Reputation: 34170

Python truncate a long string

How does one truncate a string to 75 characters in Python?

This is how it is done in JavaScript:

const data = "saddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddsaddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddsadddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd";
const info = (data.length > 75) ? data.substring(0, 75) + '..' : data;
console.log(info);

Upvotes: 369

Answers (23)

Jack

Reputation: 1

Big fan of ljust (and rjust) so it deserves to be mentioned as an option:

info = data[:75].ljust(75)

Simple, inline, and you can combine it with any of the other things if you want the ellipsis, but it's not required. Good for if you're writing to output with multiline, and want it to space evenly like a proper table.

Though for your specific example,

if(len(data)>75):
     info = data[:72]+"..."
else:
     info = data.ljust(75)

Upvotes: 0

Marcelo Cantos

Reputation: 185852

info = (data[:75] + '..') if len(data) > 75 else data

This code matches the JavaScript, but you should consider using data[:73] so that the total result including the .. fits in 75 characters.

Upvotes: 612

Noam-N

Reputation: 914

Here I use textwrap.shorten and handle more edge cases. also include part of the last word in case this word is more than 50% of the max width.

import textwrap


def shorten(text: str, width=30, placeholder="..."):
    """Collapse and truncate the given text to fit in the given width.

    The text first has its whitespace collapsed. If it then fits in the *width*, it is returned as is.
    Otherwise, as many words as possible are joined and then the placeholder is appended.
    """
    if not text or not isinstance(text, str):
        return str(text)
    t = text.strip()
    if len(t) <= width:
        return t

    # textwrap.shorten also throws ValueError if placeholder too large for max width
    shorten_words = textwrap.shorten(t, width=width, placeholder=placeholder)

    # textwrap.shorten doesn't split words, so if the text contains a long word without spaces, the result may be too short without this word.
    # Here we use a different way to include the start of this word in case shorten_words is less than 50% of `width`
    if len(shorten_words) - len(placeholder) < (width - len(placeholder)) * 0.5:
        return t[:width - len(placeholder)].strip() + placeholder
    return shorten_words

Tests:

>>> shorten("123 456", width=7, placeholder="...")
'123 456'
>>> shorten("1 23 45 678 9", width=12, placeholder="...")
'1 23 45...'
>>> shorten("1 23 45 678 9", width=10, placeholder="...")
'1 23 45...'
>>> shorten("01 23456789", width=10, placeholder="...")
'01 2345...'
>>> shorten("012 3 45678901234567", width=17, placeholder="...")
'012 3 45678901...'
>>> shorten("1 23 45 678 9", width=9, placeholder="...")
'1 23...'
>>> shorten("1 23456", width=5, placeholder="...")
'1...'
>>> shorten("123 456", width=5, placeholder="...")
'12...'
>>> shorten("123 456", width=6, placeholder="...")
'123...'
>>> shorten("12 3456789", width=9, placeholder="...")
'12 345...'
>>> shorten("   12 3456789    ", width=9, placeholder="...")
'12 345...'
>>> shorten('123 45', width=4, placeholder="...")
'1...'
>>> shorten('123 45', width=3, placeholder="...")
'...'
>>> shorten("123456", width=3, placeholder="...")
'...'
>>> shorten([1], width=9, placeholder="...")
'[1]'
>>> shorten(None, width=5, placeholder="...")
'None'
>>> shorten("", width=9, placeholder="...")
''

Upvotes: 1

Hagai Drory

Reputation: 301

If you wish to do some more sophisticated string truncate you can adopt sklearn approach as implement by:

sklearn.base.BaseEstimator.__repr__ (See Original full code at: https://github.com/scikit-learn/scikit-learn/blob/f3f51f9b6/sklearn/base.py#L262)

It adds benefits such as avoiding truncate in the middle of the word.

def truncate_string(data, N_CHAR_MAX=70):
    # N_CHAR_MAX is the (approximate) maximum number of non-blank
    # characters to render. We pass it as an optional parameter to ease
    # the tests.

    lim = N_CHAR_MAX // 2  # apprx number of chars to keep on both ends
    regex = r"^(\s*\S){%d}" % lim
    # The regex '^(\s*\S){%d}' % n
    # matches from the start of the string until the nth non-blank
    # character:
    # - ^ matches the start of string
    # - (pattern){n} matches n repetitions of pattern
    # - \s*\S matches a non-blank char following zero or more blanks
    left_lim = re.match(regex, data).end()
    right_lim = re.match(regex, data[::-1]).end()
    if "\n" in data[left_lim:-right_lim]:
        # The left side and right side aren't on the same line.
        # To avoid weird cuts, e.g.:
        # categoric...ore',
        # we need to start the right side with an appropriate newline
        # character so that it renders properly as:
        # categoric...
        # handle_unknown='ignore',
        # so we add [^\n]*\n which matches until the next \n
        regex += r"[^\n]*\n"
        right_lim = re.match(regex, data[::-1]).end()
    ellipsis = "..."
    if left_lim + len(ellipsis) < len(data) - right_lim:
        # Only add ellipsis if it results in a shorter repr
        data = data[:left_lim] + "..." + data[-right_lim:]
    return data

