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I have a list:
L = ['a', 'b']
I need to create a new list by concatenating an original list which range goes from 1 to k. Example:
k = 4
L1 = ['a1','b1', 'a2','b2','a3','b3','a4','b4']
I try:
l1 = L * k
print l1
#['a', 'b', 'a', 'b', 'a', 'b', 'a', 'b']
l = [ [x] * 2 for x in range(1, k + 1) ]
print l
#[[1, 1], [2, 2], [3, 3], [4, 4]]
l2 = [item for sublist in l for item in sublist]
print l2
#[1, 1, 2, 2, 3, 3, 4, 4]
print zip(l1,l2)
#[('a', 1), ('b', 1), ('a', 2), ('b', 2), ('a', 3), ('b', 3), ('a', 4), ('b', 4)]
print [x+ str(y) for x,y in zip(l1,l2)]
#['a1', 'b1', 'a2', 'b2', 'a3', 'b3', 'a4', 'b4']
But I think it is very complicated. What is the fastest and most generic solution?
464
Python doesn't have a built-in function to merge the result of two range() output. However, we can still be able to do it. There is a module named 'itertools' which has a chain() function to combine two range objects.
📢 TLDR: Use + In almost all simple situations, using list1 + list2 is the way you want to concatenate lists. The edge cases below are better in some situations, but + is generally the best choice. All options covered work in Python 2.3, Python 2.7, and all versions of Python 31.
First, flatten the nested lists. Take Intersection using filter() and save it to 'lst3'. Now find elements either not in lst1 or in lst2, and save them to 'temp'. Finally, append 'temp' to 'lst3'.
You can use a list comprehension:
L = ['a', 'b']
k = 4
L1 = ['{}{}'.format(x, y) for y in range(1, k+1) for x in L]
print(L1)
Output
['a1', 'b1', 'a2', 'b2', 'a3', 'b3', 'a4', 'b4']
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