#[[4]] + 1*(1 - #[[4]])] &, ImageData@img, {2}]

– kglr
Jan 26 at 13:56

@kglr Thanks, it's much more elegant! I even forgot the Map's levelspec option...

– Voldikss
Jan 26 at 13:59

for an arbitrary color in the second argument,say Orange, RemoveAlphaChannel[img, Orange] and ImageApply[Most[#]*#[[4]] + {1, 0.5, 0}*(1 - #[[4]]) &, img] gives the same image as ImageApply[Most[#]*#[[4]] + {1, 0.5, 0}*(1 - #[[4]]) &, img] (that is,

Max@(ImageData@     ImageApply[Most[#]*#[[4]] + {1, 0.5, 0}*(1 - #[[4]]) &, img] -     ImageData[RemoveAlphaChannel[img, Orange]])

.) Note that Orange is RGBColor[1,.5,0].

– kglr
Jan 26 at 14:26

Max@(ImageData@     ImageApply[Most[#]*#[[4]] + {1, 0.5, 0}*(1 - #[[4]]) &, img] -     ImageData[RemoveAlphaChannel[img, Orange]])

.) Note that Orange is RGBColor[1,.5,0].

– kglr
Jan 26 at 14:26

|
show 6 more comments

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