Indexing a pandas dataframe by integer
pandas, python
Solution
You get an error with `df['A'].ix[0]` because your indexing doesn't start at 0, it starts at 10. You can get the value you want with either of the following
df['A'].ix[10]
df['A'].irow(0)
The first uses by the correct index. The second command, which I suspect is what you want, finds the value by the row number, rather than by index value, and is technically only two characters longer than if `df['A'].ix[0]` worked.
Alternatively, you can reset the indices so that they will respond the way you expect for `df['A'].ix[0]`:
df2=df.reset_index()
This will preserve your old indices (10, 20, etc.) by moving them into a column called "index" in the df2 data frame. Then `df2['A'].ix[0]` will return 'a'. If you want to remove the old 10-based indices, you can insert the flag `drop=True` into the parenthesis of the reset_index function.
Problem
I can't seem to find an elegant way to index a `pandas.DataFrame` by an integer index. In the following example I want to get the value 'a' from the first element of the `'A'` column. ``` import pandas df = pandas.DataFrame( {'A':['a','b', 'c'], 'B':['f', 'g', 'h']}, index=[10,20,30] ) ``` I would expect `df['A'].ix[0]` and `df['A'][10]` both to return `'a'`. The `df['A'][10]` does return `'a'`, but `df['A'].ix[0]` throws a `KeyError: 0`. The only way I could think of to get the value `'a'` based on the index 0 is to use the following approach. ``` df['A'][df['A'].index[0]] ``` Is there a shorter way to get `'a'` out of the dataframe, using the 0 index? Update As of pandas 0.11 there is a another way to index by integer. ``` df.iloc[0] # integer based, gives the first row df.loc[10] # label based, gives the row with label 10 ``` This supersedes the `irow` approach .