Python 列表列表的总和；返回总和列表
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Sum of list of lists; returns sum list
提问by Albert
Let data = [[3,7,2],[1,4,5],[9,8,7]]
让 data = [[3,7,2],[1,4,5],[9,8,7]]
Let's say I want to sum the elements for the indices of each list in the list, like adding numbers in a matrix column to get a single list. I am assuming that all lists in data are equal in length.
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相等。
print foo(data)
[[3,7,2],
[1,4,5],
[9,8,7]]
_______
>>>[13,19,14]
How can I iterate over the list of lists without getting an index out of range error? Maybe lambda? Thanks!
如何遍历列表列表而不会出现索引超出范围错误？也许拉姆达？谢谢！
采纳答案by RocketDonkey
You could try this:
你可以试试这个：
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 zipand *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.
这里使用的组合zip和*解压的列表，然后根据自己的索引压缩的项目。然后，您使用列表推导式遍历相似索引的组，将它们相加并返回到它们的“原始”位置。
To hopefully make it a bit more clear, here is what happens when you iterate through zip(*l):
为了希望让它更清楚一点，以下是您迭代时会发生的情况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_longestwith a fillvalueof 0- this basically fills missing indices with 0, allowing you to sum all 'columns':
对于长度不等的列表，您可以使用itertools.izip_longesta fillvalueof 0- 这基本上用 填充缺失的索引0，允许您对所有“列”求和：
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_longestwould look like:
在这种情况下，以下是迭代的izip_longest样子：
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)
回答by Marius
This does depend on your assumption that all the inner lists (or rows) are of the same length, but it should do what you want:
这确实取决于您假设所有内部列表（或行）的长度相同，但它应该做您想做的：
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]
回答by Theuni
For any matrix (or other ambitious numerical) operations I would recommend looking into NumPy.
对于任何矩阵（或其他雄心勃勃的数值）运算，我建议研究 NumPy。
The sample for solving the sum of an array along the axis shown in your question would be:
沿着您的问题中显示的轴求解数组总和的示例是：
>>> 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])
Here's numpy's documentation for its sum function: http://docs.scipy.org/doc/numpy/reference/generated/numpy.sum.html#numpy.sum
这是 numpy 的 sum 函数文档：http: //docs.scipy.org/doc/numpy/reference/generated/numpy.sum.html#numpy.sum
Especially the second argument is interesting as it allows easily specify what should be summed up: all elements or only a specific axis of a potentially n-dimensional array(like).
特别是第二个参数很有趣，因为它可以轻松指定应总结的内容：所有元素或仅潜在 n 维数组（如）的特定轴。
回答by Nathan
>>> 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
回答by MySchizoBuddy
This will give you the sum for each sublist
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]
If you want to sum over all elements and get just one sum then use this
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
回答by user3481919
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
回答by Daniel Marmelshtein
numArr = [[12, 4], [1], [2, 3]] sumArr = 0 sumArr = sum(sum(row) for row in numArr) print(sumArr) the answere: 22
numArr = [[12, 4], [1], [2, 3]] sumArr = 0 sumArr = sum(sum(row) for row in numArr) print(sumArr) the answere: 22
what I did: when you do "for" like this for example: [row.append(1) for row in numArr] the list will change to: [[12, 4, 1], [1, 1], [2, 3, 1]] I used the function sum() from python, the function takes the list and do iteration on it and bring the sum of all the numbers in the list. when I did sum(sum()) I got the sum of all the lists in the big list.
我做了什么：当你像这样执行“for”时，例如：[row.append(1) for row in numArr] 列表将更改为：[[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，并得出列表中所有数字的总和。当我做 sum(sum()) 时，我得到了大列表中所有列表的总和。
回答by Mokogwu Chiedu
This solution assumes a square matrix and uses two for loops to loop over the columns and rows, adding column-wise in the process. The result is returned in a list.
此解决方案假设一个方阵，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
回答by pablokimon
The simplest solution that will sum a list of lists of different or identical lengths is:
对不同或相同长度的列表求和的最简单解决方案是：
total = 0
for d in data:
total += sum(d)
Once you understand list comprehension you could shorten it:
一旦你理解了列表理解，你就可以缩短它：
sum([sum(d) for d in data])
回答by wwii
For the case that the data is a list of lists of strings. Sum or concatenate a list of lists of strings elementwise.
对于数据是字符串列表列表的情况。按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']
>>>
