1
a = [1, 2, 4, 5, 7, 8, 10]
n = len(a)
d = 3
c = []
for i in range(n):
  for j in range(i,n):
    for k in range(j,n):
        for x,y,z in zip(a[i],a[j],a[k]):
            print(x,y,z)

错误:回溯(最近一次调用):
文件“”,第 8 行,类型错误:'int' 对象不可迭代

当我将列表对象转换为字符串但不能在 int 中工作时,它可以工作。

4

1 回答 1

3

因为索引将返回对象而不是此处的可迭代容器。除非你这样称呼它:zip([a[i]], [a[j]], [a[k]]).

a = [1, 2, 4, 5, 7, 8, 10]
n = len(a)
d = 3
c = []
for i in range(n):
  for j in range(i,n):
    for k in range(j,n):
        for x,y,z in zip([a[i]], [a[j]], [a[k]]):
            print(x,y,z)
于 2021-04-10T06:45:49.880 回答