python 对应元素相乘_如何在Python中执行两个列表的元素相乘？

使用与zip()混合的列表理解：。

[a*b for a,b in zip(lista,listb)]

由于您已经在使用numpy ，所以将数据存储在一个numpy数组而不是列表中是很有意义的。 一旦你做到这一点，你可以免费获得元素明智的产品：

In [1]: import numpy as np In [2]: a = np.array([1,2,3,4]) In [3]: b = np.array([2,3,4,5]) In [4]: a * b Out[4]: array([ 2, 6, 12, 20])

使用np.multiply(a，b)：

import numpy as np a = [1,2,3,4] b = [2,3,4,5] np.multiply(a,b)

你可以尝试在循环中乘以每个元素。 做这个的简短手段是

ab = [a[i]*b[i] for i in range(len(a))]

相当直观的做法是：

a = [1,2,3,4] b = [2,3,4,5] ab = [] #Create empty list for i in range(0, len(a)): ab.append(a[i]*b[i]) #Adds each element to the list

还有一个答案：

-1 …需要导入

+1 …非常可读

import operator a = [1,2,3,4] b = [10,11,12,13] list(map(operator.mul, a, b))

输出[10,22,36,52]

对于大型列表，我们可以这样做：

product_iter_object = itertools.imap(operator.mul, [1,2,3,4], [2,3,4,5])

product_iter_object.next()给出输出列表中的每个元素。

输出将是两个input列表中较短的一个的长度。

创build一个数组; 将每个列表乘以数组; 将数组转换为列表

import numpy as np a = [1,2,3,4] b = [2,3,4,5] c = (np.ones(len(a))*a*b).tolist() [2.0, 6.0, 12.0, 20.0]

你可以用lambda乘法

foo=[1,2,3,4] bar=[1,2,5,55] l=map(lambda x,y:x*y,foo,bar)

gahooa的答案是正确的，如标题中所表述的问题，但如果列表已经是numpy格式或大于十，它会快得多(3个数量级)，以及更可读，做简单的numpy乘法NPE。 我得到这些时机：

0.0049ms -> N = 4, a = [i for i in range(N)], c = [a*b for a,b in zip(a, b)] 0.0075ms -> N = 4, a = [i for i in range(N)], c = a * b 0.0167ms -> N = 4, a = np.arange(N), c = [a*b for a,b in zip(a, b)] 0.0013ms -> N = 4, a = np.arange(N), c = a * b 0.0171ms -> N = 40, a = [i for i in range(N)], c = [a*b for a,b in zip(a, b)] 0.0095ms -> N = 40, a = [i for i in range(N)], c = a * b 0.1077ms -> N = 40, a = np.arange(N), c = [a*b for a,b in zip(a, b)] 0.0013ms -> N = 40, a = np.arange(N), c = a * b 0.1485ms -> N = 400, a = [i for i in range(N)], c = [a*b for a,b in zip(a, b)] 0.0397ms -> N = 400, a = [i for i in range(N)], c = a * b 1.0348ms -> N = 400, a = np.arange(N), c = [a*b for a,b in zip(a, b)] 0.0020ms -> N = 400, a = np.arange(N), c = a * b

即从下面的testing程序。

import timeit init = [''' import numpy as np N = {} a = {} b = np.linspace(0.0, 0.5, len(a)) '''.format(i, j) for i in [4, 40, 400] for j in ['[i for i in range(N)]', 'np.arange(N)']] func = ['''c = [a*b for a,b in zip(a, b)]''', '''c = a * b'''] for i in init: for f in func: lines = i.split('\n') print('{:6.4f}ms -> {}, {}, {}'.format( timeit.timeit(f, setup=i, number=1000), lines[2], lines[3], f))

可以使用枚举。

[val*b[i] for i, val in enumerate(a)]