blocks|key|1843854|text|要删除列表中元素的第一个匹配项，只需使用list.remove|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1843855|>>>+a+=+['a',+'b',+'c',+'d']
>>>+a.remove('b')
>>>+print(a)
['a',+'c',+'d']|code-block|syntax|javascript|1843856|注意，它不会删除元素的所有匹配项。对此使用列表理解。|1843857|>>>+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]|1843858|entityMap^0|K|B|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

To remove an element's first occurrence in a list, simply use <code>list.remove</code>:

<pre><code>&gt;&gt;&gt; a = ['a', 'b', 'c', 'd']
&gt;&gt;&gt; a.remove('b')
&gt;&gt;&gt; print(a)
['a', 'c', 'd']
</code></pre>

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

<pre><code>&gt;&gt;&gt; a = [10, 20, 30, 40, 20, 30, 40, 20, 70, 20]
&gt;&gt;&gt; a = [x for x in a if x != 20]
&gt;&gt;&gt; print(a)
[10, 30, 40, 30, 40, 70]
</code></pre>

blocks|key|3251472|text|你可以做到|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3251473|a=[1,2,3,4]
if+6+in+a:
++++a.remove(6)|code-block|syntax|javascript|3251474|但上面需要在列表a中搜索6个2次，所以尝试一下，除了会更快|3251475|try:
++++a.remove(6)
except:
++++pass|3251476|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

You can do

<pre><code>a=[1,2,3,4]
if 6 in a:
 a.remove(6)
</code></pre>

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

<pre><code>try:
 a.remove(6)
except:
 pass
</code></pre>

blocks|key|3251537|text|只需使用list的remove方法，就可以更容易地在列表中查找一个值，然后删除该索引(如果存在)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3251538|>>>+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]|code-block|syntax|javascript|3251539|如果您经常这样做，您可以将其封装在一个函数中：|3251540|def+remove_if_exists(L,+value):
++try:
++++L.remove(value)
++except+ValueError:
++++pass|3251541|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

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

<pre><code>&gt;&gt;&gt; a = [1, 2, 3, 4]
&gt;&gt;&gt; try:
... a.remove(6)
... except ValueError:
... pass
... 
&gt;&gt;&gt; print a
[1, 2, 3, 4]
&gt;&gt;&gt; try:
... a.remove(3)
... except ValueError:
... pass
... 
&gt;&gt;&gt; print a
[1, 2, 4]
</code></pre>

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

<pre><code>def remove_if_exists(L, value):
 try:
 L.remove(value)
 except ValueError:
 pass
</code></pre>

blocks|key|1188688|text|通常，如果你告诉Python做一些它不能做的事情，它会抛出一个异常，所以你必须这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1188689|if+c+in+a:
++++a.remove(c)|code-block|syntax|javascript|1188690|或者：|1188691|try:
++++a.remove(c)
except+ValueError:
++++pass|1188692|异常并不一定是一件坏事，只要它是您所期望的并得到正确处理。|1188693|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

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

<pre><code>if c in a:
 a.remove(c)
</code></pre>

or:

<pre><code>try:
 a.remove(c)
except ValueError:
 pass
</code></pre>

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

blocks|key|1844045|text|下面是如何就地完成它(不需要列表理解)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1844046|def+remove_all(seq,+value):
++++pos+=+0
++++for+item+in+seq:
++++++++if+item+!=+value:
+++++++++++seq[pos]+=+item
+++++++++++pos+%2B=+1
++++del+seq[pos:]|code-block|syntax|javascript|1844047|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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

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

blocks|key|1186139|text|如果你知道要删除什么值，这里有一个简单的方法(无论如何，就像我能想到的那样简单)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1186140|a+=+[0,+1,+1,+0,+1,+2,+1,+3,+1,+4]
while+a.count(1)+>+0:
++++a.remove(1)|code-block|syntax|javascript|1186141|你会得到[0,+0,+2,+3,+4]|offset|length|style|CODE|1186142|entityMap^0|0|0|4|F|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

