Assign different colors to different biological cells in an image

I need to assign different colors to different biological cells in an image.

To be more specific, the image is only black and white (so a matrix of 0 and 255 only). The content of the cells (excluding boundaries) is represented by white color, whereas cell boundaries are represented by black color. Each cell is enclosed by some cell boundaries or image edges. I hope to assign different colors to different cells, such that I can immediately tell which cells I am currently at by simply looking at the value of its entry.

Edit: It is biological cells. I have found something similar online:

biological cells

Source: http://brainiac2.mit.edu/isbi_challenge/

edited Nov 19 '18 at 19:52

Dan Mašek

8,84932546

asked Nov 19 '18 at 14:59

NeverBe

3417

Are you talking about biological cells? Or Excel spreadsheets? I think a picture would help.
– Mark Setchell
Nov 19 '18 at 17:54

It is biological cells. I have just edited my question.
– NeverBe
Nov 19 '18 at 18:44

1

Besides the solution posted by @Dan Mašek, I have found that skimage.measure.label from skimage package may also work. Perhaps this additional information may help some people in the future. Source: scikit-image.org/docs/dev/api/…
– NeverBe
Nov 19 '18 at 19:40

add a comment |

I need to assign different colors to different biological cells in an image.

Edit: It is biological cells. I have found something similar online:

biological cells

Source: http://brainiac2.mit.edu/isbi_challenge/

edited Nov 19 '18 at 19:52

Dan Mašek

8,84932546

asked Nov 19 '18 at 14:59

NeverBe

3417

Are you talking about biological cells? Or Excel spreadsheets? I think a picture would help.
– Mark Setchell
Nov 19 '18 at 17:54

It is biological cells. I have just edited my question.
– NeverBe
Nov 19 '18 at 18:44

1

Besides the solution posted by @Dan Mašek, I have found that skimage.measure.label from skimage package may also work. Perhaps this additional information may help some people in the future. Source: scikit-image.org/docs/dev/api/…
– NeverBe
Nov 19 '18 at 19:40

add a comment |

I need to assign different colors to different biological cells in an image.

Edit: It is biological cells. I have found something similar online:

biological cells

Source: http://brainiac2.mit.edu/isbi_challenge/

edited Nov 19 '18 at 19:52

Dan Mašek

8,84932546

asked Nov 19 '18 at 14:59

NeverBe

3417

I need to assign different colors to different biological cells in an image.

Edit: It is biological cells. I have found something similar online:

biological cells

Source: http://brainiac2.mit.edu/isbi_challenge/

python image opencv image-processing cv2

edited Nov 19 '18 at 19:52

Dan Mašek

8,84932546

asked Nov 19 '18 at 14:59

NeverBe

3417

edited Nov 19 '18 at 19:52

Dan Mašek

8,84932546

asked Nov 19 '18 at 14:59

NeverBe

3417

edited Nov 19 '18 at 19:52

Dan Mašek

8,84932546

edited Nov 19 '18 at 19:52

Dan Mašek

8,84932546

edited Nov 19 '18 at 19:52

Dan Mašek

8,84932546

asked Nov 19 '18 at 14:59

NeverBe

3417

asked Nov 19 '18 at 14:59

NeverBe

3417

asked Nov 19 '18 at 14:59

NeverBe

3417

Are you talking about biological cells? Or Excel spreadsheets? I think a picture would help.
– Mark Setchell
Nov 19 '18 at 17:54

It is biological cells. I have just edited my question.
– NeverBe
Nov 19 '18 at 18:44

1

Besides the solution posted by @Dan Mašek, I have found that skimage.measure.label from skimage package may also work. Perhaps this additional information may help some people in the future. Source: scikit-image.org/docs/dev/api/…
– NeverBe
Nov 19 '18 at 19:40

add a comment |

Are you talking about biological cells? Or Excel spreadsheets? I think a picture would help.
– Mark Setchell
Nov 19 '18 at 17:54

It is biological cells. I have just edited my question.
– NeverBe
Nov 19 '18 at 18:44

1

Besides the solution posted by @Dan Mašek, I have found that skimage.measure.label from skimage package may also work. Perhaps this additional information may help some people in the future. Source: scikit-image.org/docs/dev/api/…
– NeverBe
Nov 19 '18 at 19:40

Are you talking about biological cells? Or Excel spreadsheets? I think a picture would help.
– Mark Setchell
Nov 19 '18 at 17:54

It is biological cells. I have just edited my question.
– NeverBe
Nov 19 '18 at 18:44

Besides the solution posted by @Dan Mašek, I have found that skimage.measure.label from skimage package may also work. Perhaps this additional information may help some people in the future. Source: scikit-image.org/docs/dev/api/…
– NeverBe
Nov 19 '18 at 19:40

add a comment |

1 Answer
1

active

oldest

votes

A simple approach comes to mind:

Threshold the image to binarize it, since the one you provided contains more than just 0s and 255s.

