使用 Python 计算元组的出现次数 [英] Count occurrence of tuples with Python

问题描述

我正在尝试将显示产品和成本的 Python 元组列表转换为显示给定成本下的成本和产品数量的元组列表.

例如，给定以下输入:

[('Product1', 9), ('Product2', 1),('产品3', 1), ('产品4', 2),('Product5', 3), ('Product6', 4),('Product7', 5), ('Product8', 6),('Product9', 7), ('Product10', 8),('Product11', 3), ('Product12', 1),('Product13', 2), ('Product14', 3),('Product15', 4), ('Product16', 5),('Product17', 6), ('Product18', 7)]

我正在尝试在 Python 中创建一个函数来呈现以下内容.即对于三种不同的产品，值 1 被渲染了 3 次，因此是 (1, 3).

[(1, 3), (2, 1), (3, 2), (4, 1), (5, 2), (6, 2), (7, 2), (8, 1) (9, 1)]

解决方案

也许 collections.Counter 可以解决您的问题:

<预><代码>>>>从集合导入计数器>>>c = Counter(elem[1] for elem in given_list)

输出将如下所示:

计数器({1: 3, 3: 3, 2: 2, 4: 2, 5: 2, 6: 2, 7: 2, 8: 1, 9: 1})

如果您希望将其放在问题中指定的列表中，那么您可以这样做:

<预><代码>>>>列表(c.iteritems())[(1, 3), (2, 2), (3, 3), (4, 2), (5, 2), (6, 2), (7, 2), (8, 1), (9, 1)]

I'm trying to convert a list of Python tuples that display product and cost to a list of tuples that display the cost and the count of products at a given cost.

For example, given the below input:

[('Product1', 9), ('Product2', 1),
 ('Product3', 1), ('Product4', 2),
 ('Product5', 3), ('Product6', 4),
 ('Product7', 5), ('Product8', 6), 
 ('Product9', 7), ('Product10', 8), 
 ('Product11', 3), ('Product12', 1), 
 ('Product13', 2), ('Product14', 3), 
 ('Product15', 4), ('Product16', 5), 
 ('Product17', 6), ('Product18', 7)]

I'm trying to create a function in Python that would render the below. i.e. The value 1 is rendered 3 times for three different products, hence (1, 3).

[(1, 3), (2, 1), (3, 2), (4, 1), (5, 2), (6, 2), (7, 2), (8, 1) (9, 1)]

解决方案

Maybe collections.Counter could solve your problem:

>>> from collections import Counter
>>> c = Counter(elem[1] for elem in given_list)

Output will look like this:

Counter({1: 3, 3: 3, 2: 2, 4: 2, 5: 2, 6: 2, 7: 2, 8: 1, 9: 1})

If you want it in a list like you've specified in the question, then you can do this:

>>> list(c.iteritems())
[(1, 3), (2, 2), (3, 3), (4, 2), (5, 2), (6, 2), (7, 2), (8, 1), (9, 1)]

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