CODEWITHSUNDEEP
pythonarraysnumpymatrix
Delta_Fore
Delta_Fore

Reputation: 3271

Numpy where() on a 2D matrix

I have a matrix like this

t = np.array([[1,2,3,'foo'],
 [2,3,4,'bar'],
 [5,6,7,'hello'],
 [8,9,1,'bar']])

I want to get the indices where the rows contain the string 'bar'

In a 1d array

rows = np.where(t == 'bar')

should give me the indices [0,3] followed by broadcasting:-

results = t[rows]

should give me the right rows

But I can't figure out how to get it to work with 2d arrays.

Upvotes: 31

Views: 103537

Answers (2)

Delta_Fore
Delta_Fore

Reputation: 3271

You have to slice the array to the col you want to index:

rows = np.where(t[:,3] == 'bar')
result = t[rows]

This returns:

 [[2,3,4,'bar'],
  [8,9,1,'bar']]

Upvotes: 31

Jaime
Jaime

Reputation: 67427

For the general case, where your search string can be in any column, you can do this:

>>> rows, cols = np.where(t == 'bar')
>>> t[rows]
array([['2', '3', '4', 'bar'],
       ['8', '9', '1', 'bar']],
      dtype='|S11')

Upvotes: 31

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