Find all the contours and organize them into a two-level hierarchy. This can be accomplished by calling cv2.threshold with the flag RETR_CCOMP:

At the top level, there are external boundaries of the components. At the second level, there are boundaries of the holes. If there is another contour inside a hole of a connected component, it is still put at the top level.

N.B.: "Components" are the white parts of image, "holes" are the black parts.

Iterate over contours. For each contour of a component (since such contours are placed at the top level of the hierarchy, they have no parent contour) draw the contour polygon filled with a random colour.

Sample Code:

import cv2

import numpy as np



img = cv2.imread('cells.png', cv2.IMREAD_GRAYSCALE)

thresh = cv2.threshold(img, thresh=128, maxval=255, type=cv2.THRESH_BINARY)[1]

_, contours, hierarchy = cv2.findContours(thresh, cv2.RETR_CCOMP, cv2.CHAIN_APPROX_SIMPLE)



output = cv2.cvtColor(thresh, cv2.COLOR_GRAY2BGR)

for i, contour in enumerate(contours):

    if hierarchy[0][i][3] == -1:

        colour = cv2.randu(np.zeros(3, np.uint8), 0, 256)

        cv2.drawContours(output, contours, i, colour.tolist(), -1)



cv2.imwrite('cells_colour.png', output)

Result:

answered Nov 19 '18 at 19:36

Dan Mašek

8,84932546

Thank you! This is what I am looking for.
– NeverBe
Nov 19 '18 at 19:38

Excellent solution!
– Mark Setchell
Nov 20 '18 at 10:01

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

A simple approach comes to mind:

Threshold the image to binarize it, since the one you provided contains more than just 0s and 255s.

Find all the contours and organize them into a two-level hierarchy. This can be accomplished by calling cv2.threshold with the flag RETR_CCOMP:

At the top level, there are external boundaries of the components. At the second level, there are boundaries of the holes. If there is another contour inside a hole of a connected component, it is still put at the top level.

N.B.: "Components" are the white parts of image, "holes" are the black parts.

Iterate over contours. For each contour of a component (since such contours are placed at the top level of the hierarchy, they have no parent contour) draw the contour polygon filled with a random colour.

Sample Code:

import cv2

import numpy as np



img = cv2.imread('cells.png', cv2.IMREAD_GRAYSCALE)

thresh = cv2.threshold(img, thresh=128, maxval=255, type=cv2.THRESH_BINARY)[1]

_, contours, hierarchy = cv2.findContours(thresh, cv2.RETR_CCOMP, cv2.CHAIN_APPROX_SIMPLE)



output = cv2.cvtColor(thresh, cv2.COLOR_GRAY2BGR)

for i, contour in enumerate(contours):

    if hierarchy[0][i][3] == -1:

        colour = cv2.randu(np.zeros(3, np.uint8), 0, 256)

        cv2.drawContours(output, contours, i, colour.tolist(), -1)



cv2.imwrite('cells_colour.png', output)

Result:

answered Nov 19 '18 at 19:36

Dan Mašek

8,84932546

Thank you! This is what I am looking for.
– NeverBe
Nov 19 '18 at 19:38

Excellent solution!
– Mark Setchell
Nov 20 '18 at 10:01

add a comment |

A simple approach comes to mind:

Threshold the image to binarize it, since the one you provided contains more than just 0s and 255s.

Find all the contours and organize them into a two-level hierarchy. This can be accomplished by calling cv2.threshold with the flag RETR_CCOMP:

At the top level, there are external boundaries of the components. At the second level, there are boundaries of the holes. If there is another contour inside a hole of a connected component, it is still put at the top level.

N.B.: "Components" are the white parts of image, "holes" are the black parts.

Iterate over contours. For each contour of a component (since such contours are placed at the top level of the hierarchy, they have no parent contour) draw the contour polygon filled with a random colour.

