How can I reorder a list? [closed]

Answer 1

You can do it like this

mylist = ['a', 'b', 'c', 'd', 'e']
myorder = [3, 2, 0, 1, 4]
mylist = [mylist[i] for i in myorder]
print(mylist)         # prints: ['d', 'c', 'a', 'b', 'e']

Answer 2

This is what I used when I stumbled upon this problem.

def order(list_item, i): # reorder at index i
    order_at = list_item.index(i)
    ordered_list = list_item[order_at:] + list_item[:order_at]
    return ordered_list

EX: for the the lowercase letters

order(string.ascii_lowercase, 'h'):
>>> 'hijklmnopqrstuvwxyzabcdefg'

It simply just shifts the list to a specified index

Answer 3

If you use numpy there's a neat way to do it:

items = np.array(["a","b","c","d"])
indices = np.arange(items.shape[0])
np.random.shuffle(indices)
print(indices)
print(items[indices])

This code returns:

[1 3 2 0]
['b' 'd' 'c' 'a']

Answer 4

>>> a=["a","b","c","d","e"]
>>> a[0],a[3] = a[3],a[0]
>>> a
['d', 'b', 'c', 'a', 'e']

Answer 5

From what I understand of your question, it appears that you want to apply a permutation that you specify on a list. This is done by specifying another list (lets call it p) that holds the indices of the elements of the original list that should appear in the permuted list. You then use p to make a new list by simply substituting the element at each position by that whose index is in that position in p.

def apply_permutation(lst, p):
    return [lst[x] for x in p]

arr=list("abcde")
new_order=[3,2,0,1,4]

print apply_permutation(arr,new_order)

This prints ['d', 'c', 'a', 'b', 'e'].

This actually creates a new list, but it can be trivially modified to permute the original "in place".

Answer 6

>>> import random
>>> x = [1,2,3,4,5]
>>> random.shuffle(x)
>>> x
[5, 2, 4, 3, 1]