Finding the minimum value in a numpy array and the corresponding values for the rest of that array's row

Question

Consider the following NumPy array:

a = np.array([[1,4], [2,1],(3,10),(4,8)])

This gives an array that looks like the following:

array([[ 1,  4],
       [ 2,  1],
       [ 3, 10],
       [ 4,  8]])

What I'm trying to do is find the minimum value of the second column (which in this case is 1), and then report the other value of that pair (in this case 2). I've tried using something like argmin, but that gets tripped up by the 1 in the first column.

Is there a way to do this easily? I've also considered sorting the array, but I can't seem to get that to work in a way that keeps the pairs together. The data is being generated by a loop like the following, so if there's a easier way to do this that isn't a numpy array, I'd take that as an answer too:

results = np.zeros((100,2))

# Loop over search range, change kappa each time
for i in range(100):
    results[i,0] = function1(x)
    results[i,1] = function2(y)

YXD · Accepted Answer · 2016-03-06 10:45:28Z

21

How about

a[np.argmin(a[:, 1]), 0]

Break-down

a. Grab the second column

>>> a[:, 1]
array([ 4,  1, 10,  8])

b. Get the index of the minimum element in the second column

>>> np.argmin(a[:, 1])
1

c. Index a with that to get the corresponding row

>>> a[np.argmin(a[:, 1])]
array([2, 1])

d. And take the first element

>>> a[np.argmin(a[:, 1]), 0]
2

edited Mar 6, 2016 at 10:45

answered Feb 19, 2013 at 11:46

YXD

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2 Comments

Fomite Over a year ago

Though for the sake of my own sanity, could you explain the logic of what exactly this does for me? Trying to avoid cargo-cult coding.

YXD Over a year ago

Done - hope that makes sense

wim · Accepted Answer · 2013-02-19 12:07:32Z

5

Using np.argmin is probably the best way to tackle this. To do it in pure python, you could use:

min(tuple(r[::-1]) for r in a)[::-1]

answered Feb 19, 2013 at 12:07

wim

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