[已解决]报错: TypeError: sequence item 0: expected str instance, int found

今天敲小例子,报了错TypeError: sequence item 0: expected str instance, int found

小例子

list1=[1,'two','three',4]
print(' '.join(list1))

以为会打印1 two three 4

结果报了错

Traceback (most recent call last):
  File "<pyshell#27>", line 1, in <module>
    print(" ".join(list1))
TypeError: sequence item 0: expected str instance, int found

上网查了资料,说list包含数字,不能直接转化成字符串。

解决办法:

print(" ".join('%s' %id for id in list1))

即遍历list的元素,把他转化成字符串。这样就能成功输出1 two three 4结果了。

from: https://blog.csdn.net/laochu250/article/details/67649210

posted @ 2020-06-07 22:33  hank-li  阅读(7830)  评论(0编辑  收藏  举报