create intersection from two or more 2d numpy arrays based on common value in one column

create intersection from two or more 2d numpy arrays based on common value in one column

```I have 3 numpy recarrays with following structure.
The first column is some position (Integer) and the second column is a score (Float).
Input:
a = [[1, 5.41],
[2, 5.42],
[3, 12.32],
dtype=[('position', '

Here is one approach, I believe it should be reasonably fast. I think the first thing you want to do is count the number occurrences for each position. This function will handle that:
```def count_positions(positions):
positions = np.sort(positions)
diff = np.ones(len(positions), 'bool')
diff[:-1] = positions[1:] != positions[:-1]
count = diff.nonzero()[0]
count[1:] = count[1:] - count[:-1]
count[0] += 1
uniqPositions = positions[diff]
return uniqPositions, count
```
Now using the function form above you want to take only the positions that occur 3 times:
```positions = np.concatenate((a['position'], b['position'], c['position']))
uinqPos, count = count_positions(positions)
uinqPos = uinqPos[count == 3]
```
We will be using search sorted so we sort a b and c:
```a.sort(order='position')
b.sort(order='position')
c.sort(order='position')
```
Now we can user search sorted to find where in each array to find each of our uniqPos:
```new_array = np.empty((len(uinqPos), 4))
new_array[:, 0] = uinqPos
index = a['position'].searchsorted(uinqPos)
new_array[:, 1] = a['score'][index]
index = b['position'].searchsorted(uinqPos)
new_array[:, 2] = b['score'][index]
index = c['position'].searchsorted(uinqPos)
new_array[:, 3] = c['score'][index]
```
There might be a more elegant solution using dictionaries, but I thought of this one first so I'll leave that to someone else.

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