从 Python 中的嵌套列表中删除一列
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Remove a column from a nested list in Python
提问by Crisis
I need help figuring how to work around removing a 'column' from a nested list to modify it.
我需要帮助弄清楚如何解决从嵌套列表中删除“列”以修改它的问题。
Say I have
说我有
L = [[1,2,3,4],
[5,6,7,8],
[9,1,2,3]]
and I want to remove the second column (so values 2,6,1) to get:
我想删除第二列（因此值为 2,6,1）以获得：
L = [[1,3,4],
[5,7,8],
[9,2,3]]
I'm stuck with how to modify the list with just taking out a column. I've done something sort of like this before? Except we were printing it instead, and of course it wouldn't work in this case because I believe the break conflicts with the rest of the values I want in the list.
我被困在如何通过取出一列来修改列表。我以前做过类似的事情吗？除了我们打印它之外，当然在这种情况下它不起作用，因为我相信 break 与列表中我想要的其余值冲突。
def L_break(L):
i = 0
while i < len(L):
k = 0
while k < len(L[i]):
print( L[i][k] , end = " ")
if k == 1:
break
k = k + 1
print()
i = i + 1
So, how would you go about modifying this nested list? Is my mind in the right place comparing it to the code I have posted or does this require something different?
那么，您将如何修改这个嵌套列表？将它与我发布的代码进行比较，我的想法是否正确，或者这是否需要不同的东西？
回答by arshajii
回答by kojiro
If you want to extractthat column for later use, while removing it from the original list, use a list comprehension with pop:
如果您想提取该列以供以后使用，同时将其从原始列表中删除，请使用列表推导式pop：
>>> L = [[1,2,3,4],
... [5,6,7,8],
... [9,1,2,3]]
>>>
>>> [r.pop(1) for r in L]
[2, 6, 1]
>>> L
[[1, 3, 4], [5, 7, 8], [9, 2, 3]]
Otherwise, just loop over the list and delete the fields you no longer want, as in arshajii's answer
否则，只需遍历列表并删除您不再需要的字段，如arshajii 的回答
回答by Hai Vu
Here is one way, updated to take in kojiro's advice.
这是一种方法，更新以采纳 kojiro 的建议。
>>> L[:] = [i[:1]+i[2:] for i in L]
>>> L
[[1, 3, 4], [5, 7, 8], [9, 2, 3]]
You can generalize this to remove any column:
您可以将其概括为删除任何列：
def remove_column(matrix, column):
return [row[:column] + row[column+1:] for row in matrix]
# Remove 2nd column
copyofL = remove_column(L, 1) # Column is zero-base, so, 1=second column
回答by thefourtheye
You can use operator.itemgetter, which is created for this very purpose.
您可以使用operator.itemgetter专为此目的而创建的 。
from operator import itemgetter
getter = itemgetter(0, 2, 3) # Only indexes which are needed
print(list(map(list, map(getter, L))))
# [[1, 3, 4], [5, 7, 8], [9, 2, 3]]
You can use it in List comprehension like this
您可以像这样在列表理解中使用它
print([list(getter(item)) for item in L])
# [[1, 3, 4], [5, 7, 8], [9, 2, 3]]
You can also use nested List Comprehension, in which we skip the elements if the index is 1, like this
您还可以使用嵌套列表理解，如果索引为 1，我们跳过元素，就像这样
print([[item for index, item in enumerate(items) if index != 1] for items in L])
# [[1, 3, 4], [5, 7, 8], [9, 2, 3]]
Note:All these suggested in this answer will not affect the original list. They will generate new lists without the unwanted elements.
注意：此答案中建议的所有这些都不会影响原始列表。他们将生成没有不需要的元素的新列表。
回答by idanshmu
Use map-lambda:
使用map-lambda：
print map(lambda x: x[:1]+x[2:], L)
回答by Shinto Joseph
when you do the delit will delete that index and reset the index, so you have to reduce that index. Here I use the count to reduce and reset the same from the index list we have. Hope this helps. Thanks
当您执行del 时，它将删除该索引并重置该索引，因此您必须减少该索引。在这里，我使用计数来减少和重置我们拥有的索引列表中的计数。希望这可以帮助。谢谢
nested_list = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
remove_cols_index = [1,2]
count = 0
for i in remove_cols_index:
i = i-count
count = count+1
del nested_list[i]
print (nested_list)
