blocks|key|1315311|text|itertools模块包含一个用于迭代器的过滤器函数。过滤迭代器的第一个元素可以通过对其调用next()来获得：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1315312|from+itertools+import+ifilter

print+ifilter((lambda+i:+i+>+3),+range(10)).next()|code-block|syntax|javascript|1315313|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/itertools.html^0|0|9|1A|6|0|9|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]]]]

The <a href="http://docs.python.org/library/itertools.html" rel="noreferrer"><code>itertools</code></a> module contains a filter function for iterators. The first element of the filtered iterator can be obtained by calling <code>next()</code> on it:

<pre><code>from itertools import ifilter

print ifilter((lambda i: i &gt; 3), range(10)).next()
</code></pre>

blocks|key|1315377|text|与使用ifilter类似，您可以使用生成器表达式：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1315378|>>>+(x+for+x+in+xrange(10)+if+x+>+5).next()
6|code-block|syntax|javascript|1315379|在这两种情况下，您都可能希望捕获StopIteration，以防没有元素满足您的条件。|1315380|从技术上讲，我认为你可以这样做：|1315381|>>>+foo+=+None
>>>+for+foo+in+(x+for+x+in+xrange(10)+if+x+>+5):+break
...+
>>>+foo
6|1315382|它可以避免创建try/except块。但这对语法来说似乎有些晦涩和滥用。|1315383|entityMap^0|3|7|0|0|G|D|0|0|0|7|A|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|11|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|S|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|T|$]]

Similar to using <code>ifilter</code>, you could use a generator expression:

<pre><code>&gt;&gt;&gt; (x for x in xrange(10) if x &gt; 5).next()
6
</code></pre>

In either case, you probably want to catch <code>StopIteration</code> though, in case no elements satisfy your condition.

Technically speaking, I suppose you could do something like this:

<pre><code>&gt;&gt;&gt; foo = None
&gt;&gt;&gt; for foo in (x for x in xrange(10) if x &gt; 5): break
... 
&gt;&gt;&gt; foo
6
</code></pre>

It would avoid having to make a <code>try/except</code> block. But that seems kind of obscure and abusive to the syntax.

blocks|key|1100480|text|我会写这篇文章|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1100481|next(x+for+x+in+xrange(10)+if+x+>+3)|code-block|syntax|javascript|1100482|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|$]]

I would write this 

<pre><code>next(x for x in xrange(10) if x &gt; 3)
</code></pre>

blocks|key|1315467|text|线条：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1315468|thefirst+=+[i+for+i+in+range(10)+if+i+>+3][0]|code-block|syntax|javascript|1315469|如果您不确定根据条件是否有任何元素是有效的，则应该使用try/except将其包含在内，因为该[0]可以引发IndexError。|offset|length|style|CODE|1315470|entityMap^0|0|0|R|A|1B|3|1I|A|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]|$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|N|$]]

Oneliner:

<pre><code>thefirst = [i for i in range(10) if i &gt; 3][0]
</code></pre>

If youre not sure that any element will be valid according to the criteria, you should enclose this with <code>try/except</code> since that <code>[0]</code> can raise an <code>IndexError</code>.

blocks|key|1315520|text|对于不存在下一个内置组件的旧版本的Python：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1315521|(x+for+x+in+range(10)+if+x+>+3).next()|code-block|syntax|javascript|1315522|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|$]]

For older versions of Python where the next built-in doesn't exist:

<pre><code>(x for x in range(10) if x &gt; 3).next()
</code></pre>

blocks|key|1315599|text|作为一个可重用的、有文档记录和测试过的功能|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1315600|def+first(iterable,+condition+=+lambda+x:+True):
++++"""
++++Returns+the+first+item+in+the+`iterable`+that
++++satisfies+the+`condition`.

++++If+the+condition+is+not+given,+returns+the+first+item+of
++++the+iterable.

++++Raises+`StopIteration`+if+no+item+satysfing+the+condition+is+found.

++++>>>+first(+(1,2,3),+condition=lambda+x:+x+%25+2+==+0)
++++2
++++>>>+first(range(3,+100))
++++3
++++>>>+first(+()+)
++++Traceback+(most+recent+call+last):
++++...
++++StopIteration
++++"""

++++return+next(x+for+x+in+iterable+if+condition(x))|code-block|syntax|javascript|1315601|带有默认参数的版本|1315602|@zorf建议了此函数的一个版本，如果可迭代为空或没有匹配条件的项，则可以使用预定义的返回值：|1315603|def+first(iterable,+default+=+None,+condition+=+lambda+x:+True):
++++"""
++++Returns+the+first+item+in+the+`iterable`+that
++++satisfies+the+`condition`.