Upvotes: 1

m3trik

Reputation: 333

Here's a simple function that will truncate a given string from either side:

def truncate(string, length=75, beginning=True, insert='..'):
    '''Shorten the given string to the given length.
    An ellipsis will be added to the section trimmed.

    :Parameters:
        length (int) = The maximum allowed length before trunicating.
        beginning (bool) = Trim starting chars, else; ending.
        insert (str) = Chars to add at the trimmed area. (default: ellipsis)

    :Return:
        (str)

    ex. call: truncate('12345678', 4)
        returns: '..5678'
    '''
    if len(string)>length:
        if beginning: #trim starting chars.
            string = insert+string[-length:]
        else: #trim ending chars.
            string = string[:length]+insert
    return string

Upvotes: 0

Toothpick Anemone

Reputation: 4644

Suppose that stryng is a string which we wish to truncate and that nchars is the number of characters desired in the output string.

stryng = "sadddddddddddddddddddddddddddddddddddddddddddddddddd"
nchars = 10

We can truncate the string as follows:

def truncate(stryng:str, nchars:int):
    return (stryng[:nchars - 6] + " [...]")[:min(len(stryng), nchars)]

The results for certain test cases are shown below:

s = "sadddddddddddddddddddddddddddddd!"
s = "sa" + 30*"d" + "!"

truncate(s, 2)                ==  sa
truncate(s, 4)                ==  sadd
truncate(s, 10)               ==  sadd [...]
truncate(s, len(s)//2)        ==  sadddddddd [...]

My solution produces reasonable results for the test cases above.

However, some pathological cases are shown below:

Some Pathological Cases!

truncate(s, len(s) - 3)()       ==  sadddddddddddddddddddddd [...]
truncate(s, len(s) - 2)()       ==  saddddddddddddddddddddddd [...]
truncate(s, len(s) - 1)()       ==  sadddddddddddddddddddddddd [...]
truncate(s, len(s) + 0)()       ==  saddddddddddddddddddddddddd [...]
truncate(s, len(s) + 1)()       ==  sadddddddddddddddddddddddddd [...
truncate(s, len(s) + 2)()       ==  saddddddddddddddddddddddddddd [..
truncate(s, len(s) + 3)()       ==  sadddddddddddddddddddddddddddd [.
truncate(s, len(s) + 4)()       ==  saddddddddddddddddddddddddddddd [
truncate(s, len(s) + 5)()       ==  sadddddddddddddddddddddddddddddd 
truncate(s, len(s) + 6)()       ==  sadddddddddddddddddddddddddddddd!
truncate(s, len(s) + 7)()       ==  sadddddddddddddddddddddddddddddd!
truncate(s, 9999)()             ==  sadddddddddddddddddddddddddddddd!

Notably,

When the string contains new-line characters (\n) there could be an issue.
When nchars > len(s) we should print string s without trying to print the "[...]"

Below is some more code:

import io

class truncate:
    """
        Example of Code Which Uses truncate:
        ```
            s = "\r<class\n 'builtin_function_or_method'>"
            s = truncate(s, 10)()
            print(s)
                    ```
                Examples of Inputs and Outputs:
                        truncate(s, 2)()   ==  \r
                        truncate(s, 4)()   ==  \r<c
                        truncate(s, 10)()  ==  \r<c [...]
                        truncate(s, 20)()  ==  \r<class\n 'bu [...]
                        truncate(s, 999)() ==  \r<class\n 'builtin_function_or_method'>
                    ```
                Other Notes:
                    Returns a modified copy of string input
                    Does not modify the original string
            """
    def __init__(self, x_stryng: str, x_nchars: int) -> str:
        """
        This initializer mostly exists to sanitize function inputs
        """
        try:
            stryng = repr("".join(str(ch) for ch in x_stryng))[1:-1]
            nchars = int(str(x_nchars))
        except BaseException as exc:
            invalid_stryng =  str(x_stryng)
            invalid_stryng_truncated = repr(type(self)(invalid_stryng, 20)())

            invalid_x_nchars = str(x_nchars)
            invalid_x_nchars_truncated = repr(type(self)(invalid_x_nchars, 20)())

            strm = io.StringIO()
            print("Invalid Function Inputs", file=strm)
            print(type(self).__name__, "(",
                  invalid_stryng_truncated,
                  ", ",
                  invalid_x_nchars_truncated, ")", sep="", file=strm)
            msg = strm.getvalue()

            raise ValueError(msg) from None

        self._stryng = stryng
        self._nchars = nchars

    def __call__(self) -> str:
        stryng = self._stryng
        nchars = self._nchars
        return (stryng[:nchars - 6] + " [...]")[:min(len(stryng), nchars)]

