类型错误：列表索引必须是整数，而不是 str Python
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/27667531/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
TypeError: list indices must be integers, not str Python
提问by kerschi
list[s]is a string. Why doesn't this work?
list[s]是一个字符串。为什么这不起作用？
The following error appears:
出现以下错误：
TypeError: list indices must be integers, not str
类型错误：列表索引必须是整数，而不是 str
list = ['abc', 'def']
map_list = []
for s in list:
t = (list[s], 1)
map_list.append(t)
采纳答案by GLHF
list1 = ['abc', 'def']
list2=[]
for t in list1:
for h in t:
list2.append(h)
map_list = []
for x,y in enumerate(list2):
map_list.append(x)
print (map_list)
Output:
输出：
>>>
[0, 1, 2, 3, 4, 5]
>>>
This is what you want exactly.
这正是你想要的。
If you dont want to reach each element then:
如果您不想到达每个元素，则：
list1 = ['abc', 'def']
map_list=[]
for x,y in enumerate(list1):
map_list.append(x)
print (map_list)
Output:
输出：
>>>
[0, 1]
>>>
回答by Hackaholic
it should be:
它应该是：
for s in my_list: # here s is element of list not index of list
t = (s, 1)
map_list.append(t)
i think you want:
我想你想要：
for i,s in enumerate(my_list): # here i is the index and s is the respective element
t = (s, i)
map_list.append(t)
enumerategive index and element
enumerate给出索引和元素
Note: using list as variable name is bad practice. its built in function
注意：使用列表作为变量名是不好的做法。它的内置功能
回答by Ujjwal
Do not use the name listfor a list. I have used mylistbelow.
不要将名称list用于列表。我在mylist下面用过。
for s in mylist:
t = (mylist[s], 1)
for s in mylist:assigns elements of mylistto si.e stakes the value 'abc' in the first iteration and 'def' in the second iteration. Thus, scan't be used as an index in mylist[s].
for s in mylist:将元素分配mylist给sies在第一次迭代中取值 'abc'，在第二次迭代中取值 'def'。因此，s不能用作mylist[s].
Instead, simply do:
相反，只需执行以下操作：
for s in lists:
t = (s, 1)
map_list.append(t)
print map_list
#[('abc', 1), ('def', 1)]
回答by NPE
When you iterate over a list, the loop variable receives the actual list elements, not their indices. Thus, in your example sis a string (first abc, then def).
当您迭代一个列表时，循环变量接收实际的列表元素，而不是它们的索引。因此，在您的示例中s是一个字符串（首先abc，然后def）。
It looks like what you're trying to do is essentially this:
看起来您要做的基本上是这样的：
orig_list = ['abc', 'def']
map_list = [(el, 1) for el in orig_list]
This is using a Python construct called list comprehension.
这是使用名为list comprehension的 Python 构造。
回答by Danny Staple
for s in listwill produce the items of the list and not their indexes. So swill be 'abc'for the first loop, and then
'def'. 'abc'could only be a key to a dict, not a list index.
for s in list将生成列表的项目而不是它们的索引。所以s将'abc'第一个循环，然后
'def'。'abc'只能是字典的键，而不是列表索引。
In the line with tfetching the item by index is redundant in python.
在tpython 中，通过索引获取项目是多余的。