++++If+the+condition+is+not+given,+returns+the+first+item+of
++++the+iterable.

++++If+the+`default`+argument+is+given+and+the+iterable+is+empty,
++++or+if+it+has+no+items+matching+the+condition,+the+`default`+argument
++++is+returned+if+it+matches+the+condition.

++++The+`default`+argument+being+None+is+the+same+as+it+not+being+given.

++++Raises+`StopIteration`+if+no+item+satisfying+the+condition+is+found
++++and+default+is+not+given+or+doesn't+satisfy+the+condition.

++++>>>+first(+(1,2,3),+condition=lambda+x:+x+%25+2+==+0)
++++2
++++>>>+first(range(3,+100))
++++3
++++>>>+first(+()+)
++++Traceback+(most+recent+call+last):
++++...
++++StopIteration
++++>>>+first([],+default=1)
++++1
++++>>>+first([],+default=1,+condition=lambda+x:+x+%25+2+==+0)
++++Traceback+(most+recent+call+last):
++++...
++++StopIteration
++++>>>+first([1,3,5],+default=1,+condition=lambda+x:+x+%25+2+==+0)
++++Traceback+(most+recent+call+last):
++++...
++++StopIteration
++++"""

++++try:
++++++++return+next(x+for+x+in+iterable+if+condition(x))
++++except+StopIteration:
++++++++if+default+is+not+None+and+condition(default):
++++++++++++return+default
++++++++else:
++++++++++++raise|1315604|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|$]]

<h3>As a reusable, documented and tested function</h3>

<pre><code>def first(iterable, condition = lambda x: True):
 """
 Returns the first item in the `iterable` that
 satisfies the `condition`.

 If the condition is not given, returns the first item of
 the iterable.

 Raises `StopIteration` if no item satysfing the condition is found.

 &gt;&gt;&gt; first( (1,2,3), condition=lambda x: x % 2 == 0)
 2
 &gt;&gt;&gt; first(range(3, 100))
 3
 &gt;&gt;&gt; first( () )
 Traceback (most recent call last):
 ...
 StopIteration
 """

 return next(x for x in iterable if condition(x))
</code></pre>

<h3>Version with default argument</h3>

@zorf suggested a version of this function where you can have a predefined return value if the iterable is empty or has no items matching the condition:

<pre><code>def first(iterable, default = None, condition = lambda x: True):
 """
 Returns the first item in the `iterable` that
 satisfies the `condition`.

 If the condition is not given, returns the first item of
 the iterable.

 If the `default` argument is given and the iterable is empty,
 or if it has no items matching the condition, the `default` argument
 is returned if it matches the condition.

 The `default` argument being None is the same as it not being given.

 Raises `StopIteration` if no item satisfying the condition is found
 and default is not given or doesn't satisfy the condition.

 &gt;&gt;&gt; first( (1,2,3), condition=lambda x: x % 2 == 0)
 2
 &gt;&gt;&gt; first(range(3, 100))
 3
 &gt;&gt;&gt; first( () )
 Traceback (most recent call last):
 ...
 StopIteration
 &gt;&gt;&gt; first([], default=1)
 1
 &gt;&gt;&gt; first([], default=1, condition=lambda x: x % 2 == 0)
 Traceback (most recent call last):
 ...
 StopIteration
 &gt;&gt;&gt; first([1,3,5], default=1, condition=lambda x: x % 2 == 0)
 Traceback (most recent call last):
 ...
 StopIteration
 """

 try:
 return next(x for x in iterable if condition(x))
 except StopIteration:
 if default is not None and condition(default):
 return default
 else:
 raise
</code></pre>

blocks|key|1315652|text|在Python+3中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1315653|a+=+(None,+False,+0,+1)
assert+next(filter(None,+a))+==+1|code-block|syntax|javascript|1315654|在Python+2.6中：|1315655|a+=+(None,+False,+0,+1)
assert+next(iter(filter(None,+a)))+==+1|1315656|编辑:我认为这很明显，但显然不是:你可以传递一个函数(或lambda)来检查条件，而不是None：|offset|length|style|CODE|1315657|a+=+[2,3,4,5,6,7,8]
assert+next(filter(lambda+x:+x%252,+a))+==+3|1315658|entityMap^0|0|0|0|0|S|6|18|4|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]]|$1|K|3|L|5|6|7|Y|8|@$M|Z|N|10|O|P]|$M|11|N|12|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|T|$]]

