How to concatenate element-wise two lists in Python

Than can be done elegantly with map and zip:

map(lambda (x,y): x+y, zip(list1, list2))

Example:

In [1]: map(lambda (x,y): x+y, zip([1,2,3,4],[4,5,6,7]))
Out[1]: [5, 7, 9, 11]

not using zip. I dunno, I think this is the obvious way to do it. Maybe I just learnt C first 🙂

c=[]
for i in xrange(len(a)):
    c.append("%s%02d" % (b[i],a[i]))

Using zip

[m+str(n) for m,n in zip(b,a)]

output

['asp10', 'asp11', 'asp15', 'asp16', 'asp210', 'asp211']

b = ['asp1', 'asp1', 'asp1', 'asp1', 'asp2', 'asp2']
aa = [0, 1, 5, 6, 10, 11]
new_list =[]
if len(aa) != len(b):
     print 'list length mismatch'
else:
    for each in range(0,len(aa)):
        new_list.append(b[each] + str(aa[each]))
print new_list

Other solution (preferring printf formating style over .format() usage), it’s also smaller:

>>> ["%s%02d" % t for t in zip(b, a)]
['asp100', 'asp101', 'asp105', 'asp106', 'asp210', 'asp211']

inputs:

a = [0, 1, 5, 6, 10, 11] 
b = ['asp1', 'asp1', 'asp1', 'asp1', 'asp2', 'asp2']

concat_func = lambda x,y: x + "" + str(y)

list(map(concat_func,b,a)) # list the map function

output:

['asp10', 'asp11', 'asp15', 'asp16', 'asp210', 'asp211']

If you wanted to concatenate arbitrary number of lists, you could do this:

In [1]: lists = [["a", "b", "c"], ["m", "n", "o"], ["p", "q", "r"]] # Or more

In [2]: lists
Out[2]: [['a', 'b', 'c'], ['m', 'n', 'o'], ['p', 'q', 'r']]    

In [4]: list(map("".join, zip(*lists)))
Out[4]: ['amp', 'bnq', 'cor']

I ended up using a temporary DataFrame, it’s readable and fast:

a = ["a", "b", "c"]
b = ["1", "2", "3"]

df = pd.DataFrame({"a": a, "b": b})
df["c"] = df.a + df.b
result = df.c.values

Output:

$ result 
["a1", "b2", "c3"]

Behind the scenes, DataFrames use numpy, so the result is efficient.

And the same thing as a function:

import pandas as pd
from typing import List
def _elementwise_concat(self, a: List[str], b: List[str]) -> List[str]:
    """
    Elementwise concatenate.
    :param a: List of strings.
    :param b: List of strings.
    :return: List, same length, strings concatenated.
    """
    df = pd.DataFrame({"a": a, "b": b})
    df["c"] = df.a + df.b
    return df.c.values

List comprehension / zip() / use of zfill() to format.

print ([y+str(x).zfill(2) for x,y in zip(a,b)])

Output:

['asp100', 'asp101', 'asp105', 'asp106', 'asp210', 'asp211']

Using lambda and format:

b = ['asp1', 'asp1', 'asp1', 'asp1', 'asp2', 'asp2']
a = [0, 1, 5, 6, 10, 11]

list(map(lambda x: x[0] + "{0:0=2d}".format(x[1]), zip(b, a)))

Out:

['asp100', 'asp101', 'asp105', 'asp106', 'asp210', 'asp211']