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Python – Itertools.chain()

  • Difficulty Level : Basic
  • Last Updated : 25 Aug, 2021

The itertools is a module in Python having a collection of functions that are used for handling iterators. They make iterating through the iterables like lists and strings very easily. One such itertools function is chain().
Note: For more information, refer to Python Itertools

chain() function

It is a function that takes a series of iterables and returns one iterable. It groups all the iterables together and produces a single iterable as output. Its output cannot be used directly and thus explicitly converted into iterables. This function come under the category iterators terminating iterators.
Syntax :  

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chain (*iterables)

The internal working of chain can be implemented as given below :  

def chain(*iterables):
     for it in iterables:
       for each in it:
           yield each

Example 1: The odd numbers and even numbers are in separate lists. Combine them to form a new single list.


from itertools import chain
# a list of odd numbers
odd = [1, 3, 5, 7, 9]
# a list of even numbers
even = [2, 4, 6, 8, 10]
# chaining odd and even numbers
numbers = list(chain(odd, even))
[1, 3, 5, 7, 9, 2, 4, 6, 8, 10]


Example 2: Some of the consonants are in a list. The vowels are given in a list. Combine them and also sort them.


from itertools import chain
# some consonants
consonants = ['d', 'f', 'k', 'l', 'n', 'p']
# some vowels
vowels = ['a', 'e', 'i', 'o', 'u']
# resultatnt list
res = list(chain(consonants, vowels))
# sorting the list
['a', 'd', 'e', 'f', 'i', 'k', 'l', 'n', 'o', 'p', 'u']


Example 3: In the example below, Each String is considered to be an iterable and each character in it is considered to be an element in the iterator. Thus every character is yielded


from itertools import chain
res = list(chain('ABC', 'DEF', 'GHI', 'JKL'))
['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L']


Example 4: 


from itertools import chain
st1 = "Geeks"
st2 = "for"
st3 = "Geeks"
res = list(chain(st1, st2, st3))
print("before joining :", res)
ans = ''.join(res)
print("After joining :", ans)
before joining : [‘G’, ‘e’, ‘e’, ‘k’, ‘s’, ‘f’, ‘o’, ‘r’, ‘G’, ‘e’, ‘e’, ‘k’, ‘s’] 
After joining : GeeksforGeeks

chain.from_iterable() function
It is similar to chain, but it can be used to chain items from a single iterable. The difference is demonstrated in the example given below:
Example 5:


from itertools import chain
li = ['ABC', 'DEF', 'GHI', 'JKL']
# using chain-single iterable.
res1 = list(chain(li))
res2 = list(chain.from_iterable(li))
print("using chain :", res1, end ="\n\n")
print("using chain.from_iterable :", res2)
using chain : [‘ABC’, ‘DEF’, ‘GHI’, ‘JKL’]
using chain.from_iterable : [‘A’, ‘B’, ‘C’, ‘D’, ‘E’, ‘F’, ‘G’, ‘H’, ‘I’, ‘J’, ‘K’, ‘L’]

Example 6: Now consider a list like the one given below : 

li=['123', '456', '789']

You are supposed to calculate the sum of the list taking every single digit into account. So the answer should be : 

1+2+3+5+6+7+8+9 = 45

This can be achieved easily using the code below :


from itertools import chain
li = ['123', '456', '789']
res = list(chain.from_iterable(li))
print("res =", res, end ="\n\n")
new_res = list(map(int, res))
print("new_res =", new_res)
sum_of_li = sum(new_res)
print("\nsum =", sum_of_li)
res = ['1', '2', '3', '4', '5', '6', '7', '8', '9']

new_res = [1, 2, 3, 4, 5, 6, 7, 8, 9]

sum = 45


To simplify it, we combine the steps. An optimized approach is given below:


from itertools import chain
li = ['123', '456', '789']
res = list(map(int, list(chain.from_iterable(li))))
sum_of_li = sum(res)
print("res =", res, end ="\n\n")
print("sum =", sum_of_li)
res = [1, 2, 3, 4, 5, 6, 7, 8, 9]

sum = 45


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