Is there a NumPy function to return the first index of something in an array

Question

I know there is a method for a Python list to return the first index of something    gt  gt  gt  l    1  2  3   gt  gt  gt  l index 2  1   Is there something like that for NumPy arrays

User · Answer

There are lots of operations in NumPy that could perhaps be put together to accomplish this. This will return indices of elements equal to item:

numpy.nonzero(array - item)

You could then take the first elements of the lists to get a single element.

User · Answer

l index x  returns the smallest i such that i is the index of the first occurrence of x in the list   One can safely assume that the index   function in Python is implemented so that it stops after finding the first match  and this results in an optimal average performance   For finding an element stopping after the first match in a NumPy array use an iterator  ndenumerate    In  67   l range 100   In  68   l index 2  Out 68   2   NumPy array   In  69   a   np arange 100   In  70   next  idx for idx  val in np ndenumerate a  if val  2   Out 70    2L     Note that both methods index   and next return an error if the element is not found  With next  one can use a second argument to return a special value in case the element is not found  e g   In  77   next  idx for idx  val in np ndenumerate a  if val  400  None    There are other functions in NumPy  argmax  where  and nonzero  that can be used to find an element in an array  but they all have the drawback of going through the whole array looking for all occurrences  thus not being optimized for finding the first element  Note also that where and nonzero return arrays  so you need to select the first element to get the index   In  71   np argmax a  2  Out 71   2  In  72   np where a  2  Out 72    array  2   dtype int64     In  73   np nonzero a  2  Out 73    array  2   dtype int64      Time comparison  Just checking that for large arrays the solution using an iterator is faster when the searched item is at the beginning of the array  using  timeit in the IPython shell    In  285   a   np arange 100000   In  286    timeit next  idx for idx  val in np ndenumerate a  if val  0   100000 loops  best of 3  17 6   s per loop  In  287    timeit np argmax a  0  1000 loops  best of 3  254   s per loop  In  288    timeit np where a  0  0  0  1000 loops  best of 3  314   s per loop     This is an open NumPy GitHub issue   See also  Numpy  find first index of value fast

User · Answer

Note  this is for python 2 7 version  You can use a lambda function to deal with the problem  and it works both on NumPy array and list   your list    11  22  23  44  55  result   filter lambda x your list x  gt 30  range len your list     result   3  4   import numpy as np your numpy array   np array  11  22  23  44  55   result   filter lambda x your numpy array  x  gt 30  range len your list     result   3  4    And you can use  result 0    to get the first index of the filtered elements   For python 3 6  use  list result    instead of   result

User · Answer

An alternative to selecting the first element from np where   is to use a generator expression together with enumerate  such as     gt  gt  gt  import numpy as np  gt  gt  gt  x   np arange 100      x   array  0  1  2  3      99    gt  gt  gt  next i for i  x i in enumerate x  if x i    2  2   For a two dimensional array one would do     gt  gt  gt  x   np arange 100  reshape 10 10      x   array   0  1  2     9    10   19      gt  gt  gt  next  i j  for i  x i in enumerate x                  for j  x ij in enumerate x i  if x ij    2   0  2    The advantage of this approach is that it stops checking the elements of the array after the first match is found  whereas np where checks all elements for a match  A generator expression would be faster if there s  match early in the array

User · Answer

There are lots of operations in NumPy that could perhaps be put together to accomplish this. This will return indices of elements equal to item:

numpy.nonzero(array - item)

You could then take the first elements of the lists to get a single element.

User · Answer

For 1D arrays  I d recommend np flatnonzero array    value  0   which is equivalent to both np nonzero array    value  0  0  and np where array    value  0  0  but avoids the ugliness of unboxing a 1-element tuple

User · Answer

If you re going to use this as an index into something else  you can use boolean indices if the arrays are broadcastable  you don t need explicit indices   The absolute simplest way to do this is to simply index based on a truth value   other array first array    item    Any boolean operation works   a   numpy arange 100  other array first array  gt  50    The nonzero method takes booleans  too   index   numpy nonzero first array    item  0  0    The two zeros are for the tuple of indices  assuming first array is 1D  and then the first item in the array of indices

