将列表的元素重复 n 次

Question

How do I repeat each element of a list n times and form a new list? For example:

x = [1,2,3,4]
n = 3

x1 = [1,1,1,2,2,2,3,3,3,4,4,4]

x * n doesn't work

for i in x[i]:
    x1 = n * x[i]

There must be a simple and smart way.

Answer 1

The ideal way is probably numpy.repeat :

In [16]:

x1=[1,2,3,4]
In [17]:

np.repeat(x1,3)
Out[17]:
array([1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4])

Answer 2

In case you really want result as list, and generator is not sufficient:

import itertools
lst = range(1,5)
list(itertools.chain.from_iterable(itertools.repeat(x, 3) for x in lst))

Out[8]: [1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]

Answer 3

You can use list comprehension:

[item for item in x for i in range(n)]

---

>>> x = [1, 2, 3, 4]
>>> n = 3
>>> new = [item for item in x for i in range(n)]
#[1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]

Answer 4

A simpler way to achieve this to multiply the list x with n and sort the resulting list. eg

>>> x = [1,2,3,4]
>>> n = 3
>>> a = sorted(x*n)
>>> a
>>> [1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]

Answer 5

A nested list-comp works here:

>>> [i for i in range(10) for _ in xrange(3)]
[0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9]

Or to use your example:

>>> x = [1, 2, 3, 4]
>>> n = 3
>>> [i for i in x for _ in xrange(n)]
[1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]

Answer 6

 [myList[i//n] for i in range(n*len(myList))]

Answer 7

This will solve your issue:

x=[1,2,3,4]
n = 3
x = sorted(x * n)

Answer 8

import itertools

def expand(lst, n):
    lst = [[i]*n for i in lst]
    lst = list(itertools.chain.from_iterable(lst))
    return lst

x=[1,2,3,4]
n=3
x1 = expand(x,3)

print(x1)

Gives:

[1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]

Explanation:

Doing, [3]*3 gives the result of [3,3,3] , replacing this with n we get [3,3,3,...3] (n times) Using a list comprehension we can go through each elem of the list and perform this operation, finally we need to flatten the list, which we can do by list(itertools.chain.from_iterable(lst))

Answer 9

zAxe=[]
for i in range(5):
    zAxe0 =[i] * 3
    zAxe +=(zAxe0) # append allows accimulation of data 

Answer 10

way 1:

def foo():
    for j in [1, 3, 2]:
        yield from [j]*5

way 2:

from itertools import chain
l= [3, 1, 2]
chain(*zip(*[l]*3))

way 3:

sum(([i]*5 for i in [2, 1, 3]), [])

Answer 11

If you want to modify the list in-place, the best way is to iterate from the back and assign a slice of what was previously one item to a list of that item n times.

This works because of slice assignment:

>>> ls = [1, 2, 3]
>>> ls[0: 0+1]
[1]
>>> ls[0: 0+1] = [4, 5, 6]
>>> ls
>>> [4, 5, 6, 2, 3]

def repeat_elements(ls, times):
    for i in range(len(ls) - 1, -1, -1):
        ls[i: i+1] = [ls[i]] * times

Demo usage:

>>> a = [1, 2, 3]
>>> b = a
>>> b
[1, 2, 3]
>>> repeat_elements(b, 3)
>>> b
[1, 1, 1, 2, 2, 2, 3, 3, 3]
>>> a
[1, 1, 1, 2, 2, 2, 3, 3, 3]

(If you don't want to modify it in-place, you can copy the list and return the copy, which won't modify the original. This would also work for other sequences, like tuple s, but is not lazy like the itertools.chain.from_iterable and itertools.repeat method)

def repeat_elements(ls, times):
    ls = list(ls)  # Makes a copy
    for i in range(len(ls) - 1, -1, -1):
        ls[i: i+1] = [ls[i]] * times
    return ls

Answer 12

x=[1,2,3,4]
def f11(x,n):  
    l=[]
    for item in x:
        for i in range(n):
            l.append(item)
            
    return l

f11(x,2)

Answer 13

If working with array is okay,

np.array([[e]*n for e in x]).reshape(-1)

In my opinion it is very readable.

Answer 14

For base Python 2.7:

    from itertools import repeat
    def expandGrid(**kwargs):
        # Input is a series of lists as named arguments
        # output is a dictionary defining each combination, preserving names
        #
        # lengths of each input list
        listLens = [len(e) for e in kwargs.itervalues()] 
        # multiply all list lengths together to get total number of combinations
        nCombos = reduce((lambda x, y: x * y), listLens) 
        iDict = {}
        nTimesRepEachValue=1 #initialize as repeating only once
        for key in kwargs.keys():
            nTimesRepList=nCombos/(len(kwargs[key])*nTimesRepEachValue)
            tempVals=[] #temporary list to store repeated
            for v in range(nTimesRepList):
                indicesToAdd=reduce((lambda x,y: list(x)+list(y)),[repeat(x, nTimesRepEachValue) for x in kwargs[key]])
                tempVals=tempVals+indicesToAdd
            iDict[key] = tempVals
            # Accumulating the number of times needed to repeat each value
            nTimesRepEachValue=len(kwargs[key])*nTimesRepEachValue
        return iDict

    #Example usage:
    expandedDict=expandGrid(letters=["a","b","c","d"],nums=[1,2,3],both=["v",3])