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

<pre><code>a = [0, 1, 1, 0, 1, 2, 1, 3, 1, 4]
while a.count(1) &gt; 0:
 a.remove(1)
</code></pre>

You'll get
<code>[0, 0, 2, 3, 4]</code>

blocks|key|1844148|text|考虑一下：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1844149|a+=+[1,2,2,3,4,5]|code-block|syntax|javascript|1844150|要删除所有匹配项，可以使用python中的filter函数。例如，它将如下所示：|1844151|a+=+list(filter(lambda+x:+x!=+2,+a))|1844152|因此，它将保留a+!=+2的所有元素。|offset|length|style|CODE|1844153|要删除其中的一项，请使用|1844154|a.remove(2)|1844155|entityMap^0|0|0|0|0|7|6|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@$M|11|N|12|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

Consider:

<pre><code>a = [1,2,2,3,4,5]
</code></pre>

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

<pre><code>a = list(filter(lambda x: x!= 2, a))
</code></pre>

So, it would keep all elements of <code>a != 2</code>.

To just take out one of the items use 

<pre><code>a.remove(2)
</code></pre>

blocks|key|3251759|text|此示例速度很快，将从列表中删除某个值的所有实例：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3251760|a+=+[1,2,3,1,2,3,4]
while+True:
++++try:
++++++++a.remove(3)
++++except:
++++++++break
print+a
>>>+[1,+2,+1,+2,+4]|code-block|syntax|javascript|3251761|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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

<pre><code>a = [1,2,3,1,2,3,4]
while True:
 try:
 a.remove(3)
 except:
 break
print a
&gt;&gt;&gt; [1, 2, 1, 2, 4]
</code></pre>

blocks|key|440850|text|另一种可能是使用集合而不是列表，如果集合适用于您的应用程序。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|440851|例如，如果您的数据未排序，并且没有重复项，则|440852|my_set=set([3,4,2])
my_set.discard(1)|code-block|syntax|javascript|440853|是没有错误的。|440854|通常，列表只是一个方便的容器，用于存放实际上是无序的项。有一些问题在问如何从列表中删除元素的所有匹配项。如果你一开始就不想上当受骗，那么再一次使用set是很方便的。|440855|my_set.add(3)|440856|不会从上面更改my_set。|440857|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|X|8|@]|9|@]|A|$G|H]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

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

<pre><code>my_set=set([3,4,2])
my_set.discard(1)
</code></pre>

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.

<pre><code>my_set.add(3)
</code></pre>

doesn't change my_set from above.

blocks|key|3251822|text|如果你的元素是不同的，那么一个简单的集合差异就可以了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3251823|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]|code-block|syntax|javascript|3251824|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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

<pre><code>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]
</code></pre>

blocks|key|1844234|text|我们也可以使用.pop：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1844235|>>>+lst+=+[23,34,54,45]
>>>+remove_element+=+23
>>>+if+remove_element+in+lst:
...+++++lst.pop(lst.index(remove_element))
...+
23
>>>+lst
[34,+54,+45]
>>>+|code-block|syntax|javascript|1844236|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

We can also use .pop:

<pre><code>&gt;&gt;&gt; lst = [23,34,54,45]
&gt;&gt;&gt; remove_element = 23
&gt;&gt;&gt; if remove_element in lst:
... lst.pop(lst.index(remove_element))
... 
23
&gt;&gt;&gt; lst
[34, 54, 45]
&gt;&gt;&gt; 
</code></pre>