User · Answer

l index x  returns the smallest i such that i is the index of the first occurrence of x in the list   One can safely assume that the index   function in Python is implemented so that it stops after finding the first match  and this results in an optimal average performance   For finding an element stopping after the first match in a NumPy array use an iterator  ndenumerate    In  67   l range 100   In  68   l index 2  Out 68   2   NumPy array   In  69   a   np arange 100   In  70   next  idx for idx  val in np ndenumerate a  if val  2   Out 70    2L     Note that both methods index   and next return an error if the element is not found  With next  one can use a second argument to return a special value in case the element is not found  e g   In  77   next  idx for idx  val in np ndenumerate a  if val  400  None    There are other functions in NumPy  argmax  where  and nonzero  that can be used to find an element in an array  but they all have the drawback of going through the whole array looking for all occurrences  thus not being optimized for finding the first element  Note also that where and nonzero return arrays  so you need to select the first element to get the index   In  71   np argmax a  2  Out 71   2  In  72   np where a  2  Out 72    array  2   dtype int64     In  73   np nonzero a  2  Out 73    array  2   dtype int64      Time comparison  Just checking that for large arrays the solution using an iterator is faster when the searched item is at the beginning of the array  using  timeit in the IPython shell    In  285   a   np arange 100000   In  286    timeit next  idx for idx  val in np ndenumerate a  if val  0   100000 loops  best of 3  17 6   s per loop  In  287    timeit np argmax a  0  1000 loops  best of 3  254   s per loop  In  288    timeit np where a  0  0  0  1000 loops  best of 3  314   s per loop     This is an open NumPy GitHub issue   See also  Numpy  find first index of value fast

User · Answer

There are lots of operations in NumPy that could perhaps be put together to accomplish this. This will return indices of elements equal to item:

numpy.nonzero(array - item)

You could then take the first elements of the lists to get a single element.

User · Answer

For one-dimensional sorted arrays  it would be much more simpler and efficient O log n   to use numpy searchsorted which returns a NumPy integer  position   For example   arr   np array  1  1  1  2  3  3  4   i   np searchsorted arr  3    Just make sure the array is already sorted  Also check if returned index i actually contains the searched element  since searchsorted s main objective is to find indices where elements should be inserted to maintain order   if arr i     3      print  present   else      print  not present

User · Answer

To index on any criteria  you can so something like the following   In  1   from numpy import   In  2   x   arange 125  reshape  5 5 5   In  3   y   indices x shape  In  4   locs   y   x  gt   120    put whatever you want in place of x  gt   120 In  5   pts   hsplit locs  len locs 0    In  6   for pt in pts                    print      join str p 0   for p in pt   4  4  0 4  4  1 4  4  2 4  4  3 4  4  4   And here s a quick function to do what list index   does  except doesn t raise an exception if it s not found  Beware -- this is probably very slow on large arrays   You can probably monkey patch this on to arrays if you d rather use it as a method   def ndindex ndarray  item       if len ndarray shape     1          try              return  ndarray tolist   index item           except              pass     else          for i  subarray in enumerate ndarray               try                  return  i    ndindex subarray  item              except                  pass  In  1   ndindex x  103  Out 1    4  0  3

User · Answer

If you need the index of the first occurrence of only one value  you can use nonzero  or where  which amounts to the same thing in this case     gt  gt  gt  t   array  1  1  1  2  2  3  8  3  8  8    gt  gt  gt  nonzero t    8   array  6  8  9      gt  gt  gt  nonzero t    8  0  0  6   If you need the first index of each of many values  you could obviously do the same as above repeatedly  but there is a trick that may be faster   The following finds the indices of the first element of each subsequence    gt  gt  gt  nonzero r  1  diff t   -1     array  0  3  5  6  7  8       Notice that it finds the beginning of both subsequence of 3s and both subsequences of 8s    1  1  1  2  2  3  8  3  8  8   So it s slightly different than finding the first occurrence of each value   In your program  you may be able to work with a sorted version of t to get what you want    gt  gt  gt  st   sorted t   gt  gt  gt  nonzero r  1  diff st   -1     array  0  3  5  7

