复制列表中的元素x次

Question

I want to duplicate each element individually right after each other while preserving the order. For example the function would look like

def duplicate(testList, n)
    for i in l:
        ll.append(i)
        ll.append(i)

testList being the input list and n being the number of times you duplicate each element within. With what I came up with I can't loop it for a certain number of times. Not sure how to do it.

Answer 1

how bout

def n_times(v,n):
    for i in range(n):
        yield v

[i for j in testList for i in n_times(j,2)]

Answer 2

You can simply add second loop which will iterate n times adding your values. However I advice you to learn generators as well and do it on one line like Joran Beasley suggested.

def duplicate(testList, n)
    for i in l:
        for x in range(n):
            ll.append(i)

Answer 3

If your are looking for a result like Ashwini Chaudhary described , you can go like this:

from itertools import chain  # Hinted by Ashwini Chaudhary

def repeat(lst, n):
    return list(chain(*zip(*[x for _ in range(n)])))

>>> repeat([1, 2, 3], 3)
[1, 1, 1, 2, 2, 2, 3, 3, 3]

Otherwise, if you are looking for just repeating the list n times, as a general solution you could go with the following function:

def repeat(lst, n):
    result = []
    for _ in range(n):
        result += lst[:]
    return result

>>> repeat([1, 2, 3], 2)
[1, 2, 3, 1, 2, 3]

However, if the elements are guaranteed to be immutable objects , there is a pretty nice way for a list to do that in Python:

>>> [1, 2, 3] * 2
[1, 2, 3, 1, 2, 3]

Answer 4

As a one-liner:

def n_plicate(testList, n):
    return reduce(lambda x,y: x+y, map(lambda x: [x]*n, testList))

or a little bit easier to read and faster:

    return [tm for sblst in [[x]*n for x in testList] for tm in sblst]

Answer 5

Long and Easier, without any imports,

    def duplicate(listt,number):
        """Input a list and a num evry vlue in the list is replctd by Num of times """
        temp=[]
        for i in listt:
              temp.append(i*number)
       output=[]
       for j in temp:
              output.extend(list(j))
       return output