find an element in a list with 'in' keyword and return that element

Question

using the 'in' keyword i can find if an element is in an array:

a=[1,2,3]
if 1 in a:
    print "1 present in a"

is there a way to return the index of where 'in' found that element?

That's not really a 3d array. That's a 1d array with 3 elements. — Falmarri
– Falmarri, Commented May 16, 2012 at 20:49
No I didn't mean a list with 3 elements. I was going to ask a question about 3d lists but I suppose I decided while writing the question that what I could learn about 1d lists could easily be applied to 3d lists. I fixed the title in an edit there, and changed the word array to the standard python terminology (list, not array). — fergusdawson
– fergusdawson, Commented May 17, 2012 at 0:45

Ashwini Chaudhary · Accepted Answer · 2012-05-16 21:07:09Z

4

1-d list:

  a=[1,2,3]
    def indd(val):
     try:   
       ind=a.index(val)
       return ind,val
     except:
       return 'not found'  

    print(indd(1))  
    print(indd(5))   

    (0,1)
    not found

for a 3-d list the function function will return a tuple of the x,y,z , if the number is repeated inside the 3d list then it'll return the lowest index :

def ind_3d(lis,val):
    ind=[(j,i,k) for j,x in enumerate(lis) for i,y in enumerate(x) for k,z in enumerate(y) if z==val]
    if len(ind)==0:
        return 'not found'
    else:
        return ind[0]

a=[[[1,2,3],[4,5,6],[7,8,9]],[[11,12,13],[14,15,16],[17,18,19]]]

print(ind_3d(a,11))
print(ind_3d(a,5))
print(ind_3d(a,25))

output:

(1, 0, 0)
(0, 1, 1)
not found

edited May 16, 2012 at 21:07

answered May 16, 2012 at 20:37

Ashwini Chaudhary

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Comments

Ansari · Accepted Answer · 2012-05-16 20:46:39Z

1

Use a.index(1) to get the position of 1 in the list if you know it exists for sure. Otherwise use a try/catch block.

Also, they're called "lists" in Python :) The more accurate you are with terminology the easier it is to keep track later of how they're implemented and what functions they support.

EDIT: For a multi-dimensional list (i.e. list of lists in case of 2-D), you'll have to iterate over the top n-1 levels, where n is the dimensionality. For 2-D, assuming you're searching for 1, something like this will work:

for idx, val in enumerate(a):
    if 1 in val:
        return idx, val.index(1)

edited May 16, 2012 at 20:46

answered May 16, 2012 at 20:36

Ansari

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1 Comment

fergusdawson Over a year ago

Yes thanks for pointing that out. I'll edit and take the 3d bit out :p

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