how can i sort a zip list with list index from another list? At the moment i use two loops and its not very efficient.
L1 = ['eins','zwei','drei','vier']
L2 = ['zwei','eins','vier','drei']
L3 = ['apfel','birne','banane','kirsche']
zipped = zip(L2,L3)
L4 = []
for i in L1:
for e,g in zipped:
if e == i:
L4.append(g)
print L4
To sort zipped lists by index of another list you can use the function sorted():
l1 = ['eins', 'zwei', 'drei', 'vier']
l2 = ['zwei', 'eins', 'vier', 'drei']
l3 = ['apfel', 'birne', 'banane', 'kirsche']
l = sorted(zip(l2, l3), key=lambda x: l1.index(x[0]))
# [('eins', 'birne'), ('zwei', 'apfel'), ('drei', 'kirsche'), ('vier', 'banane')]
[i for _, i in l]
# ['birne', 'apfel', 'kirsche', 'banane']
Use a python dictionary:
d = dict(zip(L2,L3))
L4 = [d[key] for key in L1]
Following your original logic, I guess, you can change a bit to make it work:
L4 = []
for e in L1:
i2 = L2.index(e) # looks for the index (i2) of the element e of L1 in L2
L4.append(L3[i2]) # append lo L4 the element at index i2 in L3
print(L4)
#=> ['birne', 'apfel', 'kirsche', 'banane']
Which can be written as a one liner:
[ L3[L2.index(e)] for e in L1 ]
I like @syltruong answer, but enumerate is one more option:
for item1 in L1:
for index, item2 in enumerate(L2):
if item1 == item2:
L4.append(L3[index])
