CODEWITHSUNDEEP
pythonnested-listssublist
john johns
john johns

Reputation: 167

python: locate elements in sublists

given these sublists

lst=[['a', 'b', 'c', 'd', 'e'], ['f', 'g', 'h']]

I am trying to find the location of its elements, for instance, the letter 'a' is located at 0,0 but this line

print(lst.index('a'))

instead produces the following error: ValueError: 'a' is not in list

Upvotes: 5

Views: 551

Answers (6)

lpgraph
lpgraph

Reputation: 31

You can use list comprehension:

>>> lst=[['a', 'b', 'a', 'd', 'a'], ['f', 'g', 'a'], ['a','a','b']]
>>> [(i,j) for i in range(len(lst)) for j in range(len(lst[i])) if lst[i][j]=='a']
[(0, 0), (0, 2), (0, 4), (1, 2), (2, 0), (2, 1)]

Upvotes: 0

I'mahdi
I'mahdi

Reputation: 24049

if your list have depth=2, you can use this:

lst=[['a', 'b', 'c', 'd', 'e'], ['f', 'g', 'h']]
def fnd_idx(char, lst):
    for x in range(len(lst)):
        try:
            idx = lst[x].index(char)
            return [x,idx]
        except ValueError:
            pass
    return None

Output:

>>> print(fnd_idx('a', lst))
[0, 0]

>>> print(fnd_idx('g', lst))
[1, 1]

>>> print(fnd_idx('z', lst))
None

Upvotes: 2

Metapod
Metapod

Reputation: 416

If 'a' can appear in multiple sublists, and you want the index in each sublist:

def GetIndexes(lst, val):
    pos = []
    for sublist in lst:
        try:
            idx = sublist.index(val)
            pos.append(idx)
        except:
            pos.append(None)
    return pos

In your example : [0, None] Meaning: In sublist 0, the first 'a' is at the index 0. In sublist 1, there is no 'a'.

Upvotes: 0

daydraze
daydraze

Reputation: 36

lst=[['a', 'b', 'c', 'd', 'c'], ['f', 'g', 'h']]
searchvalue = 'f'
counter = 0
for index in lst:
    if searchvalue in index:
       print(counter, index.index(searchvalue))
    counter+=1

Upvotes: 0

Justin.Arnold
Justin.Arnold

Reputation: 144

you can do this with numpy, the benefit is you don't have to hardcode anything for the size of nested lists. you can have hundreds or 3 and this will work!

lst=[['a', 'b', 'c', 'd', 'e'], ['f', 'g', 'h']]
arr = np.array(lst, dtype=object)

for x in arr:
    try:
        print (x.index('a'), x)
    except:
        pass

Upvotes: -1

Riccardo Bucco
Riccardo Bucco

Reputation: 15364

Try this function (just one line of code!):

def idx(lst, el):
    return next(((i, sublst.index(el))
                 for i, sublst in enumerate(lst)
                 if el in sublst),
                None)

So for example:

>>> idx(lst, 'a')
(0, 0)
>>> idx(lst, 'c')
(0, 2)

Upvotes: 0

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