Upvotes: 0

NicoHood

Reputation: 1093

Coming very late to the party I want to add my solution to trim text at character level that also handles whitespaces properly.

def trim_string(s: str, limit: int, ellipsis='…') -> str:
    s = s.strip()
    if len(s) > limit:
        return s[:limit-1].strip() + ellipsis
    return s

Simple, but it will make sure you that hello world with limit=6 will not result in an ugly hello … but hello… instead.

It also removes leading and trailing whitespaces, but not spaces inside. If you also want to remove spaces inside, checkout this stackoverflow post

Upvotes: 3

lacroixDj

Reputation: 101

Simple and short helper function:

def truncate_string(value, max_length=255, suffix='...'):
    string_value = str(value)
    string_truncated = string_value[:min(len(string_value), (max_length - len(suffix)))]
    suffix = (suffix if len(string_value) > max_length else '')
    return string_truncated+suffix

Usage examples:

# Example 1 (default):

long_string = ""
for number in range(1, 1000): 
    long_string += str(number) + ','    

result = truncate_string(long_string)
print(result)


# Example 2 (custom length):

short_string = 'Hello world'
result = truncate_string(short_string, 8)
print(result) # > Hello... 


# Example 3 (not truncated):

short_string = 'Hello world'
result = truncate_string(short_string)
print(result) # > Hello world

Upvotes: 1

Jesse

Reputation: 77

info = data[:min(len(data), 75)

Upvotes: 5

HelloGoodbye

Reputation: 3902

limit = 75
info = data[:limit] + '..' * (len(data) > limit)

Upvotes: 16

HelloGoodbye

Reputation: 3902

info = data[:75] + ('..' if len(data) > 75 else '')

Upvotes: 7

Acecool

Reputation: 754

Here's a function I made as part of a new String class... It allows adding a suffix ( if the string is size after trimming and adding it is long enough - although you don't need to force the absolute size )

I was in the process of changing a few things around so there are some useless logic costs ( if _truncate ... for instance ) where it is no longer necessary and there is a return at the top...

But, it is still a good function for truncating data...

##
## Truncate characters of a string after _len'nth char, if necessary... If _len is less than 0, don't truncate anything... Note: If you attach a suffix, and you enable absolute max length then the suffix length is subtracted from max length... Note: If the suffix length is longer than the output then no suffix is used...
##
## Usage: Where _text = 'Testing', _width = 4
##      _data = String.Truncate( _text, _width )                        == Test
##      _data = String.Truncate( _text, _width, '..', True )            == Te..
##
## Equivalent Alternates: Where _text = 'Testing', _width = 4
##      _data = String.SubStr( _text, 0, _width )                       == Test
##      _data = _text[  : _width ]                                      == Test
##      _data = ( _text )[  : _width ]                                  == Test
##
def Truncate( _text, _max_len = -1, _suffix = False, _absolute_max_len = True ):
    ## Length of the string we are considering for truncation
    _len            = len( _text )

    ## Whether or not we have to truncate
    _truncate       = ( False, True )[ _len > _max_len ]

    ## Note: If we don't need to truncate, there's no point in proceeding...
    if ( not _truncate ):
        return _text

    ## The suffix in string form
    _suffix_str     = ( '',  str( _suffix ) )[ _truncate and _suffix != False ]

    ## The suffix length
    _len_suffix     = len( _suffix_str )

    ## Whether or not we add the suffix
    _add_suffix     = ( False, True )[ _truncate and _suffix != False and _max_len > _len_suffix ]

    ## Suffix Offset
    _suffix_offset = _max_len - _len_suffix
    _suffix_offset  = ( _max_len, _suffix_offset )[ _add_suffix and _absolute_max_len != False and _suffix_offset > 0 ]

    ## The truncate point.... If not necessary, then length of string.. If necessary then the max length with or without subtracting the suffix length... Note: It may be easier ( less logic cost ) to simply add the suffix to the calculated point, then truncate - if point is negative then the suffix will be destroyed anyway.
    ## If we don't need to truncate, then the length is the length of the string.. If we do need to truncate, then the length depends on whether we add the suffix and offset the length of the suffix or not...
    _len_truncate   = ( _len, _max_len )[ _truncate ]
    _len_truncate   = ( _len_truncate, _max_len )[ _len_truncate <= _max_len ]