Sample Code:

import cv2

import numpy as np



img = cv2.imread('cells.png', cv2.IMREAD_GRAYSCALE)

thresh = cv2.threshold(img, thresh=128, maxval=255, type=cv2.THRESH_BINARY)[1]

_, contours, hierarchy = cv2.findContours(thresh, cv2.RETR_CCOMP, cv2.CHAIN_APPROX_SIMPLE)



output = cv2.cvtColor(thresh, cv2.COLOR_GRAY2BGR)

for i, contour in enumerate(contours):

    if hierarchy[0][i][3] == -1:

        colour = cv2.randu(np.zeros(3, np.uint8), 0, 256)

        cv2.drawContours(output, contours, i, colour.tolist(), -1)



cv2.imwrite('cells_colour.png', output)

Result:

answered Nov 19 '18 at 19:36

Dan Mašek

8,84932546

Thank you! This is what I am looking for.
– NeverBe
Nov 19 '18 at 19:38

Excellent solution!
– Mark Setchell
Nov 20 '18 at 10:01

add a comment |

A simple approach comes to mind:

Threshold the image to binarize it, since the one you provided contains more than just 0s and 255s.

Find all the contours and organize them into a two-level hierarchy. This can be accomplished by calling cv2.threshold with the flag RETR_CCOMP:

At the top level, there are external boundaries of the components. At the second level, there are boundaries of the holes. If there is another contour inside a hole of a connected component, it is still put at the top level.

N.B.: "Components" are the white parts of image, "holes" are the black parts.

Iterate over contours. For each contour of a component (since such contours are placed at the top level of the hierarchy, they have no parent contour) draw the contour polygon filled with a random colour.

Sample Code:

import cv2

import numpy as np



img = cv2.imread('cells.png', cv2.IMREAD_GRAYSCALE)

thresh = cv2.threshold(img, thresh=128, maxval=255, type=cv2.THRESH_BINARY)[1]

_, contours, hierarchy = cv2.findContours(thresh, cv2.RETR_CCOMP, cv2.CHAIN_APPROX_SIMPLE)



output = cv2.cvtColor(thresh, cv2.COLOR_GRAY2BGR)

for i, contour in enumerate(contours):

    if hierarchy[0][i][3] == -1:

        colour = cv2.randu(np.zeros(3, np.uint8), 0, 256)

        cv2.drawContours(output, contours, i, colour.tolist(), -1)



cv2.imwrite('cells_colour.png', output)

Result:

answered Nov 19 '18 at 19:36

Dan Mašek

8,84932546

A simple approach comes to mind:

Threshold the image to binarize it, since the one you provided contains more than just 0s and 255s.

Find all the contours and organize them into a two-level hierarchy. This can be accomplished by calling cv2.threshold with the flag RETR_CCOMP:

At the top level, there are external boundaries of the components. At the second level, there are boundaries of the holes. If there is another contour inside a hole of a connected component, it is still put at the top level.

N.B.: "Components" are the white parts of image, "holes" are the black parts.

Iterate over contours. For each contour of a component (since such contours are placed at the top level of the hierarchy, they have no parent contour) draw the contour polygon filled with a random colour.

Sample Code:

import cv2

import numpy as np



img = cv2.imread('cells.png', cv2.IMREAD_GRAYSCALE)

thresh = cv2.threshold(img, thresh=128, maxval=255, type=cv2.THRESH_BINARY)[1]

_, contours, hierarchy = cv2.findContours(thresh, cv2.RETR_CCOMP, cv2.CHAIN_APPROX_SIMPLE)



output = cv2.cvtColor(thresh, cv2.COLOR_GRAY2BGR)

for i, contour in enumerate(contours):

    if hierarchy[0][i][3] == -1:

        colour = cv2.randu(np.zeros(3, np.uint8), 0, 256)

        cv2.drawContours(output, contours, i, colour.tolist(), -1)



cv2.imwrite('cells_colour.png', output)

Result:

answered Nov 19 '18 at 19:36

Dan Mašek

8,84932546

answered Nov 19 '18 at 19:36

Dan Mašek

8,84932546

answered Nov 19 '18 at 19:36

Dan Mašek

8,84932546

answered Nov 19 '18 at 19:36

Dan Mašek

8,84932546

Thank you! This is what I am looking for.
– NeverBe
Nov 19 '18 at 19:38

Excellent solution!
– Mark Setchell
Nov 20 '18 at 10:01

add a comment |

Thank you! This is what I am looking for.
– NeverBe
Nov 19 '18 at 19:38

Excellent solution!
– Mark Setchell
Nov 20 '18 at 10:01

Thank you! This is what I am looking for.
– NeverBe
Nov 19 '18 at 19:38

Excellent solution!
– Mark Setchell
Nov 20 '18 at 10:01

add a comment |

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