Python：如何从列表中删除空列表？

I have a list with empty lists in it:

我有一个包含空列表的列表：

list1 = [[], [], [], [], [], 'text', 'text2', [], 'moreText']

How can I remove the empty lists so that I get:

如何删除空列表以便我得到：

list2 = ['text', 'text2', 'moreText']

I tried list.remove('') but that doesn't work.

我尝试了list.remove（''），但这不起作用。

9 个解决方案

#1

Try

尝试

list2 = [x for x in list1 if x != []]

If you want to get rid of everything that is "falsy", e.g. empty strings, empty tuples, zeros, you could also use

如果你想摆脱所有“麻痹”的东西，例如空字符串，空元组，零，你也可以使用

list2 = [x for x in list1 if x]

#2

You can use filter() instead of a list comprehension:

您可以使用filter（）而不是列表推导：

list2 = filter(None, list1)

If None is used as first argument to filter(), it filters out every value in the given list, which is False in a boolean context. This includes empty lists.

如果None用作filter（）的第一个参数，它会过滤掉给定列表中的每个值，在布尔上下文中为False。这包括空列表。

It might be slightly faster than the list comprehension, because it only executes a single function in Python, the rest is done in C.

它可能比列表理解稍快，因为它只在Python中执行单个函数，其余的在C中完成。

#3

Calling filter with None will filter out all falsey values from the list (which an empty list is)

使用None调用过滤器将过滤掉列表中的所有falsey值（空列表为）

list2 = filter(None, list1)

#4

>>> list1 = [[], [], [], [], [], 'text', 'text2', [], 'moreText']
>>> list2 = [e for e in list1 if e]
>>> list2
['text', 'text2', 'moreText']

#5

I found this question because I wanted to do the same as the OP. I would like to add the following observation:

我发现了这个问题，因为我想和OP一样。我想补充以下观察：

The iterative way (user225312, Sven Marnach):

迭代方式（user225312，Sven Marnach）：

list2 = [x for x in list1 if x]

Will return a list object in python3 and python2 . Instead the filter way (lunaryorn, Imran) will differently behave over versions:

将在python3和python2中返回一个列表对象。相反，过滤方式（lunaryorn，Imran）将在版本上表现不同：

list2 = filter(None, list1)

It will return a filter object in python3 and a list in python2 (see this question found at the same time). This is a slight difference but it must be take in account when developing compatible scripts.

它将返回python3中的过滤器对象和python2中的列表（同时查看此问题）。这是一个细微的差别，但在开发兼容脚本时必须考虑到这一点。

This does not make any assumption about performances of those solutions. Anyway the filter object can be reverted to a list using:

这不会对这些解决方案的性能做出任何假设。无论如何，过滤器对象可以使用以下方式恢复为列表：

list3 = list(list2)

#6

A few options:

一些选择：

filter(lambda x: len(x) > 0, list1)  # Doesn't work with number types
filter(None, list1)  # Filters out int(0)
filter(lambda x: x==0 or x, list1) # Retains int(0)

sample session:

样本会话：

Python 2.7.1 (r271:86832, Nov 27 2010, 17:19:03) [MSC v.1500 64 bit (AMD64)] on
win32
Type "help", "copyright", "credits" or "license" for more information.
>>> list1 = [[], [], [], [], [], 'text', 'text2', [], 'moreText']
>>> filter(lambda x: len(x) > 0, list1)
['text', 'text2', 'moreText']
>>> list2 = [[], [], [], [], [], 'text', 'text2', [], 'moreText', 0.5, 1, -1, 0]
>>> filter(lambda x: x==0 or x, list2)
['text', 'text2', 'moreText', 0.5, 1, -1, 0]
>>> filter(None, list2)
['text', 'text2', 'moreText', 0.5, 1, -1]
>>>

#7

I needed to filter empty items from more complex nested objects. I made a recursive function to do it. See my SO answer here.

我需要从更复杂的嵌套对象中过滤掉空项。我做了一个递归函数来做到这一点。在这里看到我的答案。

#8

a = [[1,'aa',3,12,'a','b','c','s'],[],[],[1,'aa',7,80,'d','g','f',''],[9,None,11,12,13,14,15,'k']]

b=[]
for lng in range(len(a)):
       if(len(a[lng])>=1):b.append(a[lng])
a=b
print(a)

[[1,'aa',3,12,'a','b','c','s'],[1,'aa',7,80,'d','g','f',''],[9,None,11,12,13,14,15,'k']]

[[1， 'AA'，3,12， 'A'， 'B'， 'C'， 'S']，[1， 'AA'，7,80， 'd'， 'G'，'F ”， '']，[9，无，11,12,13,14,15，数k']]

#9

-2

list1 = [[], [], [], [], [], 'text', 'text2', [], 'moreText']
list2 = []
for item in list1:
    if item!=[]:
        list2.append(item)
print(list2)

output:

输出：

['text', 'text2', 'moreText']

#1

Try

尝试

list2 = [x for x in list1 if x != []]