CODEWITHSUNDEEP
pythonnumpy
Bob
Bob

Reputation: 2005

How do I select elements of an array given condition?

Suppose I have a numpy array x = [5, 2, 3, 1, 4, 5], y = ['f', 'o', 'o', 'b', 'a', 'r']. I want to select the elements in y corresponding to elements in x that are greater than 1 and less than 5.

I tried

x = array([5, 2, 3, 1, 4, 5])
y = array(['f','o','o','b','a','r'])
output = y[x > 1 & x < 5] # desired output is ['o','o','a']

but this doesn't work. How would I do this?

Upvotes: 197

Views: 374369

Answers (6)

Gautam Sreekumar
Gautam Sreekumar

Reputation: 564

For 2D arrays, you can do this. Create a 2D mask using the condition. Typecast the condition mask to int or float, depending on the array, and multiply it with the original array.

In [8]: arr
Out[8]: 
array([[ 1.,  2.,  3.,  4.,  5.],
       [ 6.,  7.,  8.,  9., 10.]])

In [9]: arr*(arr % 2 == 0).astype(np.int) 
Out[9]: 
array([[ 0.,  2.,  0.,  4.,  0.],
       [ 6.,  0.,  8.,  0., 10.]])

Upvotes: 0

Good Fit
Good Fit

Reputation: 1316

Add one detail to @J.F. Sebastian's and @Mark Mikofski's answers:
If one wants to get the corresponding indices (rather than the actual values of array), the following code will do:

For satisfying multiple (all) conditions:

select_indices = np.where( np.logical_and( x > 1, x < 5) )[0] #   1 < x <5

For satisfying multiple (or) conditions:

select_indices = np.where( np.logical_or( x < 1, x > 5 ) )[0] # x <1 or x >5

Upvotes: 25

user4340135
user4340135

Reputation:

I like to use np.vectorize for such tasks. Consider the following:

>>> # Arrays
>>> x = np.array([5, 2, 3, 1, 4, 5])
>>> y = np.array(['f','o','o','b','a','r'])

>>> # Function containing the constraints
>>> func = np.vectorize(lambda t: t>1 and t<5)

>>> # Call function on x
>>> y[func(x)]
>>> array(['o', 'o', 'a'], dtype='<U1')

The advantage is you can add many more types of constraints in the vectorized function.

Hope it helps.

Upvotes: 6

Mark Mikofski
Mark Mikofski

Reputation: 20208

IMO OP does not actually want np.bitwise_and() (aka &) but actually wants np.logical_and() because they are comparing logical values such as True and False - see this SO post on logical vs. bitwise to see the difference.

>>> x = array([5, 2, 3, 1, 4, 5])
>>> y = array(['f','o','o','b','a','r'])
>>> output = y[np.logical_and(x > 1, x < 5)] # desired output is ['o','o','a']
>>> output
array(['o', 'o', 'a'],
      dtype='|S1')

And equivalent way to do this is with np.all() by setting the axis argument appropriately.

>>> output = y[np.all([x > 1, x < 5], axis=0)] # desired output is ['o','o','a']
>>> output
array(['o', 'o', 'a'],
      dtype='|S1')

by the numbers:

>>> %timeit (a < b) & (b < c)
The slowest run took 32.97 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 1.15 µs per loop

>>> %timeit np.logical_and(a < b, b < c)
The slowest run took 32.59 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 1.17 µs per loop

>>> %timeit np.all([a < b, b < c], 0)
The slowest run took 67.47 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 5.06 µs per loop

so using np.all() is slower, but & and logical_and are about the same.

Upvotes: 44

Shuo Yang
Shuo Yang

Reputation: 131

Actually I would do it this way:

L1 is the index list of elements satisfying condition 1;(maybe you can use somelist.index(condition1) or np.where(condition1) to get L1.)

Similarly, you get L2, a list of elements satisfying condition 2;

Then you find intersection using intersect(L1,L2).

You can also find intersection of multiple lists if you get multiple conditions to satisfy.

Then you can apply index in any other array, for example, x.

Upvotes: 1

jfs
jfs

Reputation: 414845

Your expression works if you add parentheses:

>>> y[(1 < x) & (x < 5)]
array(['o', 'o', 'a'], 
      dtype='|S1')

Upvotes: 271

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