计算每种颜色的像素总数
Count total number of pixels for each color
我正在尝试计算与分割图像的每种颜色对应的像素数
以下是我写的代码
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import argparse
from ast import literal_eval
try:
from PIL import Image, ImageDraw, ImageFont
except ImportError:
exit("This script requires the PIL module. Install with pip install Pillow")
try:
import webcolors
except ImportError:
exit("This script uses webcolors for displaying named colors. Install with pip install webcolors")
def count_pixels(filename):
"""
Returns a count of pixels per a unique color
Args:
filename (str): the image to count the number of pixels of
Returns:
a key-value pairing of the rgb color value and the number of times the color was present in the image
"""
color_count = {}
with Image.open(filename) as image:
width, height = image.size
rgb_image = image.convert('RGB')
# iterate through each pixel in the image and keep a count per unique color
for x in range(width):
for y in range(height):
rgb = rgb_image.getpixel((x, y))
if rgb in color_count:
color_count[rgb] += 1
else:
color_count[rgb] = 1
return color_count
def create_legend_image(filename, colors, title, ignore_color):
"""
Create an image of color swatches paired with a color name and number
Args:
filename (str): the name of the legend image file
colors (dict): a key-value pairing of the color name and count
title (str): a title for the legend image
ignore_color (tuple): do not add this color to the legend image
Returns:
None
"""
margin = 10
rect_width = 25
rect_outline_width = 2
font_size = 20
img_width = 250
img_height = len(colors) * (rect_width + rect_outline_width + margin) + (0 if title is None else font_size)
legend_img = Image.new("RGBA", (img_width, img_height), "white")
draw = ImageDraw.Draw(legend_img)
font = ImageFont.truetype("arialbd.ttf", font_size)
# draw title for legend if applicable
text_height = 0
if title is not None:
_, text_height = draw.textsize(title, font=font)
draw.text((0, 0), title, font=font, fill="black")
color_index = 1
for color, count in colors.items():
if color == ignore_color:
continue
try:
# convert RGB color to a human readable color if applicable
color_name = webcolors.rgb_to_name(color)
except ValueError:
color_name = color
# draw square for color legend
y0 = rect_width * (color_index - 1) + margin * color_index + text_height
y1 = rect_width * color_index + margin * color_index + text_height
draw.rectangle([(0, y0), (rect_width, y1)], fill=color, outline="black", width=2)
# draw color name next and pixel count for legend colors
draw.text((rect_width + margin, y0), "{}: {}".format(color_name, count), font=font, fill="black")
color_index += 1
legend_img.save(filename, mode="w")
def main():
parser = argparse.ArgumentParser(description='Calculates the sum of pixels per a color')
parser.add_argument('image', nargs='?', default='.', help='The image to sum the pixels per a color of')
parser.add_argument('-i', '--ignore-color', help='Skip counting pixels of this color')
parser.add_argument('-t', '--title', help='Title for the image legend')
parser.add_argument('-l', '--legend-image', help='Generate an image with color swatches paired with the pixel count')
args = parser.parse_args()
ignore_color = literal_eval(args.ignore_color) if args.ignore_color is not None else None
color_count = count_pixels(args.image)
if args.legend_image is not None:
create_legend_image(args.legend_image, color_count, args.title, ignore_color)
# outputs pixel color count to console
print('Pixel Count per Unique Color:')
print('-' * 30)
color_index = 1
for color, count in color_count.items():