User · Answer

If you re going to use this as an index into something else  you can use boolean indices if the arrays are broadcastable  you don t need explicit indices   The absolute simplest way to do this is to simply index based on a truth value   other array first array    item    Any boolean operation works   a   numpy arange 100  other array first array  gt  50    The nonzero method takes booleans  too   index   numpy nonzero first array    item  0  0    The two zeros are for the tuple of indices  assuming first array is 1D  and then the first item in the array of indices

User · Answer

Note  this is for python 2 7 version  You can use a lambda function to deal with the problem  and it works both on NumPy array and list   your list    11  22  23  44  55  result   filter lambda x your list x  gt 30  range len your list     result   3  4   import numpy as np your numpy array   np array  11  22  23  44  55   result   filter lambda x your numpy array  x  gt 30  range len your list     result   3  4    And you can use  result 0    to get the first index of the filtered elements   For python 3 6  use  list result    instead of   result

User · Answer

Yes  given an array  array  and a value  item to search for  you can use np where as  itemindex   numpy where array  item   The result is a tuple with first all the row indices  then all the column indices  For example  if an array is two dimensions and it contained your item at two locations then array itemindex 0  0   itemindex 1  0    would be equal to your item and so would be  array itemindex 0  1   itemindex 1  1

User · Answer

You can also convert a NumPy array to list in the air and get its index  For example   l    1 2 3 4 5    Python list a   numpy array l    NumPy array i   a tolist   index 2    i will return index of 2 print i   It will print 1

User · Answer

Use ndindex Sample array arr   np array   1 4                     2 3    print arr        1 4        2 3      create an empty list to store the index and the element tuples  index elements       for i in np ndindex arr shape        index elements append  arr i  i       convert the list of tuples into dictionary  index elements   dict index elements   The keys are the elements and the values are their indices - use keys to access the index  index elements 4       output        0 1

User · Answer

If you re going to use this as an index into something else  you can use boolean indices if the arrays are broadcastable  you don t need explicit indices   The absolute simplest way to do this is to simply index based on a truth value   other array first array    item    Any boolean operation works   a   numpy arange 100  other array first array  gt  50    The nonzero method takes booleans  too   index   numpy nonzero first array    item  0  0    The two zeros are for the tuple of indices  assuming first array is 1D  and then the first item in the array of indices

User · Answer

To index on any criteria  you can so something like the following   In  1   from numpy import   In  2   x   arange 125  reshape  5 5 5   In  3   y   indices x shape  In  4   locs   y   x  gt   120    put whatever you want in place of x  gt   120 In  5   pts   hsplit locs  len locs 0    In  6   for pt in pts                    print      join str p 0   for p in pt   4  4  0 4  4  1 4  4  2 4  4  3 4  4  4   And here s a quick function to do what list index   does  except doesn t raise an exception if it s not found  Beware -- this is probably very slow on large arrays   You can probably monkey patch this on to arrays if you d rather use it as a method   def ndindex ndarray  item       if len ndarray shape     1          try              return  ndarray tolist   index item           except              pass     else          for i  subarray in enumerate ndarray               try                  return  i    ndindex subarray  item              except                  pass  In  1   ndindex x  103  Out 1    4  0  3

User · Answer

If you re going to use this as an index into something else  you can use boolean indices if the arrays are broadcastable  you don t need explicit indices   The absolute simplest way to do this is to simply index based on a truth value   other array first array    item    Any boolean operation works   a   numpy arange 100  other array first array  gt  50    The nonzero method takes booleans  too   index   numpy nonzero first array    item  0  0    The two zeros are for the tuple of indices  assuming first array is 1D  and then the first item in the array of indices

User · Answer

The numpy indexed package  disclaimer  I am its author  contains a vectorized equivalent of list index for numpy ndarray  that is   sequence of arrays     0  1    1  2    -5  0   arrays to query     -5  0    1  0    import numpy indexed as npi idx   npi indices sequence of arrays  arrays to query  missing -1  print idx       2  -1    This solution has vectorized performance  generalizes to ndarrays  and has various ways of dealing with missing values

User · Answer

You can also convert a NumPy array to list in the air and get its index  For example   l    1 2 3 4 5    Python list a   numpy array l    NumPy array i   a tolist   index 2    i will return index of 2 print i   It will print 1

