This seems rather obvious, but I can't seem to figure out how do I convert an index of data frame to a column?
For example:
df= gi ptt_loc 0 384444683 593 1 384444684 594 2 384444686 596 To,
df= index1 gi ptt_loc 0 0 384444683 593 1 1 384444684 594 2 2 384444686 596 either:
df['index1'] = df.index or, .reset_index:
df.reset_index(level=0, inplace=True) so, if you have a multi-index frame with 3 levels of index, like:
>>> df val tick tag obs 2016-02-26 C 2 0.0139 2016-02-27 A 2 0.5577 2016-02-28 C 6 0.0303 and you want to convert the 1st (tick) and 3rd (obs) levels in the index into columns, you would do:
>>> df.reset_index(level=['tick', 'obs']) tick obs val tag C 2016-02-26 2 0.0139 A 2016-02-27 2 0.5577 C 2016-02-28 6 0.0303 For MultiIndex you can extract its subindex using
df['si_name'] = R.index.get_level_values('si_name') where si_name is the name of the subindex.
To provide a bit more clarity, let's look at a DataFrame with two levels in its index (a MultiIndex).
index = pd.MultiIndex.from_product([['TX', 'FL', 'CA'], ['North', 'South']], names=['State', 'Direction']) df = pd.DataFrame(index=index, data=np.random.randint(0, 10, (6,4)), columns=list('abcd')) 
The reset_index method, called with the default parameters, converts all index levels to columns and uses a simple RangeIndex as new index.
df.reset_index() 
Use the level parameter to control which index levels are converted into columns. If possible, use the level name, which is more explicit. If there are no level names, you can refer to each level by its integer location, which begin at 0 from the outside. You can use a scalar value here or a list of all the indexes you would like to reset.
df.reset_index(level='State') # same as df.reset_index(level=0) 
In the rare event that you want to preserve the index and turn the index into a column, you can do the following:
# for a single level df.assign(State=df.index.get_level_values('State')) # for all levels df.assign(**df.index.to_frame())