blocks|key|1844267|text|通过索引除要删除的元素之外的所有内容来覆盖列表|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1844268|>>>+s+=+[5,4,3,2,1]
>>>+s[0:2]+%2B+s[3:]
[5,+4,+2,+1]|code-block|syntax|javascript|1844269|更一般地说，|1844270|>>>+s+=+[5,4,3,2,1]
>>>+i+=+s.index(3)
>>>+s[:i]+%2B+s[i%2B1:]
[5,+4,+2,+1]|1844271|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Overwrite the list by indexing everything except the elements you wish to remove
<pre><code>&gt;&gt;&gt; s = [5,4,3,2,1]
&gt;&gt;&gt; s[0:2] + s[3:]
[5, 4, 2, 1]
</code></pre>
More generally,
<pre><code>&gt;&gt;&gt; s = [5,4,3,2,1]
&gt;&gt;&gt; i = s.index(3)
&gt;&gt;&gt; s[:i] + s[i+1:]
[5, 4, 2, 1]
</code></pre>

blocks|key|3251884|text|使用for循环和条件：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3251885|def+cleaner(seq,+value):++++
++++temp+=+[]++++++++++++++++++++++
++++for+number+in+seq:
++++++++if+number+!=+value:
++++++++++++temp.append(number)
++++return+temp|code-block|syntax|javascript|3251886|如果你想删除一些，但不是全部：|3251887|def+cleaner(seq,+value,+occ):
++++temp+=+[]
++++for+number+in+seq:
++++++++if+number+==+value+and+occ:
++++++++++++occ+-=+1
++++++++++++continue
++++++++else:
++++++++++++temp.append(number)
++++return+temp|3251888|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

With a for loop and a condition:

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

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

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

blocks|key|3251898|text|+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)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3251899|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> 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)
</code></pre>

blocks|key|441474|text|例如，我们想要从x中删除所有的1。我会这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|441475|x+=+[1,+2,+3,+1,+2,+3]|code-block|syntax|javascript|441476|现在，这是我的方法的一个实际应用：|441477|def+Function(List,+Unwanted):
++++[List.remove(Unwanted)+for+Item+in+range(List.count(Unwanted))]
++++return+List
x+=+Function(x,+1)
print(x)|441478|这是我在一行中的方法：|441479|[x.remove(1)+for+Item+in+range(x.count(1))]
print(x)|441480|这两种方法都会产生以下输出：|441481|[2,+3,+2,+3,+2,+3]|441482|希望这能有所帮助。请注意，这是在3.6.2版本中编写的，因此您可能需要针对较旧的版本进行调整。|441483|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

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

<pre><code>x = [1, 2, 3, 1, 2, 3]
</code></pre>

Now, this is a practical use of my method:

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

And this is my method in a single line:

<pre><code>[x.remove(1) for Item in range(x.count(1))]
print(x)
</code></pre>

Both yield this as an output:

<pre><code>[2, 3, 2, 3, 2, 3]
</code></pre>

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.

blocks|key|441505|text|正如许多其他答案所指出的那样，list.remove()可以工作，但如果该项目不在列表中，则会抛出ValueError。对于python+3.4%2B，有一种有趣的方法来处理此问题，即使用suppress+contextmanager|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|441506|from+contextlib+import+suppress
with+suppress(ValueError):
++++a.remove('b')|code-block|syntax|javascript|441507|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3.6/library/contextlib.html#contextlib.suppress^0|F|D|1D|A|2K|N|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

As stated by numerous other answers, <code>list.remove()</code> will work, but throw a <code>ValueError</code> if the item wasn't in the list. With python 3.4+, there's an interesting approach to handling this, using the <a href="https://docs.python.org/3.6/library/contextlib.html#contextlib.suppress" rel="noreferrer">suppress contextmanager</a>:

<pre><code>from contextlib import suppress
with suppress(ValueError):
 a.remove('b')
</code></pre>

blocks|key|1844390|text|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]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1844391|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>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]
</code></pre>

blocks|key|1186327|text|这将从数组sys.argv中删除"-v"的所有实例，如果未找到任何实例，则不会出现任何错误：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1186328|while+"-v"+in+sys.argv:
++++sys.argv.remove('-v')|code-block|syntax|javascript|1186329|您可以在一个名为speechToText.py的文件中查看实际运行的代码|1186330|$+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']|1186331|entityMap^0|5|8|G|4|0|0|8|F|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