User · Answer

If you need the index of the first occurrence of only one value  you can use nonzero  or where  which amounts to the same thing in this case     gt  gt  gt  t   array  1  1  1  2  2  3  8  3  8  8    gt  gt  gt  nonzero t    8   array  6  8  9      gt  gt  gt  nonzero t    8  0  0  6   If you need the first index of each of many values  you could obviously do the same as above repeatedly  but there is a trick that may be faster   The following finds the indices of the first element of each subsequence    gt  gt  gt  nonzero r  1  diff t   -1     array  0  3  5  6  7  8       Notice that it finds the beginning of both subsequence of 3s and both subsequences of 8s    1  1  1  2  2  3  8  3  8  8   So it s slightly different than finding the first occurrence of each value   In your program  you may be able to work with a sorted version of t to get what you want    gt  gt  gt  st   sorted t   gt  gt  gt  nonzero r  1  diff st   -1     array  0  3  5  7

User · Answer

Yes  given an array  array  and a value  item to search for  you can use np where as  itemindex   numpy where array  item   The result is a tuple with first all the row indices  then all the column indices  For example  if an array is two dimensions and it contained your item at two locations then array itemindex 0  0   itemindex 1  0    would be equal to your item and so would be  array itemindex 0  1   itemindex 1  1

User · Answer

To index on any criteria  you can so something like the following   In  1   from numpy import   In  2   x   arange 125  reshape  5 5 5   In  3   y   indices x shape  In  4   locs   y   x  gt   120    put whatever you want in place of x  gt   120 In  5   pts   hsplit locs  len locs 0    In  6   for pt in pts                    print      join str p 0   for p in pt   4  4  0 4  4  1 4  4  2 4  4  3 4  4  4   And here s a quick function to do what list index   does  except doesn t raise an exception if it s not found  Beware -- this is probably very slow on large arrays   You can probably monkey patch this on to arrays if you d rather use it as a method   def ndindex ndarray  item       if len ndarray shape     1          try              return  ndarray tolist   index item           except              pass     else          for i  subarray in enumerate ndarray               try                  return  i    ndindex subarray  item              except                  pass  In  1   ndindex x  103  Out 1    4  0  3

User · Answer

Yes  given an array  array  and a value  item to search for  you can use np where as  itemindex   numpy where array  item   The result is a tuple with first all the row indices  then all the column indices  For example  if an array is two dimensions and it contained your item at two locations then array itemindex 0  0   itemindex 1  0    would be equal to your item and so would be  array itemindex 0  1   itemindex 1  1

User · Answer

Just to add a very performant and handy numba alternative based on np ndenumerate to find the first index   from numba import njit import numpy as np   njit def index array  item       for idx  val in np ndenumerate array           if val    item              return idx       If no item was found return None  other return types might be a problem due to       numbas type inference    This is pretty fast and deals naturally with multidimensional arrays    gt  gt  gt  arr1   np ones  100  100  100    gt  gt  gt  arr1 2  2  2    2   gt  gt  gt  index arr1  2   2  2  2    gt  gt  gt  arr2   np ones 20   gt  gt  gt  arr2 5    2   gt  gt  gt  index arr2  2   5     This can be much faster  because it s short-circuiting the operation  than any approach using np where or np nonzero     However np argwhere could also deal gracefully with multidimensional arrays  you would need to manually cast it to a tuple and it s not short-circuited  but it would fail if no match is found    gt  gt  gt  tuple np argwhere arr1    2  0    2  2  2   gt  gt  gt  tuple np argwhere arr2    2  0    5

User · Answer

The numpy indexed package  disclaimer  I am its author  contains a vectorized equivalent of list index for numpy ndarray  that is   sequence of arrays     0  1    1  2    -5  0   arrays to query     -5  0    1  0    import numpy indexed as npi idx   npi indices sequence of arrays  arrays to query  missing -1  print idx       2  -1    This solution has vectorized performance  generalizes to ndarrays  and has various ways of dealing with missing values

