Indexing Pandas data frames: integer rows, named columns

Say df is a pandas dataframe.

• df.loc[] only accepts names
• df.iloc[] only accepts integers (actual placements)
• df.ix[] accepts both names and integers:

When referencing rows, df.ix[row_idx, ] only wants to be given names. e.g.

df = pd.DataFrame({'a' : ['one', 'two', 'three','four', 'five', 'six'],
                   '1' : np.arange(6)})
df = df.ix[2:6]
print(df)

   1      a
2  2  three
3  3   four
4  4   five
5  5    six

df.ix[0, 'a']

throws an error, it doesn't give return 'two'.

When referencing columns, iloc is prefers integers, not names. e.g.

df.ix[2, 1]

returns 'three', not 2. (Although df.idx[2, '1'] does return 2).

Oddly, I'd like the exact opposite functionality. Usually my column names are very meaningful, so in my code I reference them directly. But due to a lot of observation cleaning, the row names in my pandas data frames don't usually correspond to range(len(df)).

I realize I can use:

df.iloc[0].loc['a'] # returns three

But it seems ugly! Does anyone know of a better way to do this, so that the code would look like this?

df.foo[0, 'a'] # returns three

In fact, is it possible to add on my own new method to pandas.core.frame.DataFrames, so e.g. df.idx(rows, cols) is in fact df.iloc[rows].loc[cols]?

It's a late answer, but @unutbu's comment is still valid and a great solution to this problem.

To index a DataFrame with integer rows and named columns (labeled columns):

df.loc[df.index[#], 'NAME'] where # is a valid integer index and NAME is the name of the column.

The existing answers seem short-sighted to me.

Problematic Solutions

1. df.loc[df.index[0], 'a']
   The strategy here is to get the row label of the 0th row and then use .loc as normal. I see two issues.

   1. If df has repeated row labels, df.loc[df.index[0], 'a'] could return multiple rows.
   2. .loc is slower than .iloc so you're sacrificing speed here.

2. df.reset_index(drop=True).loc[0, 'a']
   The strategy here is to reset the index so the row labels become 0, 1, 2, ... thus .loc[0] gives the same result as .iloc[0]. Still, the problem here is runtime, as .loc is slower than .iloc and you'll incur a cost for resetting the index.

Better Solution

I suggest following @Landmaster's comment:

df.iloc[0, df.columns.get_loc("a")]

Essentially, this is the same as df.iloc[0, 0] except we get the column index dynamically using df.columns.get_loc("a").

To index multiple columns such as ['a', 'b', 'c'], use:

df.iloc[0, [df.columns.get_loc(c) for c in ['a', 'b', 'c']]]

Update

This is discussed here as part of my course on Pandas.

we can reset the index and then use 0 based indexing like this

df.reset_index(drop=True).loc[0,'a']

edit: removed [] from col name index 'a' so it just outputs the value

For getting or setting a single value in a DataFrame by row/column labels, you better use DataFrame.at instead of DataFrame.loc, as it is ...

1. faster
2. you are more explicit about wanting to access only a single value.

How others have already shown, if you start out with an integer position for the row, you still have to find the row-label first with DataFrame.index as DataFrame.at only accepts labels:

df.at[df.index[0], 'a']
# Out: 'three'

Benchmark:

%timeit df.at[df.index[0], 'a']
# 7.57 µs ± 30.8 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%timeit df.loc[df.index[0], 'a']
# 10.9 µs ± 53.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%timeit df.iloc[0, df.columns.get_loc("a")]
# 13.3 µs ± 24 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

For completeness:

DataFrame.iat for accessing a single value for a row/column pair by integer position.

A very late answer but it amzed me that pandas still doesn't have such a function after all these years. If it irks you a lot, you can monkey-patch a custom indexer into the DataFrame:

class XLocIndexer:
    def __init__(self, frame):
        self.frame = frame
    
    def __getitem__(self, key):
        row, col = key
        return self.frame.iloc[row][col]

pd.core.indexing.IndexingMixin.xloc = property(lambda frame: XLocIndexer(frame))

# Usage
df.xloc[0, 'a'] # one