if color == ignore_color:
continue
try:
# convert RGB color to a human readable color if applicable
color_name = webcolors.rgb_to_name(color)
except ValueError:
color_name = color
print('{}.) {}: {}'.format(color_index, color_name, count))
color_index += 1
# Display the total number of pixels not ignored
print('-' * 30)
print('\t{} pixels'.format(sum(color_count[color] for color in color_count if color != ignore_color)))
if __name__ == '__main__':
main()
当我测试我的代码时，我没有得到准确的结果，该结果指示图片中表示的每种独特颜色的像素总数，即（红色、黄色和黑色），你能建议我一些替代方法吗计算分割图像中的像素总数
这是获取颜色及其对应计数的最简单方法：
#!/usr/bin/env python3
from PIL import Image
import numpy as np
# Open image and ensure RGB
im = Image.open('UMN9c.png').convert('RGB')
# Make into Numpy array
na = np.array(im)
# Get colours and corresponding counts
colours, counts = np.unique(na.reshape(-1,3), axis=0, return_counts=1)
print(colours, counts)
示例输出
[[ 37 36 36]
[226 44 11]
[228 239 80]] [255169 1059 5916]
如果您不想编写任何 Python 代码，您可以在终端中使用 ImageMagick
magick -verbose UMN9c.png -format %c histogram:info:-
输出
UMN9c.png PNG 512x512 512x512+0+0 8-bit sRGB 3572B 0.020u 0:00.008
255169: (37,36,36,255) #252424FF srgba(37,36,36,1)
1059: (226,44,11,255) #E22C0BFF srgba(226,44,11,1)
5916: (228,239,80,255) #E4EF50FF srgba(228,239,80,1)
UMN9c.png=>info:- PNG 512x512 512x512+0+0 8-bit sRGB 3572B 0.320u 0:00.166
我不确定绝对速度对您来说是否是个问题。如果是这样，在 答案的末尾使用 np.dot() 有一个明显更快的方法。
使用OpenCV导入的解决方案。
# Load image
img = cv2.imread("UMN9c.png")
img = cv2.cvtColor(img, cv2.COLOR_BGR2RGB)
# Flatten image into rows = number of pixels, columns = 3 (RGB)
img = img.reshape(-1, img.shape[2])
# Count number of unique colors
unique_colors, unique_count = np.unique(img, axis=0, return_counts=True)
# Print info
print(unique_colors)
print(unique_count)
print("Total number of unique colors: ", unique_colors.shape[0])
我正在尝试计算与分割图像的每种颜色对应的像素数
以下是我写的代码
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import argparse
from ast import literal_eval
try:
from PIL import Image, ImageDraw, ImageFont
except ImportError:
exit("This script requires the PIL module. Install with pip install Pillow")
try:
import webcolors
except ImportError:
exit("This script uses webcolors for displaying named colors. Install with pip install webcolors")
def count_pixels(filename):
"""
Returns a count of pixels per a unique color
Args:
filename (str): the image to count the number of pixels of
Returns:
a key-value pairing of the rgb color value and the number of times the color was present in the image
"""
color_count = {}
with Image.open(filename) as image:
width, height = image.size
rgb_image = image.convert('RGB')
# iterate through each pixel in the image and keep a count per unique color
for x in range(width):
for y in range(height):
rgb = rgb_image.getpixel((x, y))
if rgb in color_count:
color_count[rgb] += 1
else:
color_count[rgb] = 1
return color_count
def create_legend_image(filename, colors, title, ignore_color):
"""
Create an image of color swatches paired with a color name and number
Args:
filename (str): the name of the legend image file
colors (dict): a key-value pairing of the color name and count
title (str): a title for the legend image
ignore_color (tuple): do not add this color to the legend image
Returns:
None
"""
margin = 10
rect_width = 25
rect_outline_width = 2
font_size = 20
img_width = 250
img_height = len(colors) * (rect_width + rect_outline_width + margin) + (0 if title is None else font_size)
legend_img = Image.new("RGBA", (img_width, img_height), "white")
draw = ImageDraw.Draw(legend_img)
font = ImageFont.truetype("arialbd.ttf", font_size)
