Appending the same string to a list of strings in Python

Question

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.

Answer 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.

Answer 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']

Answer 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.

Answer 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']

Answer 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

Answer 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

Answer 7

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(...).

If one prefers the modulo expression

>>> list(map('%sbar'.__mod__,  ['foo', 'fob', 'faz', 'funk']))
['foobar', 'fobbar', 'fazbar', 'funkbar']

To prepend one can use __add__ method

>>> list(map('bar'.__add__,  ['foo', 'fob', 'faz', 'funk']))
['barfoo', 'barfob', 'barfaz', 'barfunk']

Answer 8

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@1', 'b@2', 'c@3', 'd@4', 'e@5'], dtype=object)

dtype odject may be further converted str

Answer 9

Since Python 3.6 using f-strings is the best practice (instead of format or concatenating with +). See PEP498.

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

list2 = [f"{s}{mystring}" for s in list1]

Answer 10

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.

Answer 11

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))

Answer 12

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

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

Answer 13

Just in case

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