I have a list of strings like this:
X = [\"a\", \"b\", \"c\", \"d\", \"e\", \"f\", \"g\", \"h\", \"i\"]
Y = [ 0, 1, 1, 0, 1, 2, 2, 0, 1 ]
You can create a pandas Series
, using the primary list as data
and the other list as index
, and then just sort by the index:
import pandas as pd
pd.Series(data=X,index=Y).sort_index().tolist()
output:
['a', 'd', 'h', 'b', 'c', 'e', 'i', 'f', 'g']
zip, sort by the second column, return the first column.
zip(*sorted(zip(X,Y), key=operator.itemgetter(1)))[0]
I actually came here looking to sort a list by a list where the values matched.
list_a = ['foo', 'bar', 'baz']
list_b = ['baz', 'bar', 'foo']
sorted(list_b, key=lambda x: list_a.index(x))
# ['foo', 'bar', 'baz']
A quick one-liner.
list_a = [5,4,3,2,1]
list_b = [1,1.5,1.75,2,3,3.5,3.75,4,5]
Say you want list a to match list b.
orderedList = sorted(list_a, key=lambda x: list_b.index(x))
This is helpful when needing to order a smaller list to values in larger. Assuming that the larger list contains all values in the smaller list, it can be done.
Also, if you don't mind using numpy arrays (or in fact already are dealing with numpy arrays...), here is another nice solution:
people = ['Jim', 'Pam', 'Micheal', 'Dwight']
ages = [27, 25, 4, 9]
import numpy
people = numpy.array(people)
ages = numpy.array(ages)
inds = ages.argsort()
sortedPeople = people[inds]
I found it here: http://scienceoss.com/sort-one-list-by-another-list/
I like having a list of sorted indices. That way, I can sort any list in the same order as the source list. Once you have a list of sorted indices, a simple list comprehension will do the trick:
X = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]
Y = [ 0, 1, 1, 0, 1, 2, 2, 0, 1]
sorted_y_idx_list = sorted(range(len(Y)),key=lambda x:Y[x])
Xs = [X[i] for i in sorted_y_idx_list ]
print( "Xs:", Xs )
# prints: Xs: ["a", "d", "h", "b", "c", "e", "i", "f", "g"]
Note that the sorted index list can also be gotten using numpy.argsort()
.