In Python 3:

<pre><code>a = (None, False, 0, 1)
assert next(filter(None, a)) == 1
</code></pre>

In Python 2.6:

<pre><code>a = (None, False, 0, 1)
assert next(iter(filter(None, a))) == 1
</code></pre>

EDIT: I thought it was obvious, but apparently not: instead of <code>None</code> you can pass a function (or a <code>lambda</code>) with a check for the condition:

<pre><code>a = [2,3,4,5,6,7,8]
assert next(filter(lambda x: x%2, a)) == 3
</code></pre>

blocks|key|1315684|text|这个问题已经有了很好的答案。我只是补充我的两点意见，因为我在这里找到了一个我自己的问题的解决方案，这与OP非常相似。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1315685|如果您想要使用生成器查找匹配条件的第一个项目的索引，您可以简单地执行以下操作：|1315686|next(index+for+index,+value+in+enumerate(iterable)+if+condition)|code-block|syntax|javascript|1315687|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

This question already has great answers. I'm only adding my two cents because I landed here trying to find a solution to my own problem, which is very similar to the OP. 

If you want to find the INDEX of the first item matching a criteria using generators, you can simply do:

<pre><code>next(index for index, value in enumerate(iterable) if condition)
</code></pre>

blocks|key|1100807|text|您还可以使用Numpy中的argwhere函数。例如：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1100808|i)在“helloworld”中找到第一个"l“：|1100809|import+numpy+as+np
l+=+list("helloworld")+#+Create+list
i+=+np.argwhere(np.array(l)=="l")+#+i+=+array([[2],[3],[8]])
index_of_first+=+i.min()|code-block|syntax|javascript|1100810|ii)找到第一个>+0.1的随机数|1100811|import+numpy+as+np
r+=+np.random.rand(50)+#+Create+random+numbers
i+=+np.argwhere(r>0.1)
index_of_first+=+i.min()|1100812|iii)找出最后一个>+0.1的随机数|1100813|import+numpy+as+np
r+=+np.random.rand(50)+#+Create+random+numbers
i+=+np.argwhere(r>0.1)
index_of_last+=+i.max()|1100814|entityMap^0|D|8|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|11|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|13|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|U|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|V|$]]

You could also use the <code>argwhere</code> function in Numpy. For example:

i) Find the first "l" in "helloworld":

<pre><code>import numpy as np
l = list("helloworld") # Create list
i = np.argwhere(np.array(l)=="l") # i = array([[2],[3],[8]])
index_of_first = i.min()
</code></pre>

ii) Find first random number > 0.1

<pre><code>import numpy as np
r = np.random.rand(50) # Create random numbers
i = np.argwhere(r&gt;0.1)
index_of_first = i.min()
</code></pre>

iii) Find the last random number > 0.1

<pre><code>import numpy as np
r = np.random.rand(50) # Create random numbers
i = np.argwhere(r&gt;0.1)
index_of_last = i.max()
</code></pre>

blocks|key|1315739|text|对于使用Python3.8或更新版本的任何人，我建议使用PEP+572+--+Assignment+Expressions中描述的“赋值表达式”。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1315740|if+any((match+:=+i)+>+3+for+i+in+range(10)):
++++print(match)|code-block|syntax|javascript|1315741|entityMap|0|LINK|mutability|MUTABLE|url|https://www.python.org/dev/peps/pep-0572/^0|S|X|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

For anyone using Python 3.8 or newer I recommend using &quot;Assignment Expressions&quot; as described in <a href="https://www.python.org/dev/peps/pep-0572/" rel="noreferrer">PEP 572 -- Assignment Expressions</a>.
<pre><code>if any((match := i) &gt; 3 for i in range(10)):
 print(match)
</code></pre>

blocks|key|1100874|text|Perl+-让简单的事情变得容易，让困难的事情成为可能。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1100875|Python+-如果你想要一些简单的东西，那就自己去写吧，但要当心那些步枪。|1100876|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Perl - make the easy things easy, and the hard things possible.
Python - if you want something easy, go write it yourself, but beware of the footguns.

