Is there a simple way to remove multiple spaces in a string?

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Suppose this string:

The   fox jumped   over    the log.

Turning into:

The fox jumped over the log.

What is the simplest (1-2 lines) to achieve this, without splitting and going into lists?

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Is there a simple way to remove multiple spaces in a string? log: Questions

log

Python"s equivalent of && (logical-and) in an if-statement

5 answers

delete By delete

Here"s my code:

def front_back(a, b):
  # +++your code here+++
  if len(a) % 2 == 0 && len(b) % 2 == 0:
    return a[:(len(a)/2)] + b[:(len(b)/2)] + a[(len(a)/2):] + b[(len(b)/2):] 
  else:
    #todo! Not yet done. :P
  return

I"m getting an error in the IF conditional.
What am I doing wrong?

934

Answer #1

You would want and instead of &&.

934

Answer #2

Python uses and and or conditionals.

i.e.

if foo == "abc" and bar == "bac" or zoo == "123":
  # do something

Is there a simple way to remove multiple spaces in a string? log: Questions

log

How do you get the logical xor of two variables in Python?

5 answers

Zach Hirsch By Zach Hirsch

How do you get the logical xor of two variables in Python?

For example, I have two variables that I expect to be strings. I want to test that only one of them contains a True value (is not None or the empty string):

str1 = raw_input("Enter string one:")
str2 = raw_input("Enter string two:")
if logical_xor(str1, str2):
    print "ok"
else:
    print "bad"

The ^ operator seems to be bitwise, and not defined on all objects:

>>> 1 ^ 1
0
>>> 2 ^ 1
3
>>> "abc" ^ ""
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for ^: "str" and "str"
794

Answer #1

If you"re already normalizing the inputs to booleans, then != is xor.

bool(a) != bool(b)

How do you split a list into evenly sized chunks?

5 answers

jespern By jespern

I have a list of arbitrary length, and I need to split it up into equal size chunks and operate on it. There are some obvious ways to do this, like keeping a counter and two lists, and when the second list fills up, add it to the first list and empty the second list for the next round of data, but this is potentially extremely expensive.

I was wondering if anyone had a good solution to this for lists of any length, e.g. using generators.

I was looking for something useful in itertools but I couldn"t find anything obviously useful. Might"ve missed it, though.

Related question: What is the most “pythonic” way to iterate over a list in chunks?

2632

Answer #1

Here"s a generator that yields the chunks you want:

def chunks(lst, n):
    """Yield successive n-sized chunks from lst."""
    for i in range(0, len(lst), n):
        yield lst[i:i + n]

import pprint
pprint.pprint(list(chunks(range(10, 75), 10)))
[[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
 [20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
 [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
 [40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
 [50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
 [60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
 [70, 71, 72, 73, 74]]

If you"re using Python 2, you should use xrange() instead of range():

def chunks(lst, n):
    """Yield successive n-sized chunks from lst."""
    for i in xrange(0, len(lst), n):
        yield lst[i:i + n]

Also you can simply use list comprehension instead of writing a function, though it"s a good idea to encapsulate operations like this in named functions so that your code is easier to understand. Python 3:

[lst[i:i + n] for i in range(0, len(lst), n)]

Python 2 version:

[lst[i:i + n] for i in xrange(0, len(lst), n)]

2632

Answer #2

If you want something super simple:

def chunks(l, n):
    n = max(1, n)
    return (l[i:i+n] for i in range(0, len(l), n))

Use xrange() instead of range() in the case of Python 2.x

2632

Answer #3

Directly from the (old) Python documentation (recipes for itertools):

from itertools import izip, chain, repeat

def grouper(n, iterable, padvalue=None):
    "grouper(3, "abcdefg", "x") --> ("a","b","c"), ("d","e","f"), ("g","x","x")"
    return izip(*[chain(iterable, repeat(padvalue, n-1))]*n)

The current version, as suggested by J.F.Sebastian:

#from itertools import izip_longest as zip_longest # for Python 2.x
from itertools import zip_longest # for Python 3.x
#from six.moves import zip_longest # for both (uses the six compat library)

def grouper(n, iterable, padvalue=None):
    "grouper(3, "abcdefg", "x") --> ("a","b","c"), ("d","e","f"), ("g","x","x")"
    return zip_longest(*[iter(iterable)]*n, fillvalue=padvalue)

I guess Guido"s time machine works—worked—will work—will have worked—was working again.

These solutions work because [iter(iterable)]*n (or the equivalent in the earlier version) creates one iterator, repeated n times in the list. izip_longest then effectively performs a round-robin of "each" iterator; because this is the same iterator, it is advanced by each such call, resulting in each such zip-roundrobin generating one tuple of n items.

We hope this article has helped you to resolve the problem. Apart from Is there a simple way to remove multiple spaces in a string?, check other log-related topics.

Want to excel in Python? See our review of the best Python online courses 2022. If you are interested in Data Science, check also how to learn programming in R.

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Frank Lehnman

Singapore | 2022-12-03

Maybe there are another answers? What Is there a simple way to remove multiple spaces in a string? exactly means?. I just hope that will not emerge anymore

Anna OConnell

Milan | 2022-12-03

log is always a bit confusing 😭 Is there a simple way to remove multiple spaces in a string? is not the only problem I encountered. Will get back tomorrow with feedback

Walter Williams

London | 2022-12-03

Maybe there are another answers? What Is there a simple way to remove multiple spaces in a string? exactly means?. I just hope that will not emerge anymore

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