How can I reorder a list? [closed]
开发者_如何学Go
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 2 years ago.
Improve this questionGiven an arbitrary array of size n
, I'd like to reorganize the elements of the array based on the array's discrete indices.
Python example:
# Unique array of size n
[ "a", "b", "c", "d", "e", ... <n> ]
# Indices of array
[ 0, 1, 2, 3, 4, ... <index_of_n> ]
# Desired re-organization function 'indexMove'
indexMove(
[ "a", "b", "c", "d", "e", ... <n> ],
[ <index_of_n>, 4, 0, 2, 3, ... 1 ]
)
# Desired output from indexMove operation
[ <n>, "e", "a", "c", "d", ... "b" ]
What is the fastest way to perform this operation (achieving the smallest time complexity)?
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']
>>> a = [1, 2, 3]
>>> a[0], a[2] = a[2], a[0]
>>> a
[3, 2, 1]
>>> import random
>>> x = [1,2,3,4,5]
>>> random.shuffle(x)
>>> x
[5, 2, 4, 3, 1]
Is the final order defined by a list of indices ?
>>> items = [1, None, "chicken", int]
>>> order = [3, 0, 1, 2]
>>> ordered_list = [items[i] for i in order]
>>> ordered_list
[<type 'int'>, 1, None, 'chicken']
edit: meh. AJ was faster... How can I reorder a list in python?
>>> a=["a","b","c","d","e"]
>>> a[0],a[3] = a[3],a[0]
>>> a
['d', 'b', 'c', 'a', 'e']
You can provide your own sort function to list.sort()
:
The sort() method takes optional arguments for controlling the comparisons.
cmp specifies a custom comparison function of two arguments (list items) which should return a negative, zero or positive number depending on whether the first argument is considered smaller than, equal to, or larger than the second argument:
cmp=lambda x,y: cmp(x.lower(), y.lower())
. The default value isNone
.key specifies a function of one argument that is used to extract a comparison key from each list element:
key=str.lower
. The default value isNone
.reverse is a boolean value. If set to True, then the list elements are sorted as if each comparison were reversed.
In general, the key and reverse conversion processes are much faster than specifying an equivalent cmp function. This is because cmp is called multiple times for each list element while key and reverse touch each element only once.
If you do not care so much about efficiency, you could rely on numpy's array indexing to make it elegant:
a = ['123', 'abc', 456]
order = [2, 0, 1]
a2 = list( np.array(a, dtype=object)[order] )
One more thing which can be considered is the other interpretation as pointed out by darkless
Code in Python 2.7
Mainly:
- Reorder by value - Already solved by AJ above
Reorder by index
mylist = ['a', 'b', 'c', 'd', 'e'] myorder = [3, 2, 0, 1, 4] mylist = sorted(zip(mylist, myorder), key=lambda x: x[1]) print [item[0] for item in mylist]
This will print ['c', 'd', 'b', 'a', 'e']
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']
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".
newList = [oldList[3]]
newList.extend(oldList[:3])
newList.extend(oldList[4:])
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
精彩评论