33

让data = [[3,7,2],[1,4,5],[9,8,7]]

假设我想对列表中每个列表的索引元素求和，例如在矩阵列中添加数字以获得单个列表。我假设数据中的所有列表的长度都是相等的。

    print foo(data)

   [[3,7,2],
    [1,4,5],
    [9,8,7]]
    _______
 >>>[13,19,14]

如何在不出现索引超出范围错误的情况下遍历列表列表？也许是拉姆达？谢谢！

4

10 回答 10

75

You could try this:

In [9]: l = [[3,7,2],[1,4,5],[9,8,7]]

In [10]: [sum(i) for i in zip(*l)]
Out[10]: [13, 19, 14]

This uses a combination of zip and * to unpack the list and then zip the items according to their index. You then use a list comprehension to iterate through the groups of similar indices, summing them and returning in their 'original' position.

To hopefully make it a bit more clear, here is what happens when you iterate through zip(*l):

In [13]: for i in zip(*l):
   ....:     print i
   ....:     
   ....:     
(3, 1, 9)
(7, 4, 8)
(2, 5, 7)

In the case of lists that are of unequal length, you can use itertools.izip_longest with a fillvalue of 0 - this basically fills missing indices with 0, allowing you to sum all 'columns':

In [1]: import itertools

In [2]: l = [[3,7,2],[1,4],[9,8,7,10]]

In [3]: [sum(i) for i in itertools.izip_longest(*l, fillvalue=0)]
Out[3]: [13, 19, 9, 10]

In this case, here is what iterating over izip_longest would look like:

In [4]: for i in itertools.izip_longest(*l, fillvalue=0):
   ...:     print i
   ...:     
(3, 1, 9)
(7, 4, 8)
(2, 0, 7)
(0, 0, 10)
于 2012-12-09T00:18:27.987 回答
13

对于任何矩阵（或其他雄心勃勃的数值）运算，我建议您研究 NumPy。

解决问题中显示的沿轴的数组总和的示例是：

>>> from numpy import array
>>> data = array([[3,7,2],
...     [1,4,5],
...     [9,8,7]])
>>> from numpy import sum
>>> sum(data, 0)
array([13, 19, 14])

这是 numpy 的 sum 函数的文档：http: //docs.scipy.org/doc/numpy/reference/generated/numpy.sum.html#numpy.sum

特别是第二个参数很有趣，因为它可以轻松指定应该总结的内容：所有元素或仅可能是 n 维数组（如）的特定轴。

于 2012-12-09T00:15:54.090 回答
11

这将为您提供每个子列表的总和

data = [[3,7,2],[1,4],[9,8,7,10]]
list(map(sum, data))
[12, 5, 34]

如果你想对所有元素求和并只得到一个总和，那么使用这个

data = [[3,7,2],[1,4],[9,8,7,10]]
sum(sum(data, []))
51
于 2013-07-14T13:04:49.420 回答
2 
>>> data = [[1, 2, 3], [1, 2, 3], [1, 2, 3]]
>>> for column in enumerate(data[0]):
...     count = sum([x[column[0]] for x in data])
...     print 'Column %s: %d' % (column[0], count)
... 
Column 0: 3
Column 1: 6
Column 2: 9
于 2012-12-09T00:17:39.070 回答
0

这确实取决于您假设所有内部列表（或行）的长度相同，但它应该做您想要的：

sum_list = []

ncols = len(data[0])

for col in range(ncols):
    sum_list.append(sum(row[col] for row in data))


sum_list
Out[9]: [13, 19, 14]
于 2012-12-09T00:15:18.783 回答
0 
def sum(L):
res = list()
for j in range(0,len(L[0])):
    tmp = 0
    for i in range(0,len(L)):
        tmp = tmp + L[i][j]
    res.append(tmp)
return res
于 2014-06-19T13:27:27.927 回答
0

将不同或相同长度的列表相加的最简单的解决方案是：

total = 0
for d in data:
    total += sum(d)

一旦你理解了列表理解，你就可以缩短它：

sum([sum(d) for d in data])
于 2019-10-11T06:13:22.687 回答
0

对于数据是字符串列表的情况。逐元素求和或连接字符串列表的列表。

>>> a = [list('abc'),list('def'),list('tyu')]
>>> a
[['a', 'b', 'c'], ['d', 'e', 'f'], ['t', 'y', 'u']]
>>> [''.join(thing) for thing in zip(*a)]
['adt', 'bey', 'cfu']
>>>
于 2020-01-05T18:04:34.070 回答
0 
numArr = [[12, 4], [1], [2, 3]]

sumArr = 0

sumArr = sum(sum(row) for row in numArr)

print(sumArr) 

the answere: 22

我做了什么：当你像这样做“for”时，例如：[row.append(1) for row in numArr] 列表将变为：[[12, 4, 1], [1, 1], [2 , 3, 1]] 我使用了 python 中的 sum() 函数，该函数获取列表并对其进行迭代，并将列表中所有数字的总和。当我执行 sum(sum()) 时，我得到了大列表中所有列表的总和。

于 2019-03-24T10:18:01.123 回答
0

此解决方案假定一个方阵并使用两个 for 循环来循环列和行，在此过程中逐列添加。结果以列表形式返回。

def foo(data):
    # initialise length of data(n) and sum_of_col variable 
    n = len(data)
    sum_of_col = []

    # iterate over column
    for col_i in range(n):
        # column sum
        col_count = 0;

        #iterate over row
        for row_i in range(n):
            col_count += data[row_i][col_i]

        # append sum of column to list    
        sum_of_col.append(col_count)

    return sum_of_col
于 2019-10-11T06:00:21.973 回答