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

<pre><code>while "-v" in sys.argv:
 sys.argv.remove('-v')
</code></pre>

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

<pre><code>$ 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']
</code></pre>

blocks|key|1844469|text|也许你的解决方案适用于It，但它不适用于我的字典。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1844470|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。|1844471|另一方面，“del”也没有正常工作。在我的例子中，使用的是python3.6:当我尝试使用'del‘命令从'for’bucle中的列表中删除一个元素时，python更改了进程中的索引，并且bucle提前停止。仅当您按reversed+order逐个删除元素时，它才起作用。这样，在遍历挂起元素数组索引时，就不会更改它|offset|length|style|BOLD|1844472|然后，使用Im：|1844473|c+=+len(list)-1
for+element+in+(reversed(list)):
++++if+condition(element):
++++++++del+list[c]
++++c+-=+1
print(list)|code-block|syntax|javascript|1844474|其中'list‘类似于{'key1':value1'}，{'key2':value2}，{'key3':value3}，...|1844475|此外，您还可以使用enumerate执行更多pythonic操作：|1844476|for+i,+element+in+enumerate(reversed(list)):
++++if+condition(element):
++++++++del+list[(i%2B1)*-1]
print(list)|1844477|entityMap^0|0|0|30|8|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|10|8|@$F|11|G|12|H|I]]|9|@]|A|$]]|$1|J|3|K|5|6|7|13|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|14|8|@]|9|@]|A|$O|P]]|$1|Q|3|R|5|6|7|15|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|16|8|@]|9|@]|A|$]]|$1|U|3|V|5|N|7|17|8|@]|9|@]|A|$O|P]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

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:

<pre><code>c = len(list)-1
for element in (reversed(list)):
 if condition(element):
 del list[c]
 c -= 1
print(list)
</code></pre>

where 'list' is like [{'key1':value1'},{'key2':value2}, {'key3':value3}, ...]

Also You can do more pythonic using enumerate:

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

blocks|key|441557|text|在一行代码中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|441558|a.remove('b')+if+'b'+in+a+else+None|code-block|syntax|javascript|441559|有时它很有用。|441560|更简单：|441561|if+'b'+in+a:+a.remove('b')|441562|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

In one line:

<pre><code>a.remove('b') if 'b' in a else None
</code></pre>

sometimes it usefull.

Even easier:

<pre><code>if 'b' in a: a.remove('b')
</code></pre>

blocks|key|1844516|text|这就是我的答案，只需使用，而和则用于|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|1844517|def+remove_all(data,+value):
++++i+=+j+=+0
++++while+j+<+len(data):
++++++++if+data[j]+==+value:
++++++++++++j+%2B=+1
++++++++++++continue
++++++++data[i]+=+data[j]
++++++++i+%2B=+1
++++++++j+%2B=+1
++++for+x+in+range(j+-+i):
++++++++data.pop()|code-block|syntax|javascript|1844518|entityMap^0|C|2|F|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

this is my answer, just use while and for

<pre><code>def remove_all(data, value):
 i = j = 0
 while j &lt; 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()
</code></pre>

blocks|key|1844552|text|许多答案都涉及到创建一个新的列表。这涉及将旧列表中的所有数据复制到新列表中(删除的项目除外)。如果你的清单很大，你可能负担不起(或者你不应该想要)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1844553|在这些情况下，它是快得多的来改变列表就地。如果你必须从列表中删除1个以上的元素，这可能是很棘手的。假设您循环遍历列表，并删除一个项，那么列表将发生变化，标准的for循环将不会考虑这一点。循环的结果可能不是您所期望的。|offset|length|style|BOLD|1844554|示例：|1844555|a+=+[0,+1,+2,+3,+4,+5]
for+i+in+a:
++++a.remove(i)++#+Remove+all+items
print(a)

