python pandas 3 smallest & 3 largest values

What have you tried? You could sort with s.sort() and then call s.head(3).index and s.tail(3).index.

With smaller Series, you're better off just sorting then taking head/tail!

This is a pandas feature request, should see in 0.14 (need to overcome some fiddly bits with different dtypes), an efficient solution for larger Series (> 1000 elements) is using kth_smallest from pandas algos (warning this function mutates the array it's applied to so use a copy!):

In [11]: s = pd.Series(np.random.randn(10))

In [12]: s
Out[12]: 
0    0.785650
1    0.969103
2   -0.618300
3   -0.770337
4    1.532137
5    1.367863
6   -0.852839
7    0.967317
8   -0.603416
9   -0.889278
dtype: float64

In [13]: n = 3

In [14]: pd.algos.kth_smallest(s.values.astype(float), n - 1)
Out[14]: -0.7703374582084163

In [15]: s[s <= pd.algos.kth_smallest(s.values.astype(float), n - 1)]
Out[15]: 
3   -0.770337
6   -0.852839
9   -0.889278
dtype: float64

If you want this in order:

In [16]: s[s <= pd.algos.kth_smallest(s.values.astype(float), n - 1)].order()
Out[16]: 
9   -0.889278
6   -0.852839
3   -0.770337
dtype: float64

If you're worried about duplicates (join nth place) you can take the head:

In [17]: s[s <= pd.algos.kth_smallest(s.values.astype(float), n - 1)].order().head(n)
Out[17]: 
9   -0.889278
6   -0.852839
3   -0.770337
dtype: float64

You want to take a look at argsort (in numpy and in pandas)

df = pd.DataFrame(np.random.randint(1,100,100).reshape(10,10))
# bottom three indexes
df[0].argsort().values[:3]    
# top three indexes
df[0].argsort().values[-3:]

import pandas as pd
import numpy as np
np.random.seed(1)
x=np.random.randint(1,100,10)
y=np.random.randint(1000,10000,10)

x
array([38, 13, 73, 10, 76,  6, 80, 65, 17,  2])
y
array([8751, 4462, 6396, 6374, 3962, 3516, 9444, 4562, 5764, 9093])

data=pd.DataFrame({"age":x,
               "salary":y})


data.nlargest(5,"age").nsmallest(5,"salary")