有没有一种简单的方法可以按值删除列表元素?

问题:有没有一种简单的方法可以按值删除列表元素?

a = [1, 2, 3, 4]
b = a.index(6)

del a[b]
print(a)

上面显示了以下错误:

Traceback (most recent call last):
  File "D:\zjm_code\a.py", line 6, in <module>
    b = a.index(6)
ValueError: list.index(x): x not in list

所以我必须这样做:

a = [1, 2, 3, 4]

try:
    b = a.index(6)
    del a[b]
except:
    pass

print(a)

但是,没有简单的方法可以做到这一点吗?

a = [1, 2, 3, 4]
b = a.index(6)

del a[b]
print(a)

The above shows the following error:

Traceback (most recent call last):
  File "D:\zjm_code\a.py", line 6, in <module>
    b = a.index(6)
ValueError: list.index(x): x not in list

So I have to do this:

a = [1, 2, 3, 4]

try:
    b = a.index(6)
    del a[b]
except:
    pass

print(a)

But is there not a simpler way to do this?


回答 0

要删除列表中元素的首次出现,只需使用list.remove

>>> a = ['a', 'b', 'c', 'd']
>>> a.remove('b')
>>> print(a)
['a', 'c', 'd']

请注意,它不会删除所有出现的元素。为此使用列表理解。

>>> a = [10, 20, 30, 40, 20, 30, 40, 20, 70, 20]
>>> a = [x for x in a if x != 20]
>>> print(a)
[10, 30, 40, 30, 40, 70]

To remove an element’s first occurrence in a list, simply use list.remove:

>>> a = ['a', 'b', 'c', 'd']
>>> a.remove('b')
>>> print(a)
['a', 'c', 'd']

Mind that it does not remove all occurrences of your element. Use a list comprehension for that.

>>> a = [10, 20, 30, 40, 20, 30, 40, 20, 70, 20]
>>> a = [x for x in a if x != 20]
>>> print(a)
[10, 30, 40, 30, 40, 70]

回答 1

通常,如果您告诉Python做一些它无法做的事情,它将抛出一个Exception,因此您必须执行以下任一操作:

if c in a:
    a.remove(c)

要么:

try:
    a.remove(c)
except ValueError:
    pass

只要它是您期望的并且可以正确处理的异常,它就不一定是一件坏事。

Usually Python will throw an Exception if you tell it to do something it can’t so you’ll have to do either:

if c in a:
    a.remove(c)

or:

try:
    a.remove(c)
except ValueError:
    pass

An Exception isn’t necessarily a bad thing as long as it’s one you’re expecting and handle properly.


回答 2

你可以做

a=[1,2,3,4]
if 6 in a:
    a.remove(6)

但以上需要在列表中搜索6 2次,因此尝试使用除外会更快

try:
    a.remove(6)
except:
    pass

You can do

a=[1,2,3,4]
if 6 in a:
    a.remove(6)

but above need to search 6 in list a 2 times, so try except would be faster

try:
    a.remove(6)
except:
    pass

回答 3

考虑:

a = [1,2,2,3,4,5]

要排除所有情况,可以在python中使用filter函数。例如,它看起来像:

a = list(filter(lambda x: x!= 2, a))

因此,它将保留的所有元素a != 2

仅取出其中一项使用

a.remove(2)

Consider:

a = [1,2,2,3,4,5]

To take out all occurrences, you could use the filter function in python. For example, it would look like:

a = list(filter(lambda x: x!= 2, a))

So, it would keep all elements of a != 2.

