Pandas `isin` 函数的更快替代方案

Question

I have a very large data frame df that looks like:

ID       Value1    Value2
1345      3.2      332
1355      2.2      32
2346      1.0      11
3456      8.9      322

And I have a list that contains a subset of IDs ID_list . I need to have a subset of df for the ID contained in ID_list .

Currently, I am using df_sub=df[df.ID.isin(ID_list)] to do it. But it takes a lot time. ID s contained in ID_list doesn't have any pattern, so it's not within certain range. (And I need to apply the same operation to many similar dataframes. I was wondering if there is any faster way to do this. Will it help a lot if make ID as the index?

Thanks!

Answer 1

EDIT 2: Here's a link to a more recent look into the performance of various pandas operations, though it doesn't seem to include merge and join to date.

https://github.com/mm-mansour/Fast-Pandas

EDIT 1: These benchmarks were for a quite old version of pandas and likely are not still relevant. See Mike's comment below on merge .

It depends on the size of your data but for large datasets DataFrame.join seems to be the way to go. This requires your DataFrame index to be your 'ID' and the Series or DataFrame you're joining against to have an index that is your 'ID_list'. The Series must also have a name to be used with join , which gets pulled in as a new field called name . You also need to specify an inner join to get something like isin because join defaults to a left join. query in syntax seems to have the same speed characteristics as isin for large datasets.

If you're working with small datasets, you get different behaviors and it actually becomes faster to use a list comprehension or apply against a dictionary than using isin .

Otherwise, you can try to get more speed with Cython .

# I'm ignoring that the index is defaulting to a sequential number. You
# would need to explicitly assign your IDs to the index here, e.g.:
# >>> l_series.index = ID_list
mil = range(1000000)
l = mil
l_series = pd.Series(l)

df = pd.DataFrame(l_series, columns=['ID'])


In [247]: %timeit df[df.index.isin(l)]
1 loops, best of 3: 1.12 s per loop

In [248]: %timeit df[df.index.isin(l_series)]
1 loops, best of 3: 549 ms per loop

# index vs column doesn't make a difference here
In [304]: %timeit df[df.ID.isin(l_series)]
1 loops, best of 3: 541 ms per loop

In [305]: %timeit df[df.index.isin(l_series)]
1 loops, best of 3: 529 ms per loop

# query 'in' syntax has the same performance as 'isin'
In [249]: %timeit df.query('index in @l')
1 loops, best of 3: 1.14 s per loop

In [250]: %timeit df.query('index in @l_series')
1 loops, best of 3: 564 ms per loop

# ID must be the index for DataFrame.join and l_series must have a name.
# join defaults to a left join so we need to specify inner for existence.
In [251]: %timeit df.join(l_series, how='inner')
10 loops, best of 3: 93.3 ms per loop

# Smaller datasets.
df = pd.DataFrame([1,2,3,4], columns=['ID'])
l = range(10000)
l_dict = dict(zip(l, l))
l_series = pd.Series(l)
l_series.name = 'ID_list'


In [363]: %timeit df.join(l_series, how='inner')
1000 loops, best of 3: 733 µs per loop

In [291]: %timeit df[df.ID.isin(l_dict)]
1000 loops, best of 3: 742 µs per loop

In [292]: %timeit df[df.ID.isin(l)]
1000 loops, best of 3: 771 µs per loop

In [294]: %timeit df[df.ID.isin(l_series)]
100 loops, best of 3: 2 ms per loop

# It's actually faster to use apply or a list comprehension for these small cases.
In [296]: %timeit df[[x in l_dict for x in df.ID]]
1000 loops, best of 3: 203 µs per loop

In [299]: %timeit df[df.ID.apply(lambda x: x in l_dict)]
1000 loops, best of 3: 297 µs per loop

Answer 2

Yes, isin is quite slow.

Instead it's faster to make ID an index then use use loc , like:

df.set_index('ID', inplace=True)
df.loc[list_of_indices]

Actually what brought me to this page was that I needed to create a label in my df based on index in another df: "if df_1's index matches df_2's index, label it a 1, otherwise NaN", which I accomplished like this:

df_2['label'] = 1  # Create a label column
df_1.join(df_2['label'])

Which is also very fast.