python：从列表（序列）中获取项目数，具有一定条件 [英] python: get number of items from list(sequence) with certain condition

问题描述

假设我有一个包含大量项目的列表。

Assuming that I have a list with huge number of items.

l = [ 1, 4, 6, 30, 2, ... ]

我想获得该列表中的项目数，应满足一定条件。我的第一个想法是：

I want to get the number of items from that list, where an item should satisfy certain condition. My first thought was:

count = len([i for i in l if my_condition(l)])

但是如果my_condition（）过滤的列表也有很多项目，我认为
创建新列表过滤结果只是浪费内存。为了效率，IMHO，上面的调用不能比：

But if the my_condition() filtered list has also great number of items, I think that creating new list for filtered result is just waste of memory. For efficiency, IMHO, above call can't be better than:

count = 0
for i in l:
    if my_condition(l):
        count += 1

函数式方式来实现获得满足某些条件而不产生临时列表的项目＃

Is there any functional-style way to achieve to get the # of items that satisfy certain condition without generating temporary list?

提前感谢。

推荐答案

您可以使用生成器表达式：

>>> l = [1, 3, 7, 2, 6, 8, 10]
>>> sum(1 for i in l if i % 4 == 3)
2

甚至

>>> sum(i % 4 == 3 for i in l)
2

事实， int（True）== 1 。

或者，您可以使用 itertools.imap （python 2） / code>（python 3）：

Alternatively, you could use itertools.imap (python 2) or simply map (python 3):

>>> def my_condition(x):
...     return x % 4 == 3
... 
>>> sum(map(my_condition, l))
2

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