Remap values in pandas column with a dict, preserve NaNs

Solution 1:

You can use .replace. For example:

>>> df = pd.DataFrame({'col2': {0: 'a', 1: 2, 2: np.nan}, 'col1': {0: 'w', 1: 1, 2: 2}})
>>> di = {1: "A", 2: "B"}
>>> df
  col1 col2
0    w    a
1    1    2
2    2  NaN
>>> df.replace({"col1": di})
  col1 col2
0    w    a
1    A    2
2    B  NaN

or directly on the Series, i.e. df["col1"].replace(di, inplace=True).

Solution 2:

map can be much faster than replace

If your dictionary has more than a couple of keys, using map can be much faster than replace. There are two versions of this approach, depending on whether your dictionary exhaustively maps all possible values (and also whether you want non-matches to keep their values or be converted to NaNs):

Exhaustive Mapping

In this case, the form is very simple:

df['col1'].map(di)       # note: if the dictionary does not exhaustively map all
                         # entries then non-matched entries are changed to NaNs

Although map most commonly takes a function as its argument, it can alternatively take a dictionary or series: Documentation for Pandas.series.map

Non-Exhaustive Mapping

If you have a non-exhaustive mapping and wish to retain the existing variables for non-matches, you can add fillna:

df['col1'].map(di).fillna(df['col1'])

as in @jpp's answer here: Replace values in a pandas series via dictionary efficiently

Benchmarks

Using the following data with pandas version 0.23.1:

di = {1: "A", 2: "B", 3: "C", 4: "D", 5: "E", 6: "F", 7: "G", 8: "H" }
df = pd.DataFrame({ 'col1': np.random.choice( range(1,9), 100000 ) })

and testing with %timeit, it appears that map is approximately 10x faster than replace.

Note that your speedup with map will vary with your data. The largest speedup appears to be with large dictionaries and exhaustive replaces. See @jpp answer (linked above) for more extensive benchmarks and discussion.