计算元组列表中的出现次数 [英] Counting the amount of occurrences in a list of tuples

问题描述

我对python还是很陌生，但是我无法在任何地方找到解决我问题的方法.

I am fairly new to python, but I haven't been able to find a solution to my problem anywhere.

我想计算一个元组列表中一个字符串的出现次数.

I want to count the occurrences of a string inside a list of tuples.

这是元组列表:

list1 = [
         ('12392', 'some string', 'some other string'),
         ('12392', 'some new string', 'some other string'),
         ('7862', None, 'some other string')
        ]

我已经尝试过了，但是它只能打印0

I've tried this but it just prints 0

for entry in list1:
    print list1.count(entry[0])

由于同一ID在列表中出现两次，因此应返回:

As the same ID occurs twice in the list, this should return:

2
1

我还尝试为每次出现相同ID的计数器增加一个计数器，但不太了解如何编写.

I also tried to increment a counter for each occurrence of the same ID but couldn't quite grasp how to write it.

*使用Eumiro的出色答案.我只是意识到我没有解释整个问题.我实际上需要值大于1的条目总数.但是，如果我尝试这样做:

* Using Eumiro's awesome answer. I just realized that I didn't explain the whole problem. I actually need the total amount of entries which has a value more than 1. But if I try doing:

for name, value in list1:

    if value > 1:
        print value

我收到此错误:

ValueError: Too many values to unpack

推荐答案

也许 collections.Counter 可以解决您的问题:

Maybe collections.Counter could solve your problem:

from collections import Counter
Counter(elem[0] for elem in list1)

返回

Counter({'12392': 2, '7862': 1})

速度很快，因为它仅对列表进行一次迭代.您遍历条目，然后尝试获取列表中这些条目的数量.不能使用 .count 完成此操作，但可以按以下步骤完成:

It is fast since it iterates over your list just once. You iterate over entries and then try to get a count of these entries within your list. That cannot be done with .count, but might be done as follows:

for entry in list1:
    print sum(1 for elem in list1 if elem[0] == entry[0])

但是认真的看一下 collections.Counter .

编辑:我实际上需要值大于1的条目总数.

您仍然可以使用 Counter :

c = Counter(elem[0] for elem in list1)
sum(v for k, v in c.iteritems() if v > 1)

返回 2 ，即大于1的计数之和.

returns 2, i.e. the sum of counts that are higher than 1.

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