【发布时间】:2012-12-28 14:28:40
【问题描述】:
我正在尝试使用 Python 2.7 重用 How to automatically generate N "distinct" colors? 末尾的解决方案。
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,似乎内部映射调用就像
gethsvs = lambda: flatten(itertools.imap(genhsv, getfracs()))
强制获取所有迭代。除了将迭代次数传递给所有 lambda 函数之外,有没有办法在主代码中进行 islice() 调用,如下所示,但在所有内部映射中也只获得前 100 次迭代?
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)))
【问题讨论】:
-
一些建议 - 您可能应该将所有这些包装到一个类中,并使用 self 参数来回传递参数和变量,这样您就可以拥有一个您想要共享的迭代次数的类变量您想要的所有切片和方法。
标签: python python-2.7 iterator