Python：TypeError：'int'对象不可下标
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Python: TypeError: 'int' object is not subscriptable
提问by itVico
I get a TypeError and I don't understand why. The error is at c = t[i][0](according to the debugger). I have 3 char groups(lists): g1, g2and g3and I'm trying to change the char's index by substracting the key's k1, k2or k3from the index. What I'm using now for testing:
我得到一个 TypeError，我不明白为什么。错误在c = t[i][0]（根据调试器）。我有3个字符组（名单）g1，g2和g3我试图通过减去关键的改变字符的索引k1，k2或者k3从索引。我现在用于测试的内容：
text = 'abcd'
l_text = [('a', 0), ('b', 1), ('c', 2), ('d', 3)]
k1, k2, k3 = 2, 3, 1
And this is the code:
这是代码：
def rotate_left(text, l_text, k1, k2, k3):
i = 0
newstr = [None]*len(text)
for t in l_text: # t = tuple
c = t[i][0]
if c in g1: # c = char
l = int(l_text[i][1]) # l = index of the char in the list
if l - k1 < 0:
newstr[l%len(text)-k1] = l_text[i][0]
else:
newstr[l-k1] = l_text[i][0]
elif c in g2:
l = l_text[i][1] # l = index of the char in the list
if l - k1 < 0:
newstr[l%len(text)-k2] = l_text[i][0]
else:
newstr[l-k2] = l_text[i][0]
else:
l = l_text[i][1] # l = index of the char in the list
if l - k1 < 0:
newstr[l%len(text)-k3] = l_text[i][0]
else:
newstr[l-k3] = l_text[i][0]
i += 1
return newstr
Can someone explain me why do I get this error and how do I fix it? It's not like I'm using an inttype there. The debugger shows it's a str type and it breaks after the 2nd iteration.
有人可以解释我为什么会收到此错误以及如何修复它？这不像我在int那里使用一种类型。调试器显示它是一个 str 类型，它在第二次迭代后中断。
PS google didn't help PPS I know there is too much repetition in the code. I did it to see in the debugger what's happening.
PS google 没有帮助 PPS 我知道代码中有太多重复。我这样做是为了在调试器中查看发生了什么。
UPDATE:
更新：
Traceback (most recent call last):
File "/hometriplerotatie.py", line 56, in <module>
print(codeer('abcd', 2, 3, 1))
File "/home/triplerotatie.py", line 47, in codeer
text = rotate_left(text, l_text, k1, k2, k3)
File "/home/triplerotatie.py", line 9, in rotate_left
c = t[i][0]
TypeError: 'int' object is not subscriptable
采纳答案by Martijn Pieters
You are indexing into each individualtuple:
您正在索引每个单独的元组：
c = t[i][0]
istarts out as 0, but you increment it each loop iteration:
i开始为0，但每次循环迭代都会增加它：
i += 1
The forloop is binding tto each individual tuple from l_text, so first tis bound to ('a', 0), then to ('b', 1), etc.
该for环被结合t到从每个单独的元组l_text，那么第一t势必('a', 0)，然后向('b', 1)等
So first you are looking at ('a', 0)[0][0]which is 'a'[0]which is 'a'. The next iteration you look at ('b', 1)[1][0]which is 1[0]which raises your exception, because integers are not sequences.
因此，首先您要查看('a', 0)[0][0]which is 'a'[0]which is 'a'。你看看下一个迭代('b', 1)[1][0]是1[0]这引起了你的例外，因为整数不序列。
You need to remove the i; you do notneed to keep a running index here as the for t in l_text:is already giving you each individual tuple.
您需要删除i; 你不是要在这里保持运行指数作为for t in l_text:已经给你每个单独的元组。
回答by Tom Dalton
The error is here:
错误在这里：
l_text = [('a', 0), ('b', 1), ('c', 2), ('d', 3)]
...
for t in l_text: # t = tuple
# t is a tuple of 2 items: ('a', 0)
c = t[i][0] # Breaks when i == 1
I think you want:
我想你想要：
c = t[0]
It doesn't break the first time round the loop because when i == 0, t[i]is 'a'and then t[i][0]is also 'a'.
它不会在第一次循环时中断，因为 when i == 0, t[i]is 'a'， then t[i][0]is 'a'。
回答by Tarun Gaba
You are doing the index part wrong. Your tuple is 1 dimensional, so you cant use a 2-D array subscript notation. Assuming that
你做的索引部分是错误的。您的元组是一维的，因此您不能使用二维数组下标表示法。假如说
t = ('a',0)
you should use t[0]or t[1]to access aand 0respectively.
您应该使用t[0]或分别t[1]访问a和0。
Hope it helps.. :)
希望能帮助到你.. ：）
回答by Johan
The problem is that t is a tuple, you access the elements in a tuple that like a list. Currently you acces the elements like a 2D list which would, given your lists result in trying to indexing a char.
问题是 t 是一个元组，您可以访问元组中类似于列表的元素。当前，您可以像 2D 列表一样访问元素，鉴于您的列表会导致尝试对字符进行索引。
for t in l_text: # t = tuple
c = t[i][0]
should be changed to
应该改为
for t in l_text: # t = tuple
c = t[0]
