Assume that I have two arrays `A`

and `B`

, where both `A`

and `B`

are `m x n`

. My goal is now, for each row of `A`

and `B`

, to find where I should insert the elements of row `i`

of `A`

in the corresponding row of `B`

. That is, I wish to apply `np.digitize`

or `np.searchsorted`

to each row of `A`

and `B`

.

My naive solution is to simply iterate over the rows. However, this is far too slow for my application. My question is therefore: is there a vectorized implementation of either algorithm that I haven’t managed to find?

## Answers:

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### Method 1

We can add each row some offset as compared to the previous row. We would use the same offset for both arrays. The idea is to use `np.searchsorted`

on flattened version of input arrays thereafter and thus each row from `b`

would be restricted to find sorted positions in the corresponding row in `a`

. Additionally, to make it work for negative numbers too, we just need to offset for the minimum numbers as well.

So, we would have a vectorized implementation like so –

def searchsorted2d(a,b): m,n = a.shape max_num = np.maximum(a.max() - a.min(), b.max() - b.min()) + 1 r = max_num*np.arange(a.shape[0])[:,None] p = np.searchsorted( (a+r).ravel(), (b+r).ravel() ).reshape(m,-1) return p - n*(np.arange(m)[:,None])

Runtime test –

In [173]: def searchsorted2d_loopy(a,b): ...: out = np.zeros(a.shape,dtype=int) ...: for i in range(len(a)): ...: out[i] = np.searchsorted(a[i],b[i]) ...: return out ...: In [174]: # Setup input arrays ...: a = np.random.randint(11,99,(10000,20)) ...: b = np.random.randint(11,99,(10000,20)) ...: a = np.sort(a,1) ...: b = np.sort(b,1) ...: In [175]: np.allclose(searchsorted2d(a,b),searchsorted2d_loopy(a,b)) Out[175]: True In [176]: %timeit searchsorted2d_loopy(a,b) 10 loops, best of 3: 28.6 ms per loop In [177]: %timeit searchsorted2d(a,b) 100 loops, best of 3: 13.7 ms per loop

### Method 2

The solution provided by @Divakar is ideal for integer data, but beware of precision issues for floating point values, especially if they span multiple orders of magnitude (e.g. `[[1.0, 2,0, 3.0, 1.0e+20],...]`

). In some cases `r`

may be so large that applying `a+r`

and `b+r`

wipes out the original values you’re trying to run `searchsorted`

on, and you’re just comparing `r`

to `r`

.

To make the approach more robust for floating-point data, you could embed the row information into the arrays as part of the values (as a structured dtype), and run searchsorted on these structured dtypes instead.

def searchsorted_2d (a, v, side='left', sorter=None): import numpy as np # Make sure a and v are numpy arrays. a = np.asarray(a) v = np.asarray(v) # Augment a with row id ai = np.empty(a.shape,dtype=[('row',int),('value',a.dtype)]) ai['row'] = np.arange(a.shape[0]).reshape(-1,1) ai['value'] = a # Augment v with row id vi = np.empty(v.shape,dtype=[('row',int),('value',v.dtype)]) vi['row'] = np.arange(v.shape[0]).reshape(-1,1) vi['value'] = v # Perform searchsorted on augmented array. # The row information is embedded in the values, so only the equivalent rows # between a and v are considered. result = np.searchsorted(ai.flatten(),vi.flatten(), side=side, sorter=sorter) # Restore the original shape, decode the searchsorted indices so they apply to the original data. result = result.reshape(vi.shape) - vi['row']*a.shape[1] return result

**Edit:** The timing on this approach is abysmal!

In [21]: %timeit searchsorted_2d(a,b) 10 loops, best of 3: 92.5 ms per loop

You would be better off just just using `map`

over the array:

In [22]: %timeit np.array(list(map(np.searchsorted,a,b))) 100 loops, best of 3: 13.8 ms per loop

For integer data, @Divakar’s approach is still the fastest:

In [23]: %timeit searchsorted2d(a,b) 100 loops, best of 3: 7.26 ms per loop

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0