问题描述

下面的代码正在运行这个列表元组的前三个元素

The below code is running for first three elements of the tuple of this list

SS1=[(1, 2, 3, 4, 5), (1, 2, 3, 4, 6), (1, 2, 3, 5, 6), (1, 2, 4, 5, 6), (1, 3, 4, 5, 6), (2, 3, 4, 5, 6)]

from collections import Counter
c = Counter(elem[0:3] for elem in SS1)

for k, v in c.items():
    if (v > 0):
        print(k,v)

输出是:

(1, 2, 3) 3
(1, 2, 4) 1
(1, 3, 4) 1
(2, 3, 4) 1

但我的期望不仅仅是前三个元组......我想要元组 (0,2,3) 或元组 (1,2,4) 的计数器代码> 同样，我可以传递元组的任何三个位置并获取它的计数......我该怎么做?

But my expectation is not just for first three tuple...i want the counter for tuple (0,2,3) or tuple (1,2,4) likewise i can pass any three position of the tuple and get the count of it... How can I do this?

推荐答案

如果我从你的问题中理解是正确的，下面的代码将解决你的问题:

If what i understood from your question is correct, the code below will solve your issue:

SS1=[(1, 2, 3, 4, 5), (1, 2, 3, 4, 6), (1, 2, 3, 5, 6), (1, 2, 4, 5, 6), (1, 3, 4, 5, 6), (2, 3, 4, 5, 6)]

from collections import Counter

def get_new_list(a, pos):
    # Check if any element in pos is > than the length of the tuples
    if any(k >= len(min(SS1, key=lambda x: len(x))) for k in pos):
        return

    for k in a:
        yield tuple(k[j] for j in pos)

def elm_counter(elm):
    if not len(elm):
        return 

    c = Counter(elm)
    for k, v in c.items():
        if v > 0:
            print(k, v)

elm = list(get_new_list(SS1, (0, 2, 4)))
elm_counter(elm)
print('---')
elm = list(get_new_list(SS1, (1, 2, 4)))
elm_counter(elm)

输出:

(1, 3, 5) 1
(1, 3, 6) 2
(1, 4, 6) 2
(2, 4, 6) 1
---
(2, 3, 6) 2
(2, 3, 5) 1
(3, 4, 6) 2
(2, 4, 6) 1

这篇关于获取您自己的元组元素的数量......不仅仅是范围或序列的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！