Why does this code snippet result in: TypeError 'int' object is not subscriptable?
return (bin(int(hexdata)[2:].zfill(16)))
hexdata is a hexadecimal string. For example, it may be 0x0101.
2 Answers
You're doing this:
>>> int('1234')[2:]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'int' object has no attribute '__getitem__'
>>> 1234[2:]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'int' object has no attribute '__getitem__'
If you intended to remove the first two character, use following:
>>> int('1234'[2:])
34
If the orignal string is hexadecimal representation, you should pass optional base argument 16
>>> int('1234'[2:], 16)
52
If [2:] is used to remove 0b generated by bin (not to remove leading characters from the original string), then following is what you want.
>>> int('0x1234', 16) # No need to remove leading `0x`
4660
>>> int('0x1234', 0) # automatically recognize the number's base using the prefix
4660
>>> int('1234', 16)
4660
>>> bin(int('1234', 16))
'0b1001000110100'
>>> bin(int('1234', 16))[2:]
'1001000110100'
>>> bin(int('1234', 16))[2:].zfill(16)
'0001001000110100'
BTW, you can use str.format or format instead of bin + str.zfill:
>>> '{:016b}'.format(int('1234', 16))
'0001001000110100'
>>> format(int('1234', 16), '016b')
'0001001000110100'
UPDATE
If you specify 0 as a base, int will automatically recognize the number's base using the prefix.
>>> int('0x10', 0)
16
>>> int('10', 0)
10
>>> int('010', 0)
8
>>> int('0b10', 0)
2
7 Comments
paxdiablo
The 2-subscript is clearly to get rid of the
0x from the sample number 0x0101. I think all the stuff about "1234"[2:] is probably superfluous.
falsetru
@paxdiablo, I don't think so. Try
int('0x12'): raises ValueError: invalid literal for int() with base 10: '0x12', which is different from what OP mentioned int he question.falsetru
@paxdiablo, IMHO,
[2:] was used to remove 0b prepended by bin.paxdiablo
Actually, you're right,
int(x,16) happily accept numbers of the form 0xWhatever. But I still think that the bits suggesting int('1234'[2:]) are not necessary in the answer, since they're not touching a 0bxxxx string.falsetru
@paxdiablo, Thank you for the comment. I updated the answer accordingly.
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Parentheses in the wrong place. Assuming you have a string like "0x0101" where you want to end up with a 16-digit binary string:
bin(int(hexdata,16))[2:].zfill(16)
The int(X,16) call interprets that as a hex number (and can accept a number of the form 0xSomething). Then bin(X) turns that into a string of the form 0b01010101.
The [2:] then gets rid of the 0b at the front and zfill(16) fills that out to 16 "bits".
>>> bin(int("0x0102",16))[2:].zfill(16)
'0000000100000010'
2 Comments
zhangxaochen
should set the base to 16:
int(hexdata, 16)
user2357112
It's hex: "hexdata is a hexadecimal string. for example, it may be 0x0101"
int("0xA", 16)