I have the following (simplified) dataframe:

```
df = pd.DataFrame({'X': [1, 2, 3, 4, 5,6,7,8,9,10],
'Y': [10,20,30,40,50,-10,-20,-30,-40,-50],
'Z': [20,18,16,14,12,10,8,6,4,2]},index=list('ABCDEFGHIJ'))
```

Which gives the following:

```
X Y Z
A 1 10 20
B 2 20 18
C 3 30 16
D 4 40 14
E 5 50 12
F 6 -10 10
G 7 -20 8
H 8 -30 6
I 9 -40 4
J 10 -50 2
```

I want to create a new dataframe that returns the index of the n smallest values, by column.

Desired output (say, 3 smallest values):

```
X Y Z
0 A J J
1 B I I
2 C H H
```

What is the best way to do this?

Faster numpy solution with `numpy.argsort`

:

```
N = 3
a = np.argsort(-df.values, axis=0)[-1:-1-N:-1]
print (a)
[[0 9 9]
[1 8 8]
[2 7 7]]
b = pd.DataFrame(df.index[a], columns=df.columns)
print (b)
X Y Z
0 A J J
1 B I I
2 C H H
```

**Timings**:

```
In [111]: %timeit (pd.DataFrame(df.index[np.argsort(-df.values, axis=0)[-1:-1-N:-1]], columns=df.columns))
159 µs ± 1.37 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [112]: %timeit (df.apply(lambda x: pd.Series(x.nsmallest(N).index)))
3.52 ms ± 49.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
```