Python 在元组内创建一个列表,该列表包含与整数成对的所有数字
我有这样一份清单:Python 在元组内创建一个列表,该列表包含与整数成对的所有数字,python,list,tuples,Python,List,Tuples,我有这样一份清单: [(0, 1), (0, 2), (0, 3), (1, 4), (1, 6), (1, 7), (1, 9)] [(0, [1, 2, 3]), (1, [4, 6, 7, 9])] 我需要将其转换为如下所示的元组: [(0, 1), (0, 2), (0, 3), (1, 4), (1, 6), (1, 7), (1, 9)] [(0, [1, 2, 3]), (1, [4, 6, 7, 9])] 这是我的代码: friends = open(file_name)
[(0, 1), (0, 2), (0, 3), (1, 4), (1, 6), (1, 7), (1, 9)]
[(0, [1, 2, 3]), (1, [4, 6, 7, 9])]
我需要将其转换为如下所示的元组:
[(0, 1), (0, 2), (0, 3), (1, 4), (1, 6), (1, 7), (1, 9)]
[(0, [1, 2, 3]), (1, [4, 6, 7, 9])]
这是我的代码:
friends = open(file_name).read().splitlines()
network = []
friends = [tuple(int(y) for y in x.split(' ')) for x in friends]
return friends
您可以尝试以下方法:
import itertools
s = [(0, 1), (0, 2), (0, 3), (1, 4), (1, 6), (1, 7), (1, 9)]
final_data = [(a, [i[-1] for i in list(b)]) for a, b in itertools.groupby(sorted(s, key=lambda x:x[0]), key=lambda x:x[0])]
输出:
[(0, [1, 2, 3]), (1, [4, 6, 7, 9])]
这不完全是你想要的,但是从你想要的输出类型来看,请让我建议你使用,它是超级可读和高效的
from collections import defaultdict
some_list = [(0, 1), (0, 2), (0, 3), (1, 4), (1, 6), (1, 7), (1, 9)]
d = defaultdict(list)
for k, v in some_list:
d[k].append(v)
输出
defaultdict(<class 'list'>, {0: [1, 2, 3], 1: [4, 6, 7, 9]})
defaultdict(,{0:[1,2,3],1:[4,6,7,9]})
带有换行符的“一行”嵌套列表理解
tpls = [(0, 1), (0, 2), (0, 3), (1, 4), (1, 6), (1, 7), (1, 9)]
[(k, [tp[1] for tp in tpls if tp[0] == k])
for k in set([*zip(*tpls)][0])]
Out[11]: [(0, [1, 2, 3]), (1, [4, 6, 7, 9])]
[*zip(*tpls)]
是一种“转置”子表的习惯用法
给出[(0,0,0,1,1,1,1,1),(1,2,3,4,6,7,9)]
所以set([*zip(*tpls)][0])
就是set((0,0,0,1,1,1,1,1))
它在tpls
:{0,1}
其中外部对k in…
进行迭代,为结果元组内的列表comp提供k
[tp[1]用于tpl中的tp,如果tp[0]==k]
用于解析文件中的列表。