python - Access multiple elements of list knowing their index

Question

I need to choose some elements from the given list, knowing their index. Let say I would like to create a new list, which contains element with index 1, 2, 5, from given list [-2, 1, 5, 3, 8, 5, 6]. What I did is:

a = [-2,1,5,3,8,5,6]
b = [1,2,5]
c = [ a[i] for i in b]

Is there any better way to do it? something like c = a[b] ?

Answer 1

You can use operator.itemgetter:

from operator import itemgetter 
a = [-2, 1, 5, 3, 8, 5, 6]
b = [1, 2, 5]
print(itemgetter(*b)(a))
# Result:
(1, 5, 5)

Or you can use numpy:

import numpy as np
a = np.array([-2, 1, 5, 3, 8, 5, 6])
b = [1, 2, 5]
print(list(a[b]))
# Result:
[1, 5, 5]

---

But really, your current solution is fine. It's probably the neatest out of all of them.

Answer 2

Alternatives:

>>> map(a.__getitem__, b)
[1, 5, 5]

---

>>> import operator
>>> operator.itemgetter(*b)(a)
(1, 5, 5)

Answer 3

Another solution could be via pandas Series:

import pandas as pd

a = pd.Series([-2, 1, 5, 3, 8, 5, 6])
b = [1, 2, 5]
c = a[b]

You can then convert c back to a list if you want:

c = list(c)

Answer 4

Basic and not very extensive testing comparing the execution time of the five supplied answers:

def numpyIndexValues(a, b):
    na = np.array(a)
    nb = np.array(b)
    out = list(na[nb])
    return out

def mapIndexValues(a, b):
    out = map(a.__getitem__, b)
    return list(out)

def getIndexValues(a, b):
    out = operator.itemgetter(*b)(a)
    return out

def pythonLoopOverlap(a, b):
    c = [ a[i] for i in b]
    return c

multipleListItemValues = lambda searchList, ind: [searchList[i] for i in ind]

using the following input:

a = range(0, 10000000)
b = range(500, 500000)

simple python loop was the quickest with lambda operation a close second, mapIndexValues and getIndexValues were consistently pretty similar with numpy method significantly slower after converting lists to numpy arrays.If data is already in numpy arrays the numpyIndexValues method with the numpy.array conversion removed is quickest.

numpyIndexValues -> time:1.38940598 (when converted the lists to numpy arrays)
numpyIndexValues -> time:0.0193445 (using numpy array instead of python list as input, and conversion code removed)
mapIndexValues -> time:0.06477512099999999
getIndexValues -> time:0.06391049500000001
multipleListItemValues -> time:0.043773591
pythonLoopOverlap -> time:0.043021754999999995

Answer 5

I'm sure this has already been considered: If the amount of indices in b is small and constant, one could just write the result like:

c = [a[b[0]]] + [a[b[1]]] + [a[b[2]]]

Or even simpler if the indices itself are constants...

c = [a[1]] + [a[2]] + [a[5]]

Or if there is a consecutive range of indices...

c = a[1:3] + [a[5]]

Answer 6

Here's a simpler way:

a = [-2,1,5,3,8,5,6]
b = [1,2,5]
c = [e for i, e in enumerate(a) if i in b]

Answer 7

My answer does not use numpy or python collections.

One trivial way to find elements would be as follows:

a = [-2, 1, 5, 3, 8, 5, 6]
b = [1, 2, 5]
c = [i for i in a if i in b]

Drawback: This method may not work for larger lists. Using numpy is recommended for larger lists.

Answer 8

Kind of pythonic way:

c = [x for x in a if a.index(x) in b]

Answer 9

Static indexes and small list?

Don't forget that if the list is small and the indexes don't change, as in your example, sometimes the best thing is to use sequence unpacking:

_,a1,a2,_,_,a3,_ = a

The performance is much better and you can also save one line of code:

 %timeit _,a1,b1,_,_,c1,_ = a
10000000 loops, best of 3: 154 ns per loop 
%timeit itemgetter(*b)(a)
1000000 loops, best of 3: 753 ns per loop
 %timeit [ a[i] for i in b]
1000000 loops, best of 3: 777 ns per loop
 %timeit map(a.__getitem__, b)
1000000 loops, best of 3: 1.42 µs per loop