To just take out one of the items use

a.remove(2)

回答 4

这是就地执行此操作的方法(无需列表理解):

def remove_all(seq, value):
    pos = 0
    for item in seq:
        if item != value:
           seq[pos] = item
           pos += 1
    del seq[pos:]

Here’s how to do it inplace (without list comprehension):

def remove_all(seq, value):
    pos = 0
    for item in seq:
        if item != value:
           seq[pos] = item
           pos += 1
    del seq[pos:]

回答 5

如果您知道要删除的值,这是一种简单的方法(无论如何,我仍然可以想到):

a = [0, 1, 1, 0, 1, 2, 1, 3, 1, 4]
while a.count(1) > 0:
    a.remove(1)

你会得到 [0, 0, 2, 3, 4]

If you know what value to delete, here’s a simple way (as simple as I can think of, anyway):

a = [0, 1, 1, 0, 1, 2, 1, 3, 1, 4]
while a.count(1) > 0:
    a.remove(1)

You’ll get [0, 0, 2, 3, 4]


回答 6

如果集合适用于您的应用程序,则另一种可能性是使用集合而不是列表。

IE,如果您的数据未排序,并且没有重复,则

my_set=set([3,4,2])
my_set.discard(1)

没有错误。

通常,列表只是一个方便存放实际未排序商品的容器。有一些问题询问如何从列表中删除所有出现的元素。如果您不想一开始就喜欢做傻瓜,那么再说一遍就很方便了。

my_set.add(3)

不会从上面改变my_set。

Another possibility is to use a set instead of a list, if a set is applicable in your application.

IE if your data is not ordered, and does not have duplicates, then

my_set=set([3,4,2])
my_set.discard(1)

is error-free.

Often a list is just a handy container for items that are actually unordered. There are questions asking how to remove all occurences of an element from a list. If you don’t want dupes in the first place, once again a set is handy.

my_set.add(3)

doesn’t change my_set from above.


回答 7

如许多其他答案所述,它list.remove()可以工作,但ValueError如果该项目不在列表中,则抛出a 。在python 3.4及更高版本中,有一种有趣的方法可以使用抑制上下文管理器来处理此问题:

from contextlib import suppress
with suppress(ValueError):
    a.remove('b')

As stated by numerous other answers, list.remove() will work, but throw a ValueError if the item wasn’t in the list. With python 3.4+, there’s an interesting approach to handling this, using the suppress contextmanager:

from contextlib import suppress
with suppress(ValueError):
    a.remove('b')

回答 8

通过使用列表的remove方法,可以更轻松地在列表中查找值,然后删除该索引(如果存在)。

>>> a = [1, 2, 3, 4]
>>> try:
...   a.remove(6)
... except ValueError:
...   pass
... 
>>> print a
[1, 2, 3, 4]
>>> try:
...   a.remove(3)
... except ValueError:
...   pass
... 
>>> print a
[1, 2, 4]

如果您经常这样做,则可以将其包装在一个函数中:

def remove_if_exists(L, value):
  try:
    L.remove(value)
  except ValueError:
    pass

Finding a value in a list and then deleting that index (if it exists) is easier done by just using list’s remove method:

>>> a = [1, 2, 3, 4]
>>> try:
...   a.remove(6)
... except ValueError:
...   pass
... 
>>> print a
[1, 2, 3, 4]
>>> try:
...   a.remove(3)
... except ValueError:
...   pass
... 
>>> print a
[1, 2, 4]

If you do this often, you can wrap it up in a function:

def remove_if_exists(L, value):
  try:
    L.remove(value)
  except ValueError:
    pass

回答 9

这个例子很快,并且将从列表中删除该值的所有实例:

a = [1,2,3,1,2,3,4]
while True:
    try:
        a.remove(3)
    except:
        break
print a
>>> [1, 2, 1, 2, 4]

This example is fast and will delete all instances of a value from the list:

a = [1,2,3,1,2,3,4]
while True:
    try:
        a.remove(3)
    except:
        break
print a
>>> [1, 2, 1, 2, 4]

回答 10

一行:

a.remove('b') if 'b' in a else None

有时很有用。

更简单:

if 'b' in a: a.remove('b')

In one line:

a.remove('b') if 'b' in a else None

sometimes it usefull.

