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Trying to strip the "0b1" from the left end of a binary number.

The following code results in stripping all of binary object. (not good)

>>> bbn = '0b1000101110100010111010001' #converted bin(2**24+**2^24/11)
>>> aan=bbn.lstrip("0b1")  #Try stripping all left-end junk at once.
>>> print aan    #oops all gone.
''

So I did the .lstrip() in two steps:

>>> bbn = '0b1000101110100010111010001' #    Same fraction expqansion
>>> aan=bbn.lstrip("0b")# Had done this before.
>>> print aan    #Extra "1" still there.
'1000101110100010111010001'
>>> aan=aan.lstrip("1")#  If at first you don't succeed...
>>> print aan    #YES!
'000101110100010111010001'

What's the deal?

Thanks again for solving this in one simple step. (see my previous question)

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7 Answers 7

Reset to default
12

The strip family treat the arg as a set of characters to be removed. The default set is "all whitespace characters".

You want:

if strg.startswith("0b1"):
   strg = strg[3:]
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2 Comments

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Yes, but why only the first "1" in the second case?
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@Harryooo: A question for you: Why does Python have both an lstrip and a rstrip and also a strip that combines lstrip and rstrip functionality?
10

No. Stripping removes all characters in the sequence passed, not just the literal sequence. Slice the string if you want to remove a fixed length.

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For example, "ababbbbbbaaccc".lstrip("ab") => "ccc"
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I sort of figured that lstrip() was using char as a list, but then I was surprised again when lstrip("1") only eliminated the first "1" and not the rest.
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That's because there was only a single 1 on the left.
8

In Python 3.9 you can use bbn.removeprefix('0b1').

(Actually this question has been mentioned as part of the rationale in PEP 616.)

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0

This is the way lstrip works. It removes any of the characters in the parameter, not necessarily the string as a whole. In the first example, since the input consisted of only those characters, nothing was left.

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Yes, but why only the first "1" in the second case?
Image
@Harryooo, in your second example lstrip("1") removes characters on the left until it reaches something that isn't a 1. If it had started with '1111' it would have removed all of them.
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Lstrip is removing any of the characters in the string. So, as well as the initial 0b1, it is removing all zeros and all ones. Hence it is all gone!

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Yes, but why only the first "1" in the second case?
0

@Harryooo: lstrip only takes the characters off the left hand end. So, because there's only one 1 before the first 0, it removes that. If the number started 0b11100101..., calling a.strip('0b').strip('1') would remove the first three ones, so you'd be left with 00101.

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>>> i = 0b1000101110100010111010001
>>> print(bin(i))
'0b1000101110100010111010001'

>>> print(format(i, '#b'))
'0b1000101110100010111010001'
>>> print(format(i, 'b'))
'1000101110100010111010001'

See Example in python tutor:

From the standard doucmentation (See standard documentation for function bin()): bin(x) Convert an integer number to a binary string prefixed with “0b”. The result is a valid Python expression. If x is not a Python int object, it has to define an index() method that returns an integer. Some examples:

>>> bin(3)
'0b11'
>>> bin(-10)
'-0b1010'

If prefix “0b” is desired or not, you can use either of the following ways.

>>> format(14, '#b'), format(14, 'b')
('0b1110', '1110')
>>> f'{14:#b}', f'{14:b}'
('0b1110', '1110')

See also format() for more information.

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