58

I have two lists and I want to concatenate them element-wise. One of the list is subjected to string-formatting before concatenation.

For example :

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

In this case, a is subjected to string-formatting. That is, new a or aa should be :

aa = [00, 01, 05, 06, 10, 11]

Final output should be :

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

Can somebody please tell me how to do that?

Trenton McKinney
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Sanchit
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    try to play with [`zip`](http://docs.python.org/2/library/functions.html#zip) and [`string.format`](http://docs.python.org/2/library/string.html#format-specification-mini-language) – Francesco Montesano Oct 24 '13 at 07:54
  • @FrancescoMontesano Thanks, answer by nightcracker is working fine! – Sanchit Oct 24 '13 at 07:56
  • @nightcracker Thanks for the answer. Yes, sure. But, I can not accept your answer just now. It is showing to wait for 5 mins. Then, I will do that :) – Sanchit Oct 24 '13 at 08:01

11 Answers11

67

Use zip:

>>> ["{}{:02}".format(b_, a_) for a_, b_ in zip(a, b)]
['asp100', 'asp101', 'asp105', 'asp106', 'asp210', 'asp211']
orlp
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    @dmedvinsky What about this? `c = map("{0[1]}{0[0]:02}".format, zip(a, b))` Let the magic commence! – orlp Oct 24 '13 at 08:08
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    @nightcracker: good. The example with `map` works in python2. For python3 you need `list(map(...))` if you want a list, while the list comprehension works for both – Francesco Montesano Oct 24 '13 at 08:10
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    You don't *need* that, you just need it if you need to have a list (i.e. a iterable is not sufficient). – dom0 Oct 24 '13 at 08:13
40

Using zip

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

output

['asp10', 'asp11', 'asp15', 'asp16', 'asp210', 'asp211']
RMcG
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9

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']
vaab
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5

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]
Edgar Klerks
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5

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']
Veneet Reddy
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    I think, this answer should get more credit. If one has variable number of lists. Let's say in data preparation you have positives and negatives (user can define the number of negatives), then the above solution will solve this issue with variable number of negatives list. – user1415536 Feb 21 '20 at 11:29
3

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']
IndPythCoder
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2

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]))
Vorsprung
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1
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
MONTYHS
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0

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
Contango
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0

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']
Synthase
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0

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']
vpz
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