问题:按元素添加2个列表?

我现在有了:

list1 = [1, 2, 3]
list2 = [4, 5, 6]

我希望有:

[1, 2, 3]
 +  +  +
[4, 5, 6]
|| || ||
[5, 7, 9]

只是两个列表的逐个元素相加。

我当然可以迭代这两个列表,但是我不想这样做。

什么是最Python的方式这样做的?

I have now:

list1 = [1, 2, 3]
list2 = [4, 5, 6]

I wish to have:

[1, 2, 3]
 +  +  +
[4, 5, 6]
|| || ||
[5, 7, 9]

Simply an element-wise addition of two lists.

I can surely iterate the two lists, but I don’t want do that.

What is the most Pythonic way of doing so?


回答 0

使用mapoperator.add

>>> from operator import add
>>> list( map(add, list1, list2) )
[5, 7, 9]

zip具有列表理解:

>>> [sum(x) for x in zip(list1, list2)]
[5, 7, 9]

时序比较:

>>> list2 = [4, 5, 6]*10**5
>>> list1 = [1, 2, 3]*10**5
>>> %timeit from operator import add;map(add, list1, list2)
10 loops, best of 3: 44.6 ms per loop
>>> %timeit from itertools import izip; [a + b for a, b in izip(list1, list2)]
10 loops, best of 3: 71 ms per loop
>>> %timeit [a + b for a, b in zip(list1, list2)]
10 loops, best of 3: 112 ms per loop
>>> %timeit from itertools import izip;[sum(x) for x in izip(list1, list2)]
1 loops, best of 3: 139 ms per loop
>>> %timeit [sum(x) for x in zip(list1, list2)]
1 loops, best of 3: 177 ms per loop

Use map with operator.add:

>>> from operator import add
>>> list( map(add, list1, list2) )
[5, 7, 9]

or zip with a list comprehension:

>>> [sum(x) for x in zip(list1, list2)]
[5, 7, 9]

Timing comparisons:

>>> list2 = [4, 5, 6]*10**5
>>> list1 = [1, 2, 3]*10**5
>>> %timeit from operator import add;map(add, list1, list2)
10 loops, best of 3: 44.6 ms per loop
>>> %timeit from itertools import izip; [a + b for a, b in izip(list1, list2)]
10 loops, best of 3: 71 ms per loop
>>> %timeit [a + b for a, b in zip(list1, list2)]
10 loops, best of 3: 112 ms per loop
>>> %timeit from itertools import izip;[sum(x) for x in izip(list1, list2)]
1 loops, best of 3: 139 ms per loop
>>> %timeit [sum(x) for x in zip(list1, list2)]
1 loops, best of 3: 177 ms per loop

回答 1

其他人给出了如何在纯python中执行此操作的示例。如果要对具有100.000个元素的数组执行此操作,则应使用numpy:

In [1]: import numpy as np
In [2]: vector1 = np.array([1, 2, 3])
In [3]: vector2 = np.array([4, 5, 6])

现在,按元素进行加法与

In [4]: sum_vector = vector1 + vector2
In [5]: print sum_vector
[5 7 9]

就像在Matlab中一样。

与Ashwini最快版本进行比较的时间:

In [16]: from operator import add
In [17]: n = 10**5
In [18]: vector2 = np.tile([4,5,6], n)
In [19]: vector1 = np.tile([1,2,3], n)
In [20]: list1 = [1,2,3]*n
In [21]: list2 = [4,5,6]*n
In [22]: timeit map(add, list1, list2)
10 loops, best of 3: 26.9 ms per loop

In [23]: timeit vector1 + vector2
1000 loops, best of 3: 1.06 ms per loop

因此这快25倍!但是请使用适合您情况的东西。对于一个简单的程序,您可能不想安装numpy,因此请使用标准python(我发现Henry的版本是最Pythonic 的版本)。如果您正认真处理数字,numpy那么就进行繁重的工作。对于速度怪胎:似乎numpy解决方案的运行速度更快n = 8

The others gave examples how to do this in pure python. If you want to do this with arrays with 100.000 elements, you should use numpy:

In [1]: import numpy as np
In [2]: vector1 = np.array([1, 2, 3])
In [3]: vector2 = np.array([4, 5, 6])

Doing the element-wise addition is now as trivial as

In [4]: sum_vector = vector1 + vector2
In [5]: print sum_vector
[5 7 9]

just like in Matlab.

Timing to compare with Ashwini’s fastest version:

In [16]: from operator import add
In [17]: n = 10**5
In [18]: vector2 = np.tile([4,5,6], n)
In [19]: vector1 = np.tile([1,2,3], n)
In [20]: list1 = [1,2,3]*n
In [21]: list2 = [4,5,6]*n
In [22]: timeit map(add, list1, list2)
10 loops, best of 3: 26.9 ms per loop

In [23]: timeit vector1 + vector2
1000 loops, best of 3: 1.06 ms per loop

So this is a factor 25 faster! But use what suits your situation. For a simple program, you probably don’t want to install numpy, so use standard python (and I find Henry’s version the most Pythonic one). If you are into serious number crunching, let numpy do the heavy lifting. For the speed freaks: it seems that the numpy solution is faster starting around n = 8.


