我正在尝试使用Python 2.7在How to automatically generate N "distinct" colors?末尾重用该解决方案。
不幸的是,下面的代码永远不会返回,因为即使islice()
请求前100次迭代,内部地图似乎也会调用
gethsvs = lambda: flatten(itertools.imap(genhsv, getfracs()))
强制进行所有迭代。有没有一种方法可以在主代码上进行islice()
调用,而不是将迭代次数传递给所有lambda函数,如下所示,但是它只能在所有内部映射中获得前100次迭代? / p>
import colorsys, itertools, numpy as np
from fractions import Fraction
def zenos_dichotomy():
'''
http://en.wikipedia.org/wiki/1/2_%2B_1/4_%2B_1/8_%2B_1/16_%2B_%C2%B7_%C2%B7_%C2%B7
'''
for k in itertools.count():
yield Fraction(1, 2 ** k)
def getfracs():
'''
[Fraction(0, 1), Fraction(1, 2), Fraction(1, 4), Fraction(3, 4), Fraction(1, 8), Fraction(3, 8), Fraction(5, 8), Fraction(7, 8), Fraction(1, 16), Fraction(3, 16), ...]
[0.0, 0.5, 0.25, 0.75, 0.125, 0.375, 0.625, 0.875, 0.0625, 0.1875, ...]
'''
yield 0
for k in zenos_dichotomy():
i = k.denominator # [1,2,4,8,16,...]
for j in range(1, i, 2):
yield Fraction(j, i)
'''Can be used for the v in hsv to map linear values 0..1 to something that looks equidistant.'''
bias = lambda x: (np.sqrt(x / 3) / Fraction(2, 3) + Fraction(1, 3)) / Fraction(6, 5)
def genhsv(h):
for s in [Fraction(6, 10)]: # optionally use range
for v in [Fraction(8, 10), Fraction(5, 10)]: # could use range too
yield (h, s, v) # use bias for v here if you use range
genrgb = lambda x: colorsys.hsv_to_rgb(*x)
flatten = itertools.chain.from_iterable
gethsvs = lambda: flatten(itertools.imap(genhsv, getfracs()))
getrgbs = lambda: itertools.imap(genrgb, gethsvs())
def genhtml(x):
uint8tuple = itertools.imap(lambda y: int(y * 255), x)
return 'rgb({},{},{})'.format(*uint8tuple)
gethtmlcolors = lambda: map(genhtml, getrgbs())
if __name__ == '__main__':
print(list(itertools.islice(gethtmlcolors(), 100)))
答案 0 :(得分:2)
您在map()
lambda中使用的是itertools.imap()
而不是gethtmlcolors
:
gethtmlcolors = lambda: map(genhtml, getrgbs())
在python 2中,map()
将尝试处理输入可迭代的所有元素。改变使用imap()
,一切正常:
gethtmlcolors = lambda: itertools.imap(genhtml, getrgbs())