User · Answer

An alternative to selecting the first element from np where   is to use a generator expression together with enumerate  such as     gt  gt  gt  import numpy as np  gt  gt  gt  x   np arange 100      x   array  0  1  2  3      99    gt  gt  gt  next i for i  x i in enumerate x  if x i    2  2   For a two dimensional array one would do     gt  gt  gt  x   np arange 100  reshape 10 10      x   array   0  1  2     9    10   19      gt  gt  gt  next  i j  for i  x i in enumerate x                  for j  x ij in enumerate x i  if x ij    2   0  2    The advantage of this approach is that it stops checking the elements of the array after the first match is found  whereas np where checks all elements for a match  A generator expression would be faster if there s  match early in the array

User · Answer

There are lots of operations in NumPy that could perhaps be put together to accomplish this. This will return indices of elements equal to item:

numpy.nonzero(array - item)

You could then take the first elements of the lists to get a single element.

User · Answer

Yes  given an array  array  and a value  item to search for  you can use np where as  itemindex   numpy where array  item   The result is a tuple with first all the row indices  then all the column indices  For example  if an array is two dimensions and it contained your item at two locations then array itemindex 0  0   itemindex 1  0    would be equal to your item and so would be  array itemindex 0  1   itemindex 1  1

User · Answer

Just to add a very performant and handy numba alternative based on np ndenumerate to find the first index   from numba import njit import numpy as np   njit def index array  item       for idx  val in np ndenumerate array           if val    item              return idx       If no item was found return None  other return types might be a problem due to       numbas type inference    This is pretty fast and deals naturally with multidimensional arrays    gt  gt  gt  arr1   np ones  100  100  100    gt  gt  gt  arr1 2  2  2    2   gt  gt  gt  index arr1  2   2  2  2    gt  gt  gt  arr2   np ones 20   gt  gt  gt  arr2 5    2   gt  gt  gt  index arr2  2   5     This can be much faster  because it s short-circuiting the operation  than any approach using np where or np nonzero     However np argwhere could also deal gracefully with multidimensional arrays  you would need to manually cast it to a tuple and it s not short-circuited  but it would fail if no match is found    gt  gt  gt  tuple np argwhere arr1    2  0    2  2  2   gt  gt  gt  tuple np argwhere arr2    2  0    5

User · Answer

Use ndindex Sample array arr   np array   1 4                     2 3    print arr        1 4        2 3      create an empty list to store the index and the element tuples  index elements       for i in np ndindex arr shape        index elements append  arr i  i       convert the list of tuples into dictionary  index elements   dict index elements   The keys are the elements and the values are their indices - use keys to access the index  index elements 4       output        0 1

User · Answer

For 1D arrays  I d recommend np flatnonzero array    value  0   which is equivalent to both np nonzero array    value  0  0  and np where array    value  0  0  but avoids the ugliness of unboxing a 1-element tuple

User · Answer

For one-dimensional sorted arrays  it would be much more simpler and efficient O log n   to use numpy searchsorted which returns a NumPy integer  position   For example   arr   np array  1  1  1  2  3  3  4   i   np searchsorted arr  3    Just make sure the array is already sorted  Also check if returned index i actually contains the searched element  since searchsorted s main objective is to find indices where elements should be inserted to maintain order   if arr i     3      print  present   else      print  not present

User · Answer

To index on any criteria  you can so something like the following   In  1   from numpy import   In  2   x   arange 125  reshape  5 5 5   In  3   y   indices x shape  In  4   locs   y   x  gt   120    put whatever you want in place of x  gt   120 In  5   pts   hsplit locs  len locs 0    In  6   for pt in pts                    print      join str p 0   for p in pt   4  4  0 4  4  1 4  4  2 4  4  3 4  4  4   And here s a quick function to do what list index   does  except doesn t raise an exception if it s not found  Beware -- this is probably very slow on large arrays   You can probably monkey patch this on to arrays if you d rather use it as a method   def ndindex ndarray  item       if len ndarray shape     1          try              return  ndarray tolist   index item           except              pass     else          for i  subarray in enumerate ndarray               try                  return  i    ndindex subarray  item              except                  pass  In  1   ndindex x  103  Out 1    4  0  3

[python] Is there a NumPy function to return the first index of something in an array?

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