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