I have two list x and y. The former is always longer than or equal to the latter.
x = ["foo","bar","qux","zox"]
y1 = ["zox", "qux","foo"]
What I want to do is to sort y1 based on the order of x.
Yielding
["foo","qux","zox"]
I tried this, but doesn't work:
In [10]: [x[1] for x in sorted(zip(x,y))]
Out[10]: ['zox', 'qux']
What's the right way to do it?
You can use the key argument of sorted to use the index from the first list.
>>> sorted(y1, key=x.index)
['foo', 'qux', 'zox']
Just iterate through each element in x and add that element to a new list only if the corresponding element exists in y1.
>>> x = ["foo","bar","qux","zox"]
>>> y1 = ["zox", "qux","foo"]
>>> l = []
>>> for i in x:
if i in y1:
l.append(i)
>>> l
['foo', 'qux', 'zox']
OR
Use list_comprehension.
>>> [i for i in x if i in y1]
['foo', 'qux', 'zox']
Note that depending how large your list is .index requires a linear scan, and can be expensive. You may just create a one off dictionary of key->last index, and then use that as a reference to sort:
x = ["foo","bar","qux","zox"]
y1 = ["zox", "qux","foo"]
y1.sort(key=lambda L, d={k:i for i,k in enumerate(x)}: d[L])
# ['foo', 'qux', 'zox']
However, you could also do it the other way around, which could well be more efficient... Convert y1 to set then only keep elements also in x, eg:
filter(set(y1).__contains__, x)
You can also define a custom comparator for when sorting, like so:
x = ["foo","bar","qux","zox"]
y1 = ["zox", "qux","foo"]
def foo_sort(a,b):
return x.index(a)-x.index(b)
y1.sort(cmp=foo_sort)
print y1
Gives:
['foo', 'qux', 'zox']