Setting to first rows of pandas DataFrame

Question

I would like to set a value in some column for the first n rows of a pandas DataFrame.

>>> example = pd.DataFrame({'number':range(10),'name':list('aaabbbcccc')},index=range(20,0,-2)) # nontrivial index
>>> example
   name  number
20    a       0
18    a       1
16    a       2
14    b       3
12    b       4
10    b       5
8     c       6
6     c       7
4     c       8
2     c       9

I would like to set "number" for the first, say, 5 rows to the number 19. What I really want is to set the lowest values of "number" to that value, so I just sort first. If my index was the trivial one, I could do

example.loc[:5-1,'number'] = 19 # -1 for inclusive indexing
# or 
example.ix[:5-1,'number'] = 19

But since it's not, this would produce the following artifact (where all index values up to 4 have been chosen):

>>> example
   name  number
20    a      19
18    a      19
16    a      19
14    b      19
12    b      19
10    b      19
8     c      19
6     c      19
4     c      19
2     c       9

Using .iloc[] would be nice, except that it doesn't accept column names.

example.iloc[:5]['number'] = 19

works but gives a SettingWithCopyWarning.

My current solution is to do:

>>> example.sort_values('number',inplace=True)
>>> example.reset_index(drop=True,inplace=True)
>>> example.ix[:5-1,'number'] = 19
>>> example
  name  number
0    a      19
1    a      19
2    a      19
3    b      19
4    b      19
5    b       5
6    c       6
7    c       7
8    c       8
9    c       9

And since I have to repeat this for several columns, I have to do this a few times and reset the index each time, which also costs me my index (but never mind that).

Does anyone have a better solution?

Answer 1

I would use .iloc as .loc might yield unexpected results if certain indexes are repeated.

example.iloc[:5, example.columns.get_loc('number')] = 19

Answer 2

example.loc[example.index[:5], 'number'] = 19