Out:+[1,+3,+5]|code-block|syntax|javascript|1844556|一个简单的解决方案是以相反的顺序遍历列表。在这种情况下，您将获得：|1844557|a+=+[0,+1,+2,+3,+4,+5]
for+i+in+reversed(a):
++++a.remove(i)++#+Remove+all+items
print(a)

Out:+[]|1844558|然后，如果您只需要删除具有某些特定值的元素，您可以简单地在循环中放置一个if+statement，结果如下：|CODE|1844559|a+=+[0,+1,+2,+3,+4,+5]
for+i+in+reversed(a):
++++if+a[i]+==+2+or+a[i]+==+3:++#+Remove+all+items+having+value+2+or+3.
++++++++a.remove(i)
print(a)

Out:+[0,+1,+4,+5]|1844560|entityMap^0|0|9|4|0|0|0|0|0|10|C|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|10|8|@$D|11|E|12|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|13|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|14|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|15|8|@]|9|@]|A|$]]|$1|Q|3|R|5|L|7|16|8|@]|9|@]|A|$M|N]]|$1|S|3|T|5|6|7|17|8|@$D|18|E|19|F|U]]|9|@]|A|$]]|$1|V|3|W|5|L|7|1A|8|@]|9|@]|A|$M|N]]|$1|X|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Y|$]]

Many of answers involve creating a new list. This involves copying all the data from the old list to the new list (except for the removed items). If your list is huge, you may not be able to afford it (or you should not want to).
In these cases, it is much faster to alter the list in place. If you have to remove more than 1 element from the list it can be tricky. Suppose you loop over the list, and remove an item, then the list changes and a standard for-loop will not take this into account. The result of the loop may not be what you expected.
Example:
<pre><code>a = [0, 1, 2, 3, 4, 5]
for i in a:
 a.remove(i) # Remove all items
print(a)

Out: [1, 3, 5]
</code></pre>
A simple solution is to loop through the list in reverse order. In this case you get:
<pre><code>a = [0, 1, 2, 3, 4, 5]
for i in reversed(a):
 a.remove(i) # Remove all items
print(a)

Out: []
</code></pre>
Then, if you need to only remove elements having some specific values, you can simply put an <code>if statement</code> in the loop resulting in:
<pre><code>a = [0, 1, 2, 3, 4, 5]
for i in reversed(a):
 if a[i] == 2 or a[i] == 3: # Remove all items having value 2 or 3.
 a.remove(i)
print(a)

Out: [0, 1, 4, 5]
</code></pre>

blocks|key|441609|text|一些最简单的方法的基准测试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|441610|import+random
from+copy+import+copy
sample+=+random.sample(range(100000),+10000)
remove+=+random.sample(range(100000),+1000)

%25%25timeit
sample1+=+copy(sample)
remove1+=+copy(remove)

for+i+in+reversed(sample1):
++++if+i+in+remove1:
++++++++sample1.remove(i)
#+271+ms+±+16+ms+per+loop+(mean+±+std.+dev.+of+7+runs,+1+loop+each)
#+remove+all+instances

%25%25timeit
sample1+=+copy(sample)
remove1+=+copy(remove)

filtered+=+list(filter(lambda+x:+x+not+in+remove1,+sample1))
#+280+ms+±+18.9+ms+per+loop+(mean+±+std.+dev.+of+7+runs,+1+loop+each)
#+remove+all+instances

%25%25timeit
sample1+=+copy(sample)
remove1+=+copy(remove)+

filtered+=+[ele+for+ele+in+sample1+if+ele+not+in+remove1]
#+293+ms+±+72.1+ms+per+loop+(mean+±+std.+dev.+of+7+runs,+1+loop+each)
#+remove+all+instances

%25%25timeit
sample1+=+copy(sample)
remove1+=+copy(remove)+

for+val+in+remove1:
++++if+val+in+sample1:
++++++++sample1.remove(val)
#+558+ms+±+40.7+ms+per+loop+(mean+±+std.+dev.+of+7+runs,+1+loop+each)
#+only+remove+first+occurrence