Even easier:

if 'b' in a: a.remove('b')

回答 11

如果您的元素是不同的,那么简单的集合差异就可以了。

c = [1,2,3,4,'x',8,6,7,'x',9,'x']
z = list(set(c) - set(['x']))
print z
[1, 2, 3, 4, 6, 7, 8, 9]

If your elements are distinct, then a simple set difference will do.

c = [1,2,3,4,'x',8,6,7,'x',9,'x']
z = list(set(c) - set(['x']))
print z
[1, 2, 3, 4, 6, 7, 8, 9]

回答 12

我们还可以使用.pop:

>>> lst = [23,34,54,45]
>>> remove_element = 23
>>> if remove_element in lst:
...     lst.pop(lst.index(remove_element))
... 
23
>>> lst
[34, 54, 45]
>>> 

We can also use .pop:

>>> lst = [23,34,54,45]
>>> remove_element = 23
>>> if remove_element in lst:
...     lst.pop(lst.index(remove_element))
... 
23
>>> lst
[34, 54, 45]
>>> 

回答 13

通过索引除您要删除的元素以外的所有内容来覆盖列表

>>> s = [5,4,3,2,1]
>>> s[0:2] + s[3:]
[5, 4, 2, 1]

Overwrite the list by indexing everything except the elements you wish to remove

>>> s = [5,4,3,2,1]
>>> s[0:2] + s[3:]
[5, 4, 2, 1]

回答 14

有一个for循环和一个条件:

def cleaner(seq, value):    
    temp = []                      
    for number in seq:
        if number != value:
            temp.append(number)
    return temp

如果要删除一些但不是全部:

def cleaner(seq, value, occ):
    temp = []
    for number in seq:
        if number == value and occ:
            occ -= 1
            continue
        else:
            temp.append(number)
    return temp

With a for loop and a condition:

def cleaner(seq, value):    
    temp = []                      
    for number in seq:
        if number != value:
            temp.append(number)
    return temp

And if you want to remove some, but not all:

def cleaner(seq, value, occ):
    temp = []
    for number in seq:
        if number == value and occ:
            occ -= 1
            continue
        else:
            temp.append(number)
    return temp

回答 15

 list1=[1,2,3,3,4,5,6,1,3,4,5]
 n=int(input('enter  number'))
 while n in list1:
    list1.remove(n)
 print(list1)
 list1=[1,2,3,3,4,5,6,1,3,4,5]
 n=int(input('enter  number'))
 while n in list1:
    list1.remove(n)
 print(list1)

回答 16

举例来说,我们要从x中删除所有1。这就是我要做的:

x = [1, 2, 3, 1, 2, 3]

现在,这是我的方法的实际用法:

def Function(List, Unwanted):
    [List.remove(Unwanted) for Item in range(List.count(Unwanted))]
    return List
x = Function(x, 1)
print(x)

这是我的方法,只需一行:

[x.remove(1) for Item in range(x.count(1))]
print(x)

两者都将其作为输出:

[2, 3, 2, 3, 2, 3]

希望这可以帮助。PS,请注意,这是在3.6.2版中编写的,因此您可能需要针对旧版本进行调整。

Say for example, we want to remove all 1’s from x. This is how I would go about it:

x = [1, 2, 3, 1, 2, 3]

Now, this is a practical use of my method:

def Function(List, Unwanted):
    [List.remove(Unwanted) for Item in range(List.count(Unwanted))]
    return List
x = Function(x, 1)
print(x)

And this is my method in a single line:

[x.remove(1) for Item in range(x.count(1))]
print(x)

Both yield this as an output:

[2, 3, 2, 3, 2, 3]

Hope this helps. PS, pleas note that this was written in version 3.6.2, so you might need to adjust it for older versions.


