将一个列表的元素除以另一个列表的元素

Question

我有两张单子

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window=2组成：

b[0] = a[0] + a[1] = 10 + 20 = 30 etc.

我怎样才能得到另一个列表，它将由给定窗口的列表的a元素和b元素的分数组成？ 在我的例子中，我想得到一个列表：

c = [10/30, 20/30, 30/70, 40/70, 50/110, 60/110]

Answer 1

两个任务都可以使用列表理解（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]

---

如果确实需要分数，可以使用^{}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)]

注意

作为@劳伦斯。注意，在注释中，以上仅适用于大小为2的块（您称之为windows）。对于更一般的方法，您可以定义一个生成器，它将yield为您定义它们：

# 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 code

例如n = 3：

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]

Answer 2

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']

Answer 3

下面是我的建议，其中一个函数用于求和，一个函数用于分数，列表和窗口宽度作为参数。它不使用外部包装，而且非常简短、易读和易懂：

# 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))

这也是一行文风格的版本，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))