python - Pandas: Modify a particular level of Multiindex

Question

I have a dataframe with Multiindex and would like to modify one particular level of the Multiindex. For instance, the first level might be strings and I may want to remove the white spaces from that index level:

df.index.levels[1] = [x.replace(' ', '') for x in df.index.levels[1]]

However, the code above results in an error:

TypeError: 'FrozenList' does not support mutable operations.

I know I can reset_index and modify the column and then re-create the Multiindex, but I wonder whether there is a more elegant way to modify one particular level of the Multiindex directly.

Answer 1

Thanks to @cxrodgers's comment, I think the fastest way to do this is:

df.index = df.index.set_levels(df.index.levels[0].str.replace(' ', ''), level=0)

---

Old, longer answer:

I found that the list comprehension suggested by @Shovalt works but felt slow on my machine (using a dataframe with >10,000 rows).

Instead, I was able to use .set_levels method, which was quite a bit faster for me.

%timeit pd.MultiIndex.from_tuples([(x[0].replace(' ',''), x[1]) for x in df.index])
1 loop, best of 3: 394 ms per loop

%timeit df.index.set_levels(df.index.get_level_values(0).str.replace(' ',''), level=0)
10 loops, best of 3: 134 ms per loop

In actuality, I just needed to prepend some text. This was even faster with .set_levels:

%timeit pd.MultiIndex.from_tuples([('00'+x[0], x[1]) for x in df.index])
100 loops, best of 3: 5.18 ms per loop

%timeit df.index.set_levels('00'+df.index.get_level_values(0), level=0)
1000 loops, best of 3: 1.38 ms per loop

%timeit df.index.set_levels('00'+df.index.levels[0], level=0)
1000 loops, best of 3: 331 μs per loop

This solution is based on the answer in the link from the comment by @denfromufa ...

python - Multiindex and timezone - Frozen list error - Stack Overflow