回答 17

也许您的解决方案适用于int,但不适用于字典。

一方面,remove()对我不起作用。但也许它适用于基本类型。我猜下面的代码也是从对象列表中删除项目的方法。

另一方面,“ del”也无法正常工作。就我而言,使用python 3.6:当我尝试使用“ del”命令从“ for”气泡中的列表中删除元素时,python会更改进程中的索引,而bucle会在时间之前过早停止。仅当您以相反的顺序删除一个元素时,它才有效。这样,您在遍历时不会更改未决元素数组索引

然后,我用了:

c = len(list)-1
for element in (reversed(list)):
    if condition(element):
        del list[c]
    c -= 1
print(list)

其中“列表”类似于[{‘key1’:value1’},{‘key2’:value2},{‘key3’:value3},…]

另外,您可以使用enumerate做更多的pythonic操作:

for i, element in enumerate(reversed(list)):
    if condition(element):
        del list[(i+1)*-1]
print(list)

Maybe your solutions works with ints, but It Doesnt work for me with dictionarys.

In one hand, remove() has not worked for me. But maybe it works with basic Types. I guess the code bellow is also the way to remove items from objects list.

In the other hand, ‘del’ has not worked properly either. In my case, using python 3.6: when I try to delete an element from a list in a ‘for’ bucle with ‘del’ command, python changes the index in the process and bucle stops prematurely before time. It only works if You delete element by element in reversed order. In this way you dont change the pending elements array index when you are going through it

Then, Im used:

c = len(list)-1
for element in (reversed(list)):
    if condition(element):
        del list[c]
    c -= 1
print(list)

where ‘list’ is like [{‘key1′:value1’},{‘key2’:value2}, {‘key3’:value3}, …]

Also You can do more pythonic using enumerate:

for i, element in enumerate(reversed(list)):
    if condition(element):
        del list[(i+1)*-1]
print(list)

回答 18

arr = [1, 1, 3, 4, 5, 2, 4, 3]

# to remove first occurence of that element, suppose 3 in this example
arr.remove(3)

# to remove all occurences of that element, again suppose 3
# use something called list comprehension
new_arr = [element for element in arr if element!=3]

# if you want to delete a position use "pop" function, suppose 
# position 4 
# the pop function also returns a value
removed_element = arr.pop(4)

# u can also use "del" to delete a position
del arr[4]
arr = [1, 1, 3, 4, 5, 2, 4, 3]

# to remove first occurence of that element, suppose 3 in this example
arr.remove(3)

# to remove all occurences of that element, again suppose 3
# use something called list comprehension
new_arr = [element for element in arr if element!=3]

# if you want to delete a position use "pop" function, suppose 
# position 4 
# the pop function also returns a value
removed_element = arr.pop(4)

# u can also use "del" to delete a position
del arr[4]

回答 19

"-v"将从数组中删除所有实例sys.argv,并且如果未找到实例,则不会发出任何投诉:

while "-v" in sys.argv:
    sys.argv.remove('-v')

您可以在名为的文件中查看运行中的代码speechToText.py

$ python speechToText.py -v
['speechToText.py']

$ python speechToText.py -x
['speechToText.py', '-x']

$ python speechToText.py -v -v
['speechToText.py']

$ python speechToText.py -v -v -x
['speechToText.py', '-x']

This removes all instances of "-v" from the array sys.argv, and does not complain if no instances were found:

while "-v" in sys.argv:
    sys.argv.remove('-v')

You can see the code in action, in a file called speechToText.py:

$ python speechToText.py -v
['speechToText.py']

$ python speechToText.py -x
['speechToText.py', '-x']

$ python speechToText.py -v -v
['speechToText.py']

$ python speechToText.py -v -v -x
['speechToText.py', '-x']

回答 20

这就是我的回答,只是使用

def remove_all(data, value):
    i = j = 0
    while j < len(data):
        if data[j] == value:
            j += 1
            continue
        data[i] = data[j]
        i += 1
        j += 1
    for x in range(j - i):
        data.pop()

this is my answer, just use while and for

def remove_all(data, value):
    i = j = 0
    while j < len(data):
        if data[j] == value:
            j += 1
            continue
        data[i] = data[j]
        i += 1
        j += 1
    for x in range(j - i):
        data.pop()