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问题:
I've been tinkering in python this week and I got stuck on something.
If I had a 2D list like this:
myList = [[1,2],[3,4],[5,6]]
and I did this
>>>myList.index([3,4])
it would return
1
However, I want the index of something in side one of the lists, like this
>>>myList.index(3)
and it would return
1, 0
Is there anything that can do this?
Cheers
回答1:
Try this:
def index_2d(myList, v):
for i, x in enumerate(myList):
if v in x:
return (i, x.index(v))
Usage:
>>> index_2d(myList, 3)
(1, 0)
回答2:
If you are doing many lookups you could create a mapping.
>>> myList = [[1,2],[3,4],[5,6]]
>>> d = dict( (j,(x, y)) for x, i in enumerate(myList) for y, j in enumerate(i) )
>>> d
{1: (0, 0), 2: (0, 1), 3: (1, 0), 4: (1, 1), 5: (2, 0), 6: (2, 1)}
>>> d[3]
(1, 0)
回答3:
There is nothing that does this already, unless it's in numpy, which I don't know much about. This means you'll have to write code that does it. And that means questions like "What does [[1, 2], [2, 3], [3, 4]].index(3)
return?" are important.
回答4:
Using simple genexpr:
def index2d(list2d, value):
return next((i, j) for i, lst in enumerate(list2d)
for j, x in enumerate(lst) if x == value)
Example
print index2d([[1,2],[3,4],[5,6]], 3)
# -> (1, 0)
回答5:
def td(l,tgt):
rtr=[]
for sub in l:
if tgt in sub:
rtr.append( (l.index(sub),sub.index(tgt)) )
return rtr
myList = [[1,2],[3,4],[5,6]]
print td(myList,3)
This will return more than one instance of the sub list, if any.
回答6:
Try this! this worked for me :)
def ret_pos(mylist,val_to_find):
for i in (len(mylist)):
for j in (len(i)):
if mylist[i][j]== val_to_find:
postn=[i,j]
return(postn);
回答7:
Based on kevpie's answer I managed to get a 2D list containing the coordinates of all occurences
myList = [[0,1],[1,1],[0,0],[1,0]]
coordsList = [[x, y] for x, li in enumerate(myList) for y, val in enumerate(li) if val==1]
Now coordsList contains all indexes for value 1 in myList :
[[0, 1], [1, 0], [1, 1], [3, 0]]