对二维列表python的每个元素求和

时间:2019-03-05 04:26:15

标签: python python-3.x

我有一个这样的二维列表:

list1 = [[2,4,6,8,9],[8,9,10,12,15],[8,9,4,20,25]]

我想将每一行的每个元素与另一行相加,结果是这样的:

outcome_list = [[10,13,16,20,24],[16,18,14,32,40],[10,13,20,28,34]]

我的代码是这样的:

d = len(list1) 

for i in range(0, d-1):
    list2 = list[i][:] + list[i+1][:]

但这不起作用。

6 个答案:

答案 0 :(得分:1)

可以这样做:

代码:

list1 = [[2, 4, 6, 8, 9], [8, 9, 10, 12, 15], [8, 9, 4, 20, 25]]

print([[sum(l) for l in zip(list1[i], list1[(i+1) % len(list1)])]
       for i in range(len(list1))])

结果:

[[10, 13, 16, 20, 24], [16, 18, 14, 32, 40], [10, 13, 10, 28, 34]]

答案 1 :(得分:0)

zip()与列表理解结合使用:

list1 = [[2,4,6,8,9],[8,9,10,12,15],[8,9,4,20,25]]

list1 = list1 + [list1[0]]
print([list(map(lambda x: sum(x), zip(x, y))) for x, y in zip(list1, list1[1:])])

# [[10, 13, 16, 20, 24], [16, 18, 14, 32, 40], [10, 13, 10, 28, 34]]

答案 2 :(得分:0)

尝试一下

Object.entries(obj)[0][1].thumbId

答案 3 :(得分:0)

您可以通过将列表本身压缩,但将项目向右旋转来对子列表进行配对,然后将d = len(list1) for i in range(0, d-1): list2 = [a + b for a,b in zip(list[i],list[i+1])] 对与列表理解中的 string[] getfilesname() { string folderPath = Path.Combine(Application.persistentDataPath, foldername); string[] filePaths = Directory.GetFiles(folderPath, "*.txt"); foreach (string file in filePaths) { var onlyFileName = Path.GetFileNameWithoutExtension(file); if (mylist.IndexOf(onlyFileName) == -1) { mylist.Add(onlyFileName); } Debug.Log(onlyFileName); } dropi.AddOptions(mylist); return filePaths; } 方法配对:

map

这将返回:

operator.add

答案 4 :(得分:0)

使用numpy.roll

In [1]: import numpy as np

In [2]: a = np.array([[2,4,6,8,9],[8,9,10,12,15],[8,9,4,20,25]])

In [3]: a + np.roll(a, -1, axis=0)
Out[3]:
array([[10, 13, 16, 20, 24],
       [16, 18, 14, 32, 40],
       [10, 13, 10, 28, 34]])

答案 5 :(得分:0)

这是使用itertools的另一种方法,该工具应适用于list1中任意数量的列表:

from itertools import combinations
list1   = [[2,4,6,8,9],[8,9,10,12,15],[8,9,4,20,25]]
outcome = [ list(map(sum,zip(*rows))) for rows in combinations(list1,2) ]