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.
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.
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])
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]
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]
A simpler way to achieve this to multiply the list x with n and sort the resulting list. e.g.
>>> x = [1,2,3,4]
>>> n = 3
>>> a = sorted(x*n)
>>> a
>>> [1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]
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]
zAxe=[]
for i in range(5):
zAxe0 =[i] * 3
zAxe +=(zAxe0) # append allows accimulation of data
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]), [])
This will solve your issue:
x=[1,2,3,4]
n = 3
x = sorted(x * n)
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))
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 tuples, 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
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])
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)
If working with array is okay,
np.array([[e]*n for e in x]).reshape(-1)
In my opinion it is very readable.