回答 2

[a + b for a, b in zip(list1, list2)]
[a + b for a, b in zip(list1, list2)]

回答 3

如其他人所述,一种快速且节省空间的解决方案是使用numpy(np)及其内置的矢量处理功能:

1.脾气暴躁

x = np.array([1,2,3])
y = np.array([2,3,4])
print x+y

2.内置

2.1 Lambda

list1=[1, 2, 3]
list2=[4, 5, 6]
print map(lambda x,y:x+y, list1, list2)

注意,map()支持多个参数。

2.2 zip和列表理解

list1=[1, 2, 3]
list2=[4, 5, 6]
print [x + y for x, y in zip(list1, list2)]

As described by others, a fast and also space efficient solution is using numpy (np) with it’s built-in vector manipulation capability:

1. With Numpy

x = np.array([1,2,3])
y = np.array([2,3,4])
print x+y

2. With built-ins

2.1 Lambda

list1=[1, 2, 3]
list2=[4, 5, 6]
print map(lambda x,y:x+y, list1, list2)

Notice that map() supports multiple arguments.

2.2 zip and list comprehension

list1=[1, 2, 3]
list2=[4, 5, 6]
print [x + y for x, y in zip(list1, list2)]

回答 4

numpy我认为使用起来更简单:

import numpy as np
list1=[1,2,3]
list2=[4,5,6]
np.add(list1,list2)

结果:

终端执行

有关详细的参数信息,请在此处检查:numpy.add

It’s simpler to use numpy from my opinion:

import numpy as np
list1=[1,2,3]
list2=[4,5,6]
np.add(list1,list2)

Results:

Terminal execution

For detailed parameter information, check here: numpy.add


回答 5

也许“最Python的方式”应该包括处理list1和list2大小不相同的情况。应用其中一些方法会悄悄地为您提供答案。numpy方法会让您知道,很可能是发生ValueError。

例:

import numpy as np
>>> list1 = [ 1, 2 ]
>>> list2 = [ 1, 2, 3]
>>> list3 = [ 1 ]
>>> [a + b for a, b in zip(list1, list2)]
[2, 4]
>>> [a + b for a, b in zip(list1, list3)]
[2]
>>> a = np.array (list1)
>>> b = np.array (list2)
>>> a+b
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: operands could not be broadcast together with shapes (2) (3)

如果这在您的问题中起作用,您可能想要哪个结果?

Perhaps “the most pythonic way” should include handling the case where list1 and list2 are not the same size. Applying some of these methods will quietly give you an answer. The numpy approach will let you know, most likely with a ValueError.

Example:

import numpy as np
>>> list1 = [ 1, 2 ]
>>> list2 = [ 1, 2, 3]
>>> list3 = [ 1 ]
>>> [a + b for a, b in zip(list1, list2)]
[2, 4]
>>> [a + b for a, b in zip(list1, list3)]
[2]
>>> a = np.array (list1)
>>> b = np.array (list2)
>>> a+b
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: operands could not be broadcast together with shapes (2) (3)

Which result might you want if this were in a function in your problem?


回答 6

这很简单 numpy.add()

import numpy

list1 = numpy.array([1, 2, 3])
list2 = numpy.array([4, 5, 6])
result = numpy.add(list1, list2) # result receive element-wise addition of list1 and list2
print(result)
array([5, 7, 9])

在这里查看文档

如果要接收python列表:

result.tolist()

This is simple with numpy.add()

import numpy

list1 = numpy.array([1, 2, 3])
list2 = numpy.array([4, 5, 6])
result = numpy.add(list1, list2) # result receive element-wise addition of list1 and list2
print(result)
array([5, 7, 9])

See doc here

If you want to receiver a python list:

result.tolist()

回答 7

这将适用于2个或更多列表;遍历列表列表,但使用numpy加法处理每个列表的元素

import numpy as np
list1=[1, 2, 3]
list2=[4, 5, 6]

lists = [list1, list2]
list_sum = np.zeros(len(list1))
for i in lists:
   list_sum += i
list_sum = list_sum.tolist()    

[5.0, 7.0, 9.0]

This will work for 2 or more lists; iterating through the list of lists, but using numpy addition to deal with elements of each list

import numpy as np
list1=[1, 2, 3]
list2=[4, 5, 6]

lists = [list1, list2]
list_sum = np.zeros(len(list1))
for i in lists:
   list_sum += i
list_sum = list_sum.tolist()    

[5.0, 7.0, 9.0]

回答 8

也许这是pythonic的,如果列表数量未知且未导入任何内容,则稍微有用。

只要列表的长度相同,就可以使用以下功能。

在这里,* args接受可变数量的列表参数(但每个参数仅求和相同数量的元素)。

在返回的列表中再次使用*来解压缩每个列表中的元素。

def sum_lists(*args):
    return list(map(sum, zip(*args)))

a = [1,2,3]
b = [1,2,3]  

sum_lists(a,b)

输出:

[2, 4, 6]

或有3个清单

sum_lists([5,5,5,5,5], [10,10,10,10,10], [4,4,4,4,4])

输出:

[19, 19, 19, 19, 19]

Perhaps this is pythonic and slightly useful if you have an unknown number of lists, and without importing anything.

