blocks|key|1220649|text|>>>+L+=+range+(11)
>>>+[+x+if+x%252+==+1+else+None+for+x+in+L+]
[None,+1,+None,+3,+None,+5,+None,+7,+None,+9,+None]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1220650|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>&gt;&gt;&gt; L = range (11)
&gt;&gt;&gt; [ x if x%2 == 1 else None for x in L ]
[None, 1, None, 3, None, 5, None, 7, None, 9, None]
</code></pre>

blocks|key|1220692|text|使用列表理解构建一个新列表：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1220693|new_items+=+[x+if+x+%25+2+else+None+for+x+in+items]|code-block|syntax|javascript|1220694|如果需要，您可以就地修改原始列表，但这实际上并不节省时间：|1220695|items+=+[0,+1,+2,+3,+4,+5,+6,+7,+8,+9,+10]
for+index,+item+in+enumerate(items):
++++if+not+(item+%25+2):
++++++++items[index]+=+None|1220696|以下是演示非timesave的(Python+3.6.3)计时：|1220697|In+[1]:+%25%25timeit
+++...:+items+=+[0,+1,+2,+3,+4,+5,+6,+7,+8,+9,+10]
+++...:+for+index,+item+in+enumerate(items):
+++...:+++++if+not+(item+%25+2):
+++...:+++++++++items[index]+=+None
+++...:
1.06+µs+±+33.7+ns+per+loop+(mean+±+std.+dev.+of+7+runs,+1000000+loops+each)

In+[2]:+%25%25timeit
+++...:+items+=+[0,+1,+2,+3,+4,+5,+6,+7,+8,+9,+10]
+++...:+new_items+=+[x+if+x+%25+2+else+None+for+x+in+items]
+++...:
891+ns+±+13.6+ns+per+loop+(mean+±+std.+dev.+of+7+runs,+1000000+loops+each)|1220698|和Python+2.7.6计时：|1220699|In+[1]:+%25%25timeit
+++...:+items+=+[0,+1,+2,+3,+4,+5,+6,+7,+8,+9,+10]
+++...:+for+index,+item+in+enumerate(items):
+++...:+++++if+not+(item+%25+2):
+++...:+++++++++items[index]+=+None
+++...:+
1000000+loops,+best+of+3:+1.27+µs+per+loop
In+[2]:+%25%25timeit
+++...:+items+=+[0,+1,+2,+3,+4,+5,+6,+7,+8,+9,+10]
+++...:+new_items+=+[x+if+x+%25+2+else+None+for+x+in+items]
+++...:+
1000000+loops,+best+of+3:+1.14+µs+per+loop|1220700|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

Build a new list with a list comprehension:

<pre><code>new_items = [x if x % 2 else None for x in items]
</code></pre>

You can modify the original list in-place if you want, but it doesn't actually save time:

<pre><code>items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
for index, item in enumerate(items):
 if not (item % 2):
 items[index] = None
</code></pre>

Here are (Python 3.6.3) timings demonstrating the non-timesave:

<pre><code>In [1]: %%timeit
 ...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
 ...: for index, item in enumerate(items):
 ...: if not (item % 2):
 ...: items[index] = None
 ...:
1.06 µs ± 33.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [2]: %%timeit
 ...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
 ...: new_items = [x if x % 2 else None for x in items]
 ...:
891 ns ± 13.6 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
</code></pre>

And Python 2.7.6 timings:

<pre><code>In [1]: %%timeit
 ...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
 ...: for index, item in enumerate(items):
 ...: if not (item % 2):
 ...: items[index] = None
 ...: 
1000000 loops, best of 3: 1.27 µs per loop
In [2]: %%timeit
 ...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
 ...: new_items = [x if x % 2 else None for x in items]
 ...: 
1000000 loops, best of 3: 1.14 µs per loop
</code></pre>

blocks|key|1220702|text|ls+=+[x+if+(condition)+else+None+for+x+in+ls]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1220703|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>ls = [x if (condition) else None for x in ls]
</code></pre>

blocks|key|1435752|text|这里有另一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1435753|>>>+L+=+range+(11)
>>>+map(lambda+x:+x+if+x%252+else+None,+L)
[None,+1,+None,+3,+None,+5,+None,+7,+None,+9,+None]|code-block|syntax|javascript|1435754|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 another way:

<pre><code>&gt;&gt;&gt; L = range (11)
&gt;&gt;&gt; map(lambda x: x if x%2 else None, L)
[None, 1, None, 3, None, 5, None, 7, None, 9, None]
</code></pre>

blocks|key|1435759|text|这可能会有帮助。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1435760|test_list+=+[5,+8]
test_list[0]+=+None
print+test_list
#prints+[None,+8]|code-block|syntax|javascript|1435761|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 might help...

<pre><code>test_list = [5, 8]
test_list[0] = None
print test_list
#prints [None, 8]
</code></pre>

blocks|key|1435819|text|如果您想要就地替换值，您可以使用列表理解中的值更新原始列表，方法是将值分配给原始列表的整个切片。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1435820|data+=+[*range(11)]+#+[0,+1,+2,+3,+4,+5,+6,+7,+8,+9,+10]
id_before+=+id(data)
data[:]+=+[x+if+x+%25+2+else+None+for+x+in+data]
data
#+Out:+[None,+1,+None,+3,+None,+5,+None,+7,+None,+9,+None]
id_before+==+id(data)++#+check+if+list+is+still+the+same
#+Out:+True|code-block|syntax|javascript|1435821|如果有多个名称指向原始列表，例如，您在更改列表之前编写了data2=data，并且跳过了分配给data的切片表示法，则data将重新绑定以指向新创建的列表，而data2仍指向原始未更改的列表。|offset|length|style|CODE|1435822|data+=+[*range(11)]+#+[0,+1,+2,+3,+4,+5,+6,+7,+8,+9,+10]
data2+=+data
id_before+=+id(data)
data+=+[x+if+x+%25+2+else+None+for+x+in+data]++#+no+[:]+here
data
#+Out:+[None,+1,+None,+3,+None,+5,+None,+7,+None,+9,+None]
id_before+==+id(data)++#+check+if+list+is+still+the+same
#+Out:+False
data2
#+Out:+[0,+1,+2,+3,+4,+5,+6,+7,+8,+9,+10]|1435823|注意:这并不是对一般情况下优先选择其中一个的建议(无论是否更改列表)，但您应该注意行为。|1435824|entityMap^0|0|0|S|A|1B|4|1N|4|27|5|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|14|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|R|$]]

In case you want to replace values in place, you can
update your original list with values from a list
comprehension by assigning to the whole slice of the original.

<pre><code>data = [*range(11)] # [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
id_before = id(data)
data[:] = [x if x % 2 else None for x in data]
data
# Out: [None, 1, None, 3, None, 5, None, 7, None, 9, None]
id_before == id(data) # check if list is still the same
# Out: True
</code></pre>

If you have multiple names pointing to the original list,
for example you wrote <code>data2=data</code> before changing the list
and you skip the slice notation for assigning to <code>data</code>,
<code>data</code> will rebind to point to the newly created list while <code>data2</code> still points to the original unchanged list.

<pre><code>data = [*range(11)] # [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
data2 = data
id_before = id(data)
data = [x if x % 2 else None for x in data] # no [:] here
data
# Out: [None, 1, None, 3, None, 5, None, 7, None, 9, None]
id_before == id(data) # check if list is still the same
# Out: False
data2
# Out: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
</code></pre>

Note: This is no recommendation for generally preferring one over the other
(changing list in place or not), but behavior you should be aware of.

I have a list where I want to replace values with None where condition() returns True.

<pre><code>[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
</code></pre>

For example, if condition checks bool(item%2) should return:

<pre><code>[None, 1, None, 3, None, 5, None, 7, None, 9, None]
</code></pre>

What is the most efficient way to do this?

Replace values in list using Python

我有一个列表，其中我希望将值替换为None，其中condition()返回True。[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]例如，如果条件检查bool(项%2)应返回：[None, 1, None, 3, None, 5, None, 7, None, 9, None]执行此操作的最有效方法...

问使用Python替换列表中的值
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