blocks|key|1315788|text|下面是三种方法的快速测试。Next()不是最快的方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1315789|from+timeit+import+default_timer+as+timer

#+Is+set+irreflexive?

def+a():
++++return+frozenset((x3,+x3)+for+x3+in+set([x1[x2]+for+x2+in+range(2)+for+x1+in+value])+if+(x3,+x3)+in+value)+==+frozenset()


def+b():
++++return+next((False+for+x1+in+value+if+(x1[0],+x1[0])+in+value+or+(x1[1],+x1[1])+in+value),+True)


def+c():
++++for+x1+in+value:
++++++++if+(x1[0],+x1[0])+in+value+or+(x1[1],+x1[1])+in+value:
++++++++++++return+False
++++return+True


times+=+1000000
value+=+frozenset({(1,+3),+(2,+1)})


start_time+=+timer()
for+x+in+range(times):
++++a()
print("a():+Calculation+ended+after+"+%2B+str(round((timer()+-+start_time)+*+1000)+/+1000.0)+%2B+"+sec")

start_time+=+timer()
for+x+in+range(times):
++++b()
print("b():+Calculation+ended+after+"+%2B+str(round((timer()+-+start_time)+*+1000)+/+1000.0)+%2B+"+sec")

start_time+=+timer()
for+x+in+range(times):
++++c()
print("c():+Calculation+ended+after+"+%2B+str(round((timer()+-+start_time)+*+1000)+/+1000.0)+%2B+"+sec")|code-block|syntax|javascript|1315790|结果为：|1315791|Calculation+ended+after+1.365+sec
Calculation+ended+after+0.685+sec
Calculation+ended+after+0.493+sec|1315792|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|$]]

here is a speedtest of three ways. Next() is not the fastest way.
<pre><code>from timeit import default_timer as timer

# Is set irreflexive?

def a():
 return frozenset((x3, x3) for x3 in set([x1[x2] for x2 in range(2) for x1 in value]) if (x3, x3) in value) == frozenset()


def b():
 return next((False for x1 in value if (x1[0], x1[0]) in value or (x1[1], x1[1]) in value), True)


def c():
 for x1 in value:
 if (x1[0], x1[0]) in value or (x1[1], x1[1]) in value:
 return False
 return True


times = 1000000
value = frozenset({(1, 3), (2, 1)})


start_time = timer()
for x in range(times):
 a()
print(&quot;a(): Calculation ended after &quot; + str(round((timer() - start_time) * 1000) / 1000.0) + &quot; sec&quot;)

start_time = timer()
for x in range(times):
 b()
print(&quot;b(): Calculation ended after &quot; + str(round((timer() - start_time) * 1000) / 1000.0) + &quot; sec&quot;)

start_time = timer()
for x in range(times):
 c()
print(&quot;c(): Calculation ended after &quot; + str(round((timer() - start_time) * 1000) / 1000.0) + &quot; sec&quot;)
</code></pre>
Results to:
<pre><code>Calculation ended after 1.365 sec
Calculation ended after 0.685 sec
Calculation ended after 0.493 sec
</code></pre>

I would like to get the first item from a list matching a condition. It's important that the resulting method not process the entire list, which could be quite large. For example, the following function is adequate:

<pre><code>def first(the_iterable, condition = lambda x: True):
 for i in the_iterable:
 if condition(i):
 return i
</code></pre>

This function could be used something like this:

<pre><code>&gt;&gt;&gt; first(range(10))
0
&gt;&gt;&gt; first(range(10), lambda i: i &gt; 3)
4
</code></pre>

However, I can't think of a good built-in / one-liner to let me do this. I don't particularly want to copy this function around if I don't have to. Is there a built-in way to get the first item matching a condition?

Get the first item from an iterable that matches a condition

Python

Perl

我想从匹配条件的列表中获取第一个项目。重要的是，产生的方法不能处理整个列表，因为列表可能非常大。例如，以下函数就足够了：def first(the_iterable, condition = lambda x: True):    for i in the_iterable:        if condition(i...

问从符合条件的迭代器中获取第一项
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问从符合条件的迭代器中获取第一项
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