As long as the lists are of the same length, you can use the below function.

Here the *args accepts a variable number of list arguments (but only sums the same number of elements in each).

The * is used again in the returned list to unpack the elements in each of the lists.

def sum_lists(*args):
    return list(map(sum, zip(*args)))

a = [1,2,3]
b = [1,2,3]  

sum_lists(a,b)

Output:

[2, 4, 6]

Or with 3 lists

sum_lists([5,5,5,5,5], [10,10,10,10,10], [4,4,4,4,4])

Output:

[19, 19, 19, 19, 19]

回答 9

将地图与lambda函数一起使用:

>>> map(lambda x, y: x + y, list1, list2)
[5, 7, 9]

Use map with lambda function:

>>> map(lambda x, y: x + y, list1, list2)
[5, 7, 9]

回答 10

我尚未计时,但我怀疑这会很快:

import numpy as np
list1=[1, 2, 3]
list2=[4, 5, 6]

list_sum = (np.add(list1, list2)).tolist()

[5, 7, 9]

I haven’t timed it but I suspect this would be pretty quick:

import numpy as np
list1=[1, 2, 3]
list2=[4, 5, 6]

list_sum = (np.add(list1, list2)).tolist()

[5, 7, 9]

回答 11

如果您需要处理不同大小的列表,请不用担心!精彩的itertools模块涵盖了:

>>> from itertools import zip_longest
>>> list1 = [1,2,1]
>>> list2 = [2,1,2,3]
>>> [sum(x) for x in zip_longest(list1, list2, fillvalue=0)]
[3, 3, 3, 3]
>>>

在Python 2中,zip_longest称为izip_longest

另请参阅此相关答案并评论另一个问题

If you need to handle lists of different sizes, worry not! The wonderful itertools module has you covered:

>>> from itertools import zip_longest
>>> list1 = [1,2,1]
>>> list2 = [2,1,2,3]
>>> [sum(x) for x in zip_longest(list1, list2, fillvalue=0)]
[3, 3, 3, 3]
>>>

In Python 2, zip_longest is called izip_longest.

See also this relevant answer and comment on another question.


回答 12

[list1[i] + list2[i] for i in range(len(list1))]
[list1[i] + list2[i] for i in range(len(list1))]

回答 13

虽然,实际的问题并不想遍历列表来生成结果,但是已经提出的所有解决方案实际上都是在后台进行的!

刷新:如果不查看所有矢量元素,则不能添加两个矢量。因此,大多数这些解决方案的算法复杂度为Big-O(n)。其中n是向量的维数。

因此,从算法的角度来看,使用for循环迭代生成结果列表也是逻辑和pythonic的。但是,此外,此方法没有调用或导入任何其他库的开销。

# Assumption: The lists are of equal length.
resultList = [list1[i] + list2[i] for i in range(len(list1))]

这里显示/讨论的时间取决于系统和实现,并且不是测量操作效率的可靠措施。在任何情况下,向量加法运算的大O复杂度都是线性的,即O(n)。

Although, the actual question does not want to iterate over the list to generate the result, but all the solutions that has been proposed does exactly that under-neath the hood!

To refresh: You cannot add two vectors without looking into all the vector elements. So, the algorithmic complexity of most of these solutions are Big-O(n). Where n is the dimension of the vector.

So, from an algorithmic point of view, using a for loop to iteratively generate the resulting list is logical and pythonic too. However, in addition, this method does not have the overhead of calling or importing any additional library.

# Assumption: The lists are of equal length.
resultList = [list1[i] + list2[i] for i in range(len(list1))]

The timings that are being showed/discussed here are system and implementation dependent, and cannot be reliable measure to measure the efficiency of the operation. In any case, the big O complexity of the vector addition operation is linear, meaning O(n).


回答 14

a_list = []
b_list = []
for i in range(1,100):
    a_list.append(random.randint(1,100))

for i in range(1,100):
    a_list.append(random.randint(101,200))
[sum(x) for x in zip(a_list , b_list )]
a_list = []
b_list = []
for i in range(1,100):
    a_list.append(random.randint(1,100))

for i in range(1,100):
    a_list.append(random.randint(101,200))
[sum(x) for x in zip(a_list , b_list )]

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