Appending the same string to a list of strings in Python

I am trying to take one string, and append it to every string contained in a list, and then have a new list with the completed strings. Example:

list1 = ['foo', 'fob', 'faz', 'funk']
string = 'bar'

*magic*

list2 = ['foobar', 'fobbar', 'fazbar', 'funkbar']

I tried for loops, and an attempt at list comprehension, but it was garbage. As always, any help, much appreciated.

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

The simplest way to do this is with a list comprehension:

[s + mystring for s in mylist]

Notice that I avoided using builtin names like list because that shadows or hides the builtin names, which is very much not good.

Also, if you do not actually need a list, but just need an iterator, a generator expression can be more efficient (although it does not likely matter on short lists):

(s + mystring for s in mylist)

These are very powerful, flexible, and concise. Every good python programmer should learn to wield them.

Method 2

my_list = ['foo', 'fob', 'faz', 'funk']
string = 'bar'
my_new_list = [x + string for x in my_list]
print my_new_list

This will print:

['foobar', 'fobbar', 'fazbar', 'funkbar']

Method 3

map seems like the right tool for the job to me.

my_list = ['foo', 'fob', 'faz', 'funk']
string = 'bar'
list2 = list(map(lambda orig_string: orig_string + string, my_list))

See this section on functional programming tools for more examples of map.

Method 4

Here is a simple answer using pandas.

import pandas as pd
list1 = ['foo', 'fob', 'faz', 'funk']
string = 'bar'

list2 = (pd.Series(list1) + string).tolist()
list2
# ['foobar', 'fobbar', 'fazbar', 'funkbar']

Method 5

Updating with more options

Below are some of the methods I have followed, and I’m sure there could be more.

Method 1:

list1 = ['foo', 'fob', 'faz', 'funk']
list2 = [ls+"bar" for ls in list1] # using list comprehension
print(list2)

Method 2:

list1 = ['foo', 'fob', 'faz', 'funk']
list2 = list(map(lambda ls: ls+"bar", list1))
print(list2)

Method 3:

list1 = ['foo', 'fob', 'faz', 'funk']
addstring = 'bar'
for index, value in enumerate(list1):
    list1[index] = addstring + value #this will prepend the string
    #list1[index] = value + addstring #this will append the string

Method 4:

list1 = ['foo', 'fob', 'faz', 'funk']
addstring = 'bar'
list2 = []
for value in list1:
    list2.append(str(value) + "bar")
print(list2)

Method 5:

list1 = ['foo', 'fob', 'faz', 'funk']
list2 = list(map(''.join, zip(list1, ["bar"]*len(list1))))
print(list2)

Avoid using keywords as variables like ‘list’, renamed ‘list’ as ‘list1’ instead

Method 6

Running the following experiment the pythonic way:

[s + mystring for s in mylist]

seems to be ~35% faster than the obvious use of a for loop like this:

i = 0
for s in mylist:
    mylist[i] = s+mystring
    i = i + 1

Experiment

import random
import string
import time

mystring = '/test/'

l = []
ref_list = []

for i in xrange( 10**6 ):
    ref_list.append( ''.join(random.choice(string.ascii_lowercase) for i in range(10)) )

for numOfElements in [5, 10, 15 ]:

    l = ref_list*numOfElements
    print 'Number of elements:', len(l)

    l1 = list( l )
    l2 = list( l )

    # Method A
    start_time = time.time()
    l2 = [s + mystring for s in l2]
    stop_time = time.time()
    dt1 = stop_time - start_time
    del l2
    #~ print "Method A: %s seconds" % (dt1)

    # Method B
    start_time = time.time()
    i = 0
    for s in l1:
        l1[i] = s+mystring
        i = i + 1
    stop_time = time.time()
    dt0 = stop_time - start_time
    del l1
    del l
    #~ print "Method B: %s seconds" % (dt0)

    print 'Method A is %.1f%% faster than Method B' % ((1 - dt1/dt0)*100)

Results

Number of elements: 5000000
Method A is 38.4% faster than Method B
Number of elements: 10000000
Method A is 33.8% faster than Method B
Number of elements: 15000000
Method A is 35.5% faster than Method B

Method 7

Extending a bit to “Appending a list of strings to a list of strings”:

    import numpy as np
    lst1 = ['a','b','c','d','e']
    lst2 = ['1','2','3','4','5']

    at = np.full(fill_value='@',shape=len(lst1),dtype=object) #optional third list
    result = np.array(lst1,dtype=object)+at+np.array(lst2,dtype=object)

Result:

array(['<a href="https://getridbug.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="45240574">[email protected]</a>', '<a href="https://getridbug.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="53311361">[email protected]</a>', '<a href="https://getridbug.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="94f7d4a7">[email protected]</a>', '<a href="https://getridbug.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="80e4c0b4">[email protected]</a>', '<a href="https://getridbug.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="bedbfe8b">[email protected]</a>'], dtype=object)

dtype odject may be further converted str

Method 8

Combining map and format:

>>> list(map('{}bar'.format,  ['foo', 'fob', 'faz', 'funk']))
['foobar', 'fobbar', 'fazbar', 'funkbar']

Thus, there is no loop variable.
It works for Python 2 and 3. (In Python 3 one can write [*map(...)], and in Python 2 just map(...).

Method 9

you can use lambda inside map in python. wrote a gray codes generator.
https://github.com/rdm750/rdm750.github.io/blob/master/python/gray_code_generator.py
# your code goes here
”’
the n-1 bit code, with 0 prepended to each word, followed by
the n-1 bit code in reverse order, with 1 prepended to each word.
”’

    def graycode(n):
        if n==1:
            return ['0','1']
        else:
            nbit=map(lambda x:'0'+x,graycode(n-1))+map(lambda x:'1'+x,graycode(n-1)[::-1])
            return nbit

    for i in xrange(1,7):
        print map(int,graycode(i))

Method 10

list2 = ['%sbar' % (x,) for x in list]

And don’t use list as a name; it shadows the built-in type.

Method 11

new_list = [word_in_list + end_string for word_in_list in old_list]

Using names such as “list” for your variable names is bad since it will overwrite/override the builtins.

Method 12

Just in case

list = ['foo', 'fob', 'faz', 'funk']
string = 'bar'
for i in range(len(list)):
    list[i] += string
print(list)


All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

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