我想从匹配条件的列表中获得第一项。产生的方法不能处理整个列表，这一点很重要，因为列表可能相当大。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):
return i
这个函数可以这样使用:
>>> first(range(10))
0
>>> first(range(10), lambda i: i > 3)
4
但是，我想不出一个好的内置/单行程序来让我这样做。如果没有必要，我不想复制这个函数。是否有一种内置的方法来获取匹配条件的第一项?
当前回答
对于任何使用Python 3.8或更新版本的人，我建议使用PEP 572—赋值表达式中描述的“赋值表达式”。
if any((match := i) > 3 for i in range(10)):
print(match)
其他回答
作为一个可重用、文档化和测试的函数
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 % 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))
带有默认参数的版本
@zorf建议这个函数的一个版本，如果可迭代对象为空或没有匹配条件的项，你可以有一个预定义的返回值:
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 % 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 % 2 == 0)
Traceback (most recent call last):
...
StopIteration
>>> 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
下面是三种方法的速度测试。Next()不是最快的方法。
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 " + str(round((timer() - start_time) * 1000) / 1000.0) + " sec")
start_time = timer()
for x in range(times):
b()
print("b(): Calculation ended after " + str(round((timer() - start_time) * 1000) / 1000.0) + " sec")
start_time = timer()
for x in range(times):
c()
print("c(): Calculation ended after " + str(round((timer() - start_time) * 1000) / 1000.0) + " sec")
结果:
Calculation ended after 1.365 sec
Calculation ended after 0.685 sec
Calculation ended after 0.493 sec
在Python 3中:
a = (None, False, 0, 1)
assert next(filter(None, a)) == 1
在Python 2.6中:
a = (None, False, 0, 1)
assert next(iter(filter(None, a))) == 1
编辑:我认为这是显而易见的，但显然不是:而不是None，你可以传递一个函数(或一个lambda)，检查条件:
a = [2,3,4,5,6,7,8]
assert next(filter(lambda x: x%2, a)) == 3
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调用next()来获得:
from itertools import ifilter
print ifilter((lambda i: i > 3), range(10)).next()
类似于使用filter，你可以使用生成器表达式:
>>> (x for x in xrange(10) if x > 5).next()
6
在任何一种情况下，您都可能希望捕获StopIteration，以防没有元素满足您的条件。
从技术上讲，我认为你可以这样做:
>>> foo = None
>>> for foo in (x for x in xrange(10) if x > 5): break
...
>>> foo
6
它将避免必须进行try/except块。但这看起来有点模糊和滥用语法。
