CODEWITHSUNDEEP
pythonlist
Ekaterina
Ekaterina

Reputation: 335

Create duplicates in the list

I have

 list = [a, b, c, d]

and

numbers = [2, 4, 3, 1]

I want to get a list of the type of:

new_list = [a, a, b, b, b, b, c, c, c, d]

This is what I have so far:

new_list=[] 
for i in numbers: 
    for x in list: 
        for i in range(1,i+1): 
            new_list.append(x)

Upvotes: 7

Views: 8913

Answers (9)

Kasravnd
Kasravnd

Reputation: 107347

As a general approach for any object (not only string) you can use itertools.repeat() within a generator expression:

def repeat_it(lst, numbers):
    return chain.from_iterable(repeat(i, j) for i, j in zip(lst, numbers))

Demo:

In [13]: from itertools import repeat, chain

In [21]: lst=[5,4,6,0]

In [22]: list(repeat_it(lst, numbers))
Out[22]: [5, 5, 4, 4, 4, 4, 6, 6, 6, 0]

In [23]: lst=['a','b','c','d']

In [24]: list(repeat_it(lst, numbers))
Out[24]: ['a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd']

Here is a benchmark on 3 main approaches. Note that the last one onley works for strings:

In [49]: lst = lst * 1000

In [50]: numbers = numbers * 1000

In [51]: %timeit list(chain.from_iterable(repeat(i, j) for i, j in zip(lst, numbers)))
1 loops, best of 3: 8.8 s per loop

In [52]: %timeit [x for x, number in zip(lst, numbers) for _ in range(number)]
1 loops, best of 3: 12.4 s per loop

In [53]: %timeit [x for i, j in zip(lst, numbers) for x in i*j]
1 loops, best of 3: 7.2 s per loop

Upvotes: 5

alec_djinn
alec_djinn

Reputation: 10819

That's my solution, just to add a different one.

l = ['a', 'b', 'c', 'd']
n = [2, 4, 3, 1]
r = []
for i,v in enumerate(l):
    r += list(v*n[i])

>>> r
    ['a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd']

Upvotes: 1

Spherical Cowboy
Spherical Cowboy

Reputation: 566

This will work regardless of whether a, b, c and d are variables or strings:

a = 1
b = 2.0
c = "cheese"
d = ["c", "a", "k", "e"]

lst = [a, b, c, d]
numbers = [2, 4, 3, 1]

# if len(lst) == len(numbers):
new_lst = [i for i, j in zip(lst, numbers) for k in range(j)]

You might want to uncomment the if statement (and indent the line underneath) to check if the lists have the same length, otherwise new_lst will only contain as many items as the shorter list.

This, this and the documentation section on nested list comprehensions are worth reading.

Upvotes: 1

Mike Müller
Mike Müller

Reputation: 85572

A nested list comprehension works:

L = ['a','b','c','d']
numbers = [2, 4, 3, 1]

>>> [x for x, number in zip(L, numbers) for _ in range(number)]
['a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd']

The "sub-loop" for _ in range(number) repeats the value number times. Here L can hold any object, not only strings.

Example:

L = [[1, 2, 3],'b','c', 'd']
numbers = [2, 4, 3, 1]
[x for x, number in zip(L, numbers) for _ in range(number)]
[[1, 2, 3], [1, 2, 3], 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd']

but this flattens the sub list:

[x for i, j in zip(L, numbers) for x in i*j]
[1, 2, 3, 1, 2, 3, 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd']

not exactly the desired result.

Upvotes: 5

hfz
hfz

Reputation: 421

If you are unsure of how list comprehensions work, 

myList=['a','b','c','d'] # not a good idea to use list as a name for your variable
numbers=[2,4,3,1]
new_list=[]
for i in range(len(myList)):     
    for j in range(numbers[i]):          
        new_list.append(myList[i])

print(new_list)

Upvotes: 1

Moses Koledoye
Moses Koledoye

Reputation: 78554

Here's one way to do it using zip, string multiplication and a list comprehension:

lst = ['a', 'b', 'c', 'd'] 
numbers = [2 , 4, 3, 1]

r = [x for i, j in zip(lst, numbers) for x in i*j]
print(r)
# ['a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd']

Pay attention to the choice of names when using Python. A name like list renders the builtin list function unusable.

If the items in lst are not strings you can simply use a nested comprehension on range, to duplicate the items in the list.

Upvotes: 12

Nick is tired
Nick is tired

Reputation: 7056

Assuming that both lists are the same length and the second is always a list of numbers, here is a solution without using zip or any imports:

lst = ['a', 'b', 'c', 'd']
numbers = [2,4,3,1]

result = sum([[lst[i]]*numbers[i] for i in range(len(lst))],[])

Upvotes: 0

Christopher Shroba
Christopher Shroba

Reputation: 7584

Another way to do it with a loop would be:

new_list = []
for number, item in zip(numbers, l):
    for i in range(number):
        new_list.append(item)

Now we have:

new_list = ['a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd']

Upvotes: 1

akuiper
akuiper

Reputation: 215117

You can use numpy.repeat() as another option:

import numpy as np
np.repeat(lst, numbers).tolist()

# ['a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd']

Upvotes: 3

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