python nested for loop when index of outer loop equals to length of inner loop you start again-CodePudding

I have

A = ['A','B','C','D','E','F','G','H']
B= ['a','b']

want to iterate through the list to get

C = ['Aa','Bb','Ca','Db','Ea','Fb','Ga','Hb']

how can I go about it in python

CodePudding user response：

You can use a list comprehension :

a = ['A','B','C','D','E','F','G','H']
b = ['a','b']

c = [a[i]   b[i%2] for i in range(len(a))]

# ['Aa', 'Bb', 'Ca', 'Db', 'Ea', 'Fb', 'Ga', 'Hb']

CodePudding user response：

You can use zip() and cycle():

from itertools import cycle

A = ['A','B','C','D','E','F','G','H']
B = ['a', 'b']

C = [a   b for a,b in zip(A, cycle(B))]
print(C)

Output as requested.

CodePudding user response：

You can do this by using the index in the A list mapped on the number of elements in the B list with the modulo operation (%) :

A = [1,2,3,4,5,6,7,8]
B= ['a','b']
C = []

for i in range(len(A)):
    C.append(f"{A[i]}{B[i%2]}")

CodePudding user response：

More variations:

from itertools import cycle
from operator import add

A = ['A','B','C','D','E','F','G','H']
B = ['a','b']
E = ['Aa','Bb','Ca','Db','Ea','Fb','Ga','Hb']

C = list(map(add, A, cycle(B)))
print(C == E)

C = [a   B[i%2] for i, a in enumerate(A)]
print(C == E)

C = list(map(''.join, zip(A, cycle(B))))
print(C == E)

C = list(map('{}{}'.format, A, cycle(B)))
print(C == E)

CodePudding user response：

You can also try this:

A = ['A','B','C','D','E','F','G','H']
B= ['a','b']
LB = len(B)
x = 0

for i in A :
    if x>=LB :
        x=0
    print(i B[x])
    x =1

CodePudding user response：

A = [1, 2, 3, 4, 5, 6, 7, 8]
B = ["a", "b", "c", "d", "f", "g", "h", "q"]
C = []
i = 0
while i < len(A):
    for j in B:
        C.append("{}{}".format(A[i], j))   
    i = i   1
        

print(C)

You can try this without any module.

You can specify your arrays, works anyway. I just tested like this.