# draw title for legend if applicable
text_height = 0
if title is not None:
_, text_height = draw.textsize(title, font=font)
draw.text((0, 0), title, font=font, fill="black")
color_index = 1
for color, count in colors.items():
if color == ignore_color:
continue
try:
# convert RGB color to a human readable color if applicable
color_name = webcolors.rgb_to_name(color)
except ValueError:
color_name = color
# draw square for color legend
y0 = rect_width * (color_index - 1) + margin * color_index + text_height
y1 = rect_width * color_index + margin * color_index + text_height
draw.rectangle([(0, y0), (rect_width, y1)], fill=color, outline="black", width=2)
# draw color name next and pixel count for legend colors
draw.text((rect_width + margin, y0), "{}: {}".format(color_name, count), font=font, fill="black")
color_index += 1
legend_img.save(filename, mode="w")
def main():
parser = argparse.ArgumentParser(description='Calculates the sum of pixels per a color')
parser.add_argument('image', nargs='?', default='.', help='The image to sum the pixels per a color of')
parser.add_argument('-i', '--ignore-color', help='Skip counting pixels of this color')
parser.add_argument('-t', '--title', help='Title for the image legend')
parser.add_argument('-l', '--legend-image', help='Generate an image with color swatches paired with the pixel count')
args = parser.parse_args()
ignore_color = literal_eval(args.ignore_color) if args.ignore_color is not None else None
color_count = count_pixels(args.image)
if args.legend_image is not None:
create_legend_image(args.legend_image, color_count, args.title, ignore_color)
# outputs pixel color count to console
print('Pixel Count per Unique Color:')
print('-' * 30)
color_index = 1
for color, count in color_count.items():
if color == ignore_color:
continue
try:
# convert RGB color to a human readable color if applicable
color_name = webcolors.rgb_to_name(color)
except ValueError:
color_name = color
print('{}.) {}: {}'.format(color_index, color_name, count))
color_index += 1
# Display the total number of pixels not ignored
print('-' * 30)
print('\t{} pixels'.format(sum(color_count[color] for color in color_count if color != ignore_color)))
if __name__ == '__main__':
main()
当我测试我的代码时，我没有得到准确的结果，该结果指示图片中表示的每种独特颜色的像素总数，即（红色、黄色和黑色），你能建议我一些替代方法吗计算分割图像中的像素总数
这是获取颜色及其对应计数的最简单方法：
#!/usr/bin/env python3
from PIL import Image
import numpy as np
# Open image and ensure RGB
im = Image.open('UMN9c.png').convert('RGB')
# Make into Numpy array
na = np.array(im)
# Get colours and corresponding counts
colours, counts = np.unique(na.reshape(-1,3), axis=0, return_counts=1)
print(colours, counts)
示例输出
[[ 37 36 36]
[226 44 11]
[228 239 80]] [255169 1059 5916]
如果您不想编写任何 Python 代码，您可以在终端中使用 ImageMagick
magick -verbose UMN9c.png -format %c histogram:info:-
输出
UMN9c.png PNG 512x512 512x512+0+0 8-bit sRGB 3572B 0.020u 0:00.008
255169: (37,36,36,255) #252424FF srgba(37,36,36,1)
1059: (226,44,11,255) #E22C0BFF srgba(226,44,11,1)
5916: (228,239,80,255) #E4EF50FF srgba(228,239,80,1)
UMN9c.png=>info:- PNG 512x512 512x512+0+0 8-bit sRGB 3572B 0.320u 0:00.166
我不确定绝对速度对您来说是否是个问题。如果是这样，在 np.dot() 有一个明显更快的方法。
使用OpenCV导入的解决方案。
# Load image
img = cv2.imread("UMN9c.png")
img = cv2.cvtColor(img, cv2.COLOR_BGR2RGB)
# Flatten image into rows = number of pixels, columns = 3 (RGB)
img = img.reshape(-1, img.shape[2])
# Count number of unique colors
unique_colors, unique_count = np.unique(img, axis=0, return_counts=True)
# Print info
print(unique_colors)
print(unique_count)
print("Total number of unique colors: ", unique_colors.shape[0])