Python: Combine then sort list

Question

I have two lists; say one of fruits and another of equal length but of unordered numbers:

eg:

Fruits = ['apple', 'banana', 'pineapple', 'kiwifruit'],
Numbers = [3, 2, 4, 1]

How can I firstly assign the number

3 to apple,
2 to banana,
4 to pineapple and
1 to kiwifruit

and secondly order them according to their new numbers?

i.e

sortedlist = ['kiwifruit', 'banana', 'apple', 'pineapple'].

My attempts so far have included the enumerate function and the sorted function, but I can't seem to assign and then sort.

The question has been answered for you, but you really should show some of you own code and attempts you've made already - this sounds like homework. — Aesthete
– Aesthete, Commented Dec 3, 2012 at 5:37
I did use the sorted function, but my attempts were quite far off from a good solution. Also, this isn't a homework question. — user1653402
– user1653402, Commented Dec 3, 2012 at 5:47
@gnibbler on the off chance you aren't just joking, en.wikipedia.org/wiki/Kiwifruit — Karl Knechtel
– Karl Knechtel, Commented Dec 3, 2012 at 7:22
@KarlKnechtel, indeed :) It was originally "kiwi" there. As an expat Kiwi (person) it drives me nuts. — John La Rooy
– John La Rooy, Commented Dec 3, 2012 at 8:59

Jesse the Game · Accepted Answer · 2012-12-03 05:35:37Z

10

sortedlist = [x[1] for x in sorted(zip(Numbers, Fruits))]

answered Dec 3, 2012 at 5:35

Jesse the Game

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Comments

avasal · Accepted Answer · 2012-12-03 05:36:55Z

0

assignment can be done like this

In [25]: d = dict(zip(Numbers, Fruits))

In [26]: d
Out[26]: {1: 'kiwi', 2: 'banana', 3: 'apple', 4: 'pineapple'}

and then you can sort based on keys of the dictionary

In [27]: [d[i] for i in sorted(d.keys())]
Out[27]: ['kiwi', 'banana', 'apple', 'pineapple']

answered Dec 3, 2012 at 5:36

avasal

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Joran Beasley · Accepted Answer · 2012-12-03 05:38:04Z

0

numpy is cool for this

import numpy as np
fruits = np.array(['apple', 'banana', 'pineapple', 'kiwi'])
sorted_list = fruits[[2, 1, 3, 0]] #note its 0 based indexing ...
print sorted_list

answered Dec 3, 2012 at 5:38

Joran Beasley

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mgilson · Accepted Answer · 2012-12-03 05:56:40Z

0

How about:

Fruits = ['apple', 'banana', 'pineapple', 'kiwi']
Numbers = [3, 2, 4, 1]

New_Fruits = [ Fruits[idx-1] for idx in Numbers ]

This only works since your index list (Numbers) is sequential integers. It would even be a little more clean if Numbers = [ 2, 1, 3, 0 ].

The advantage is that it avoids a call to zip, so it might be a little more efficient. The downside is that it is less general. You need to have an appropriate index list for it to work.

edited Dec 3, 2012 at 5:56

answered Dec 3, 2012 at 5:42

mgilson

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user3651854 · Accepted Answer · 2015-03-23 16:35:38Z

0

Fruits = ['apple', 'banana', 'pineapple', 'kiwifruit'],
Numbers = [3, 2, 4, 1]
new = []
for x in range(len(Numbers)):
    new.append([Fruits[x],Numbers[x]])
new.sort(key=lambda x: x[1])
final = []
for x in new:
     final.append(x[0])

Final is the final list

answered Mar 23, 2015 at 16:35

user3651854

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