将一个列表的元素除以另一个列表的元素
将一个列表的元素除以另一个列表的元素
提问于 2018-02-26 21:53:20
我有两个列表，比如说
a = [10, 20, 30 , 40, 50 , 60]
b = [30, 70, 110]如您所见，列表b由一个列表的元素和组成，其中window = 2：
b[0] = a[0] + a[1] = 10 + 20 = 30 etc.我如何才能获得另一个列表，它将包含给定窗口中列表的a元素和b元素的一部分？在我的示例中，我想要获取列表：
c = [10/30, 20/30, 30/70, 40/70, 50/110, 60/110]回答 3
Stack Overflow用户
发布于 2018-02-26 21:58:34
您可以对这两个任务使用列表理解(创建b和c列表)
a = [10, 20, 30, 40, 50, 60]
b = [i+j for i, j in zip(a[::2], a[1::2])]
print(b) # [30, 70, 110]
c = [x / b[i//2] for i, x in enumerate(a)]
print(c) # [0.3333333333333333, 0.6666666666666666, 0.42857142857142855, 0.5714285714285714, 0.45454545454545453, 0.5454545454545454]如果您确实需要分数，则可以使用fractions模块及其Fraction数据类型：
from fractions import Fraction
# same code as before
c = [Fraction(x, b[i//2]) for i, x in enumerate(a)]
print(c) # [Fraction(1, 3), Fraction(2, 3), Fraction(3, 7), Fraction(4, 7), Fraction(5, 11), Fraction(6, 11)]备注
作为@LaurentH。注意，上面的方法只适用于大小为2的块(你称之为窗口)。对于更一般的方法，你可以定义一个生成器，它将为你yield它们：
# taken from https://stackoverflow.com/a/312464/6162307
def yield_chunks(l, n):
"""Yield successive n-sized chunks from l."""
for i in range(0, len(l), n):
yield l[i:i + n]
b = [sum(chunk) for chunk in yield_chunks(a, 2)]
# same coden = 3示例
n = 3
b = [sum(chunk) for chunk in yield_chunks(a, n)]
print(b) # [60, 150]
c = [x / b[i//n] for i, x in enumerate(a)]
print(c) # [0.16666666666666666, 0.3333333333333333, 0.5, 0.26666666666666666, 0.3333333333333333, 0.4]Stack Overflow用户
发布于 2018-02-26 21:58:18
a = [10, 20, 30 , 40, 50 , 60]
b = [30, 70, 110]
# take each value in a (multiply by 1.0 to get a double), and divide by the value in b in the corresponding index (i.e. indices 0,1 in a correspond with index 0 in b, and so on...)
c = [val*1.0/b[idx//2] for idx,val in enumerate(a)]
# here is the calculation using strings, to get the desired output by OP
d = ['{}/{}'.format(val, b[idx//2]) for idx,val in enumerate(a)]
print '{}\n{}'.format(c, d)输出=
[0.3333333333333333, 0.6666666666666666, 0.42857142857142855, 0.5714285714285714, 0.45454545454545453, 0.5454545454545454]
['10/30', '20/30', '30/70', '40/70', '50/110', '60/110']Stack Overflow用户
发布于 2018-02-26 22:21:42
这是我的建议，一个函数用于求和，一个函数用于分数，列表和窗口宽度作为参数。它不使用外部包，并且非常简短，易于阅读和理解：
# Sum with a given window
def sumWithWindow(aList, window = 2):
res = []
mySum = 0
for i,elem in enumerate(aList):
mySum += elem
if (i+1) % window == 0:
res.append(mySum)
mySum = 0
return (res)
# Fraction with a given window
def fractionWithWindow(aList, window = 2):
res = []
b = sumWithWindow(aList, window)
for i,elem in enumerate(aList):
res.append(elem/b[int(i/window)])
return (res)
# Example
a = [10, 20, 30 , 40, 50 , 60]
print(sumWithWindow(a, 2))
print(fractionWithWindow(a,2))这也是单行风格的版本，带有列表理解，可读性稍差，但非常简短：
# Sum with a given window
def sumWithWindow(aList, window = 2):
return [sum(aList[n:n+window]) for n in range(0,len(aList),window)]
# Fraction with a given window
def fractionWithWindow(aList, window = 2):
return [elem/sumWithWindow(aList,window)[i//window] for i,elem in enumerate(aList)]
# Example
a = [10, 20, 30 , 40, 50 , 60]
print(sumWithWindow(a, 2))
print(fractionWithWindow(a,2))页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/48990152
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