从压缩对象到列表的转换不起作用

问题描述 投票:0回答:1

尽管转换为列表,但正在获取压缩的对象

import itertools as it
def combi(i,j):
    string_maps = {"1": "abc","2": "def","3": "ghi","4": "jkl","5": "mno","6": "pqrs","7": "tuv","8": "wxy","9": "z"}
    string_maps=dict((int(k),v) for k,v in string_maps.items())
    list1=[]
    list1.append(string_maps.get(i))
    list2=[]
    list2.append(string_maps.get(j))
    list_new =list([zip(x,list2) for x in it.permutations(list1,len(list2))])
    print(list_new)

combi(4,3)
python python-3.x
1个回答
0
投票

直接导入itertoolsdef combi(i,j):string_maps = {“ 1”:“ abc”,“ 2”:“ def”,“ 3”:“ ghi”,“ 4”:“ jkl”,“ 5”:“ mno”,“ 6”:“ pqrs” ,“ 7”:“ tuv”,“ 8”:“ wxy”,“ 9”:“ z”}string_maps = dict((int(k),list(v))表示string_maps.items()中的k,vlist1 = []list1.append(string_maps.get(i))list1 = [列表1中子列表中的项目,列表中子项目中的项目]list2 = []list2.append(string_maps.get(j))list2 = [用于子列表中项目的列表2中子列表的项目]list_new = [list(zip(zip(x,list2)))for x in.permutations(set(list1),len(list2))]list_new = [列表列表中子列表项中的子列表项]打印(list_new)combi(3,4)

© www.soinside.com 2019 - 2024. All rights reserved.