    ## If we add the suffix, add it... Suffix won't be added if the suffix is the same length as the text being output...
    if ( _add_suffix ):
        _text = _text[ 0 : _suffix_offset ] + _suffix_str + _text[ _suffix_offset: ]

    ## Return the text after truncating...
    return _text[ : _len_truncate ]

Upvotes: 0

dansalmo

Reputation: 11686

This just in:

n = 8
s = '123'
print  s[:n-3] + (s[n-3:], '...')[len(s) > n]
s = '12345678'
print  s[:n-3] + (s[n-3:], '...')[len(s) > n]
s = '123456789'     
print  s[:n-3] + (s[n-3:], '...')[len(s) > n]
s = '123456789012345'
print  s[:n-3] + (s[n-3:], '...')[len(s) > n]

123
12345678
12345...
12345...

Upvotes: 6

Bora M. Alper

Reputation: 3844

If you are using Python 3.4+, you can use textwrap.shorten from the standard library:

Collapse and truncate the given text to fit in the given width.

First the whitespace in text is collapsed (all whitespace is replaced by single spaces). If the result fits in the width, it is returned. Otherwise, enough words are dropped from the end so that the remaining words plus the placeholder fit within width:
>>> textwrap.shorten("Hello  world!", width=12)
'Hello world!'
>>> textwrap.shorten("Hello  world!", width=11)
'Hello [...]'
>>> textwrap.shorten("Hello world", width=10, placeholder="...")
'Hello...'

Upvotes: 159

Sassan

Reputation: 2339

This method doesn't use any if:

data[:75] + bool(data[75:]) * '..'

Upvotes: 6

Adam Nelson

Reputation: 8090

There's no need for a regular expression but you do want to use string formatting rather than the string concatenation in the accepted answer.

This is probably the most canonical, Pythonic way to truncate the string data at 75 characters.

>>> data = "saddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddsaddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddsadddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd"
>>> info = "{}..".format(data[:75]) if len(data) > 75 else data
>>> info
'111111111122222222223333333333444444444455555555556666666666777777777788888...'

Upvotes: 0

The Veritable Bugeater

Reputation: 81

Yet another solution. With True and False you get a little feedback about the test at the end.

data = {True: data[:75] + '..', False: data}[len(data) > 75]

Upvotes: 3

Risadinha

Reputation: 16666

For a Django solution (which has not been mentioned in the question):

from django.utils.text import Truncator
value = Truncator(value).chars(75)

Have a look at Truncator's source code to appreciate the problem: https://github.com/django/django/blob/master/django/utils/text.py#L66

Concerning truncation with Django: Django HTML truncation

Upvotes: 49

Jade

Reputation: 784

You can't actually "truncate" a Python string like you can do a dynamically allocated C string. Strings in Python are immutable. What you can do is slice a string as described in other answers, yielding a new string containing only the characters defined by the slice offsets and step. In some (non-practical) cases this can be a little annoying, such as when you choose Python as your interview language and the interviewer asks you to remove duplicate characters from a string in-place. Doh.

Upvotes: 4

Davide Guerri

Reputation: 2050

With regex:

re.sub(r'^(.{75}).*$', '\g<1>...', data)

Long strings are truncated:

>>> data="11111111112222222222333333333344444444445555555555666666666677777777778888888888"
>>> re.sub(r'^(.{75}).*$', '\g<1>...', data)
'111111111122222222223333333333444444444455555555556666666666777777777788888...'

Shorter strings never get truncated:

>>> data="11111111112222222222333333"
>>> re.sub(r'^(.{75}).*$', '\g<1>...', data)
'11111111112222222222333333'

This way, you can also "cut" the middle part of the string, which is nicer in some cases:

re.sub(r'^(.{5}).*(.{5})$', '\g<1>...\g<2>', data)

>>> data="11111111112222222222333333333344444444445555555555666666666677777777778888888888"
>>> re.sub(r'^(.{5}).*(.{5})$', '\g<1>...\g<2>', data)
'11111...88888'

Upvotes: 16

Spouk

Reputation: 712

       >>> info = lambda data: len(data)>10 and data[:10]+'...' or data
       >>> info('sdfsdfsdfsdfsdfsdfsdfsdfsdfsdfsdf')
           'sdfsdfsdfs...'
       >>> info('sdfsdf')
           'sdfsdf'
       >>>

Upvotes: 1