Reputation: 1681
from pandas import Index, MultiIndex, DataFrame, NA
columns = MultiIndex.from_product( (["foo", "bar"], list("abc")) )
index = Index(range(10))
df = DataFrame(index=index, columns=columns, dtype="Float32")
foo bar
a b c a b c
0 <NA> <NA> <NA> <NA> <NA> <NA>
1 <NA> <NA> <NA> <NA> <NA> <NA>
2 <NA> <NA> <NA> <NA> <NA> <NA>
3 <NA> <NA> <NA> <NA> <NA> <NA>
4 <NA> <NA> <NA> <NA> <NA> <NA>
5 <NA> <NA> <NA> <NA> <NA> <NA>
6 <NA> <NA> <NA> <NA> <NA> <NA>
7 <NA> <NA> <NA> <NA> <NA> <NA>
8 <NA> <NA> <NA> <NA> <NA> <NA>
9 <NA> <NA> <NA> <NA> <NA> <NA>
How can I update the values in position [0, "foo"]
with a dictionary foo_sample = {"b": 1.2, "c": 1.3, "a": 1.1}
? I tried, using pandas 1.4.2
:
df.loc[0, "foo"] = foo_sample # does nothing.
df.loc[0, "foo"].update(foo_sample) # does nothing
df.loc[0, "foo"].replace(foo_sample) # does nothing
df.loc[0, "foo"].map(foo_sample) # replaces <NA> with NaN
Upvotes: 2
Views: 209
Reputation: 13458
Here is one way to do it:
df.loc[0, ("foo", list(foo_sample.keys()))] = foo_sample.values()
print(df)
# Output
a b c a b c
0 1.1 1.2 1.3 <NA> <NA> <NA>
1 <NA> <NA> <NA> <NA> <NA> <NA>
2 <NA> <NA> <NA> <NA> <NA> <NA>
3 <NA> <NA> <NA> <NA> <NA> <NA>
4 <NA> <NA> <NA> <NA> <NA> <NA>
5 <NA> <NA> <NA> <NA> <NA> <NA>
6 <NA> <NA> <NA> <NA> <NA> <NA>
7 <NA> <NA> <NA> <NA> <NA> <NA>
8 <NA> <NA> <NA> <NA> <NA> <NA>
9 <NA> <NA> <NA> <NA> <NA> <NA>
See Advanced indexing with hierarchical index in Pandas documentation for additional guidance.
Upvotes: 2