英文:
How to get a count of first elements and sum of second element in a list of tuples in python?
问题
我有一个元组列表，我期望得到一个字典，其中我想要计算第一个元素重复的次数以及所有第二个元素的总和。
例如：-
input_list = [('Skills', 3), ('Wholesome', 3), ('FanArt', 3), ('Clip', 3), ('Skills', 2), ('FanArt', 2)]
期望的输出：
{"Skills": 5, "Wholesome": 3, "FanArt": 5, "Clip": 3}
我期望以一种优化的方式完成这个任务。
我已经尝试过使用简单的for循环进行编程。但我想要使用itertools或一些内置的包。
英文:
I have a list of tuples and I am expecting a dictionary where I want to count number of times the first element is repeated and sum of all the second elements.
For eg :-
input_list = [('Skills', 3), ('Wholesome', 3), ('FanArt', 3), ('Clip', 3), ('Skills', 2), ('FanArt', 2)]
Expected Output :
{"Skills": 5, "Wholesome": 3, "FanArt": 5, "Clip": 3}
I am expecting a optimised way to do this
I have tried it with simple programming with for loops. But I want to use itertools or some built in package.
答案1
得分: 2
以下是您要求的代码部分的翻译：
input_list = [('Skills', 3), ('Wholesome', 3), ('FanArt', 3), ('Clip', 3), ('Skills', 2), ('FanArt', 2)]output_dict = {}# 遍历元组值for k, v in input_list:if (k in output_dict): # 如果键存在则增加值output_dict[k] += velse:output_dict[k] = v # 如果键不存在则创建键值对print(output_dict)
from collections import defaultdictinput_list = [('Skills', 3), ('Wholesome', 3), ('FanArt', 3), ('Clip', 3), ('Skills', 2), ('FanArt', 2)]output_dict = defaultdict(int)for key, value in input_list:output_dict[key] += valueprint(dict(output_dict))
from collections import Counterinput_list = [('Skills', 3), ('Wholesome', 3), ('FanArt', 3), ('Clip', 3), ('Skills', 2), ('FanArt', 2)]output_dict = Counter()for key, value in input_list:output_dict[key] += valueprint(dict(output_dict))
输出
{'Skills': 5, 'Wholesome': 3, 'FanArt': 5, 'Clip': 3} # 适用于上述所有代码
英文:
You can use this
input_list = [('Skills', 3), ('Wholesome', 3), ('FanArt', 3), ('Clip', 3), ('Skills', 2), ('FanArt', 2)]output_dict = {}# Looping over tuple valuefor k, v in input_list:if (k in output_dict): # if key exists then add valueoutput_dict[k] += velse:output_dict[k] = v # if key not exists then key:valueprint(output_dict)
You can also use the default list
from collections import defaultdictinput_list = [('Skills', 3), ('Wholesome', 3), ('FanArt', 3), ('Clip', 3), ('Skills', 2), ('FanArt', 2)]output_dict = defaultdict(int)for key, value in input_list:output_dict[key] += valueprint(dict(output_dict))
Counter from collections will also work same as above
from collections import Counterinput_list = [('Skills', 3), ('Wholesome', 3), ('FanArt', 3), ('Clip', 3), ('Skills', 2), ('FanArt', 2)]output_dict = Counter()for key, value in input_list:output_dict[key] += valueprint(dict(output_dict))
Output
{'Skills': 5, 'Wholesome': 3, 'FanArt': 5, 'Clip': 3} # for all above code
答案2
得分: -1
使用排序后的列表：
from itertools import groupbyfrom operator import itemgetterinput_list = [("Skills", 3),("Wholesome", 3),("FanArt", 3),("Clip", 3),("Skills", 2),("FanArt", 2),]# 按每个元组的第一个元素对输入列表进行排序sorted_list = sorted(input_list, key=itemgetter(0))# 你也可以使用 lambda 函数作为键# sorted_list = sorted(input_list, key=lambda x: x[0])# 使用 groupby 将元组按第一个元素进行分组grouped_data = groupby(sorted_list, key=itemgetter(0))# 创建一个字典，其中包含每个分组的第二个元素的总和result = {key: sum(item[1] for item in group) for key, group in grouped_data}
使用 functools 中的 reduce：
result = reduce(lambda acc, tup: {**acc, tup[0]: acc.get(tup[0], 0) + tup[1]}, input_list, {})
**acc 从 acc 中获取键/值对（与 **kwargs 相同）。
英文:
Using a sorted list:
from itertools import groupbyfrom operator import itemgetterinput_list = [("Skills", 3),("Wholesome", 3),("FanArt", 3),("Clip", 3),("Skills", 2),("FanArt", 2),]# Sort the input list by the first element of each tuplesorted_list = sorted(input_list, key=itemgetter(0))# you can also use a lambda function as a key# sorted_list = sorted(input_list, key=lambda x: x[0])# Use groupby to group the tuples by the first elementgrouped_data = groupby(sorted_list, key=itemgetter(0))# Create a dictionary with the sums of the second elements for each groupresult = {key: sum(item[1] for item in group) for key, group in grouped_data}
Using reduce from functools:
result = reduce(lambda acc, t: {**acc, tup[0]: acc.get(tup[0], 0) + tup[1]}, input_list, {})
**acc gets the key/values pairs from acc(same as **kwargs).
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