Getting index of an element in array containing another element

Question

So lets say I have this array :

array = [[1,2,3],[4,5,6],[7,8,9]]

and I want to get the index of the inner array containing the 5 for example. So in this case the returned index I want is 1.

I did try ind = array.index(5) but I'm quite aware why this didnt work since the value in the brackets have to match the element in the array exactly. Another way I did this is

counter = 0
for each in array:
  if 5 in each: break
  else: counter = counter + 1

and this worked well for what I want but I wanted to check if there is an easier and cleaner way to do this. Thanks

Readability is very subjective here: a lot of programmers would argue that itertools are more clear than mishmash of cycles with episodic breaks in them. Efficiency: rather no. Itertools — Alex Yu
– Alex Yu, Commented Apr 20, 2017 at 8:50

UltraInstinct · Accepted Answer · 2017-04-20 08:30:10Z

7

There is a slightly better pythonic approach using next(..). Remember that if 5 doesn't exist in any of the sub-arrays, this will throw a StopIteration. You might want to handle that.

>>> your_list = [[1,2,3],[4,5,6],[7,8,9]]
>>> next(i for i, x in enumerate(your_list) if 5 in x)
1

edited Apr 20, 2017 at 8:30

answered Apr 20, 2017 at 8:28

UltraInstinct

44.6k12 gold badges85 silver badges108 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

L.D Over a year ago

how would this return the index though?

UltraInstinct Over a year ago

@L.D Yes, that was my mistake. I just fixed it

Jenian · Accepted Answer · 2017-04-20 08:19:47Z

2

Better avoid breaks. Here is a better way:

for ind, inner_arr in enumerate(array):
    if 5 in inner_arr:
        return ind

answered Apr 20, 2017 at 8:19

Jenian

5921 gold badge6 silver badges20 bronze badges

Comments

marmeladze · Accepted Answer · 2017-04-20 08:28:57Z

2

one liner solution

#a = [[1,2,3],[4,5,6],[7,8,9]]
[i for i, j in enumerate(a)  if 5 in j][0]

answered Apr 20, 2017 at 8:28

marmeladze

6,5863 gold badges28 silver badges49 bronze badges

2 Comments

UltraInstinct Over a year ago

Using next(..) on a generator expression might give you an equivalent of a short-circuit as in my answer without having to construct an entire list

marmeladze Over a year ago

yeah, i've totally forgotten it -))

Carles Mitjans · Accepted Answer · 2017-04-20 08:33:42Z

1

If you still want to use index method, here is a possible approach:

[x.index(5) for x in array if 5 in x]

answered Apr 20, 2017 at 8:33

Carles Mitjans

4,8663 gold badges22 silver badges39 bronze badges

1 Comment

marmeladze Over a year ago

do not forget to get zeroth item -))

quamrana · Accepted Answer · 2017-04-20 08:40:57Z

1

What you have seems the way to go, but there are some improvements possible:

def find(item, array):
    counter = 0
    for each in array:
        if item in each:
            return counter
        counter = counter + 1
    return None

But even the counter can be automated:

Update: actually don't need return None - this is the default of all functions

def find(item, array):
    for idx, each in enumerate(array):
        if item in each:
            return idx

edited Apr 20, 2017 at 8:40

answered Apr 20, 2017 at 8:17

quamrana

39.5k13 gold badges57 silver badges77 bronze badges

Comments

dr_dronych · Accepted Answer · 2017-04-20 08:53:21Z

1

As an alternative to the use of enumerate(), I'd go with the following code:

for elem in array:
     for i in elem:
             if 5 in elem: print array.index(elem)
             break

answered Apr 20, 2017 at 8:53

dr_dronych

914 bronze badges

Comments

M Shepherd · Accepted Answer · 2017-04-20 08:19:58Z

0

array = [[1,2,3],[4,5,6],[7,8,9]]
counter = 0

for item in array:
    for n in item:
        if n == 5:
            counter += 1
    print counter

Probably more efficient ways to do it, but this works for me

answered Apr 20, 2017 at 8:19

M Shepherd

12 bronze badges

1 Comment

L.D Over a year ago

I think you placed counter += 1 in the wrong for loop. If the array was array=[[1,2,3],[7,8,9],[4,5,6]] .. the index your code gives is still 1

Jacopo Lanzoni · Accepted Answer · 2017-04-20 08:27:45Z

0

No need for a counter variable really. I would go for:

for inner_array in array:
  if inner_array.index(5) > -1:
    return array.index(inner_array);
return None;

edited Apr 20, 2017 at 8:27

answered Apr 20, 2017 at 8:16

Jacopo Lanzoni

1,3362 gold badges11 silver badges26 bronze badges

2 Comments

Hara Over a year ago

you may want to return None, instead of -1; as -1 is also a valid index. :)

L.D Over a year ago

what if 5 was in index 0 of the inner array?

Alex Yu · Accepted Answer · 2017-04-20 08:36:52Z

0

I see functional examples, so I'm a bit late:

[(id,inner.index(5)) for id,inner 
 in enumerate(arr) 
 if 5 in inner]

answered Apr 20, 2017 at 8:36

Alex Yu

3,5871 gold badge27 silver badges42 bronze badges

Collectives™ on Stack Overflow

Getting index of an element in array containing another element

9 Answers 9

2 Comments

Comments

2 Comments

1 Comment

Comments

Comments

1 Comment

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

9 Answers 9

2 Comments

Comments

2 Comments

1 Comment

Comments

Comments

1 Comment

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related