%25%25timeit
sample1+=+copy(sample)
remove1+=+copy(remove)+

for+val+in+remove1:
++++try:
++++++++sample1.remove(val)
++++except:
++++++++pass
#+609+ms+±+11.7+ms+per+loop+(mean+±+std.+dev.+of+7+runs,+1+loop+each)
#+only+remove+first+occurrence|code-block|syntax|javascript|441611|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Benchmark of some of the simplest method:
<pre class="lang-py prettyprint-override"><code>import random
from copy import copy
sample = random.sample(range(100000), 10000)
remove = random.sample(range(100000), 1000)

%%timeit
sample1 = copy(sample)
remove1 = copy(remove)

for i in reversed(sample1):
 if i in remove1:
 sample1.remove(i)
# 271 ms ± 16 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# remove all instances

%%timeit
sample1 = copy(sample)
remove1 = copy(remove)

filtered = list(filter(lambda x: x not in remove1, sample1))
# 280 ms ± 18.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# remove all instances

%%timeit
sample1 = copy(sample)
remove1 = copy(remove) 

filtered = [ele for ele in sample1 if ele not in remove1]
# 293 ms ± 72.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# remove all instances

%%timeit
sample1 = copy(sample)
remove1 = copy(remove) 

for val in remove1:
 if val in sample1:
 sample1.remove(val)
# 558 ms ± 40.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# only remove first occurrence

%%timeit
sample1 = copy(sample)
remove1 = copy(remove) 

for val in remove1:
 try:
 sample1.remove(val)
 except:
 pass
# 609 ms ± 11.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# only remove first occurrence
</code></pre>

blocks|key|1186418|text|当nums为列表，c为要删除的值时：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1186419|要删除列表中第一个出现的c，只需执行以下操作：|1186420|if+c+in+nums:
++++nums.remove(c)|code-block|syntax|javascript|1186421|要从列表中删除所有出现的c，请执行以下操作：|1186422|while+c+in+nums:
++++nums.remove(c)|1186423|添加异常处理将是最佳实践，但我主要想展示如何从列表中删除所有出现的元素。|1186424|entityMap^0|1|4|9|1|0|C|1|0|0|C|1|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|16|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|17|8|@]|D|@]|E|$]]|$1|S|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|T|$]]

When <code>nums</code> is the list and <code>c</code> is the value to be removed:
To remove the first occurrence of <code>c</code> in the list, just do:
<pre><code>if c in nums:
 nums.remove(c)
</code></pre>
To remove all occurrences of <code>c</code> from the list do:
<pre><code>while c in nums:
 nums.remove(c)
</code></pre>
Adding the exception handling would be the best practice, but I mainly wanted to show how to remove all occurrences of an element from the list.

I want to remove a value from a list if it exists in the list (which it may not).
<pre><code>a = [1, 2, 3, 4]
b = a.index(6)

del a[b]
print(a)
</code></pre>
The above case (in which it does not exist) shows the following error:
<pre class="lang-none prettyprint-override"><code>Traceback (most recent call last):
 File &quot;D:\zjm_code\a.py&quot;, line 6, in &lt;module&gt;
 b = a.index(6)
ValueError: list.index(x): x not in list
</code></pre>
So I have to do this:
<pre><code>a = [1, 2, 3, 4]

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

print(a)
</code></pre>
But is there not a simpler way to do this?

Is there a simple way to delete a list element by value?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我想从列表中删除一个值，如果它存在于列表中(可能不存在)。a = [1, 2, 3, 4]b = a.index(6)del a[b]print(a)上面的情况(其中它不存在)显示了以下错误：Traceback (most recent call last):  File "D:\zjm_code\a.py", lin...

问有没有一种简单的方法可以按值删除列表元素？
EN

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问有没有一种简单的方法可以按值删除列表元素？EN