Convert elements of list in pandas series using a dict
Method 1
I am using unnesting
d={z : x for x , y in d.items() for z in y }
s=unnesting(s.to_frame().dropna(),[0])[0]\
.map(d).groupby(level=0).apply(set).reindex(s.index)
Out[260]:
0 {1}
1 {2}
2 NaN
3 {1}
4 {1, 2}
Name: 0, dtype: object
Method 2 loop it
[set(d.get(y) for y in x) if x is not None else None for x in s ]
#s=[set(d.get(y) for y in x) if x is not None else None for x in s ]
Out[265]: [{1}, {2}, None, {1}, {1, 2}]
Data input
s=pd.Series([["Apple", "Banana"],["Kiwi"],None,["Apple"],["Banana", "Kiwi"]])
d={1: ["Apple", "Banana"],
2: ["Kiwi"]}
One way would be to first unnest the dictionary and set the values as keys with their corresponding keys as values. And then you can use a list comprehension and map the values in each of the lists in the dataframe.
It'll be necessary to take a set
before returning a the result from the mapping in each iteration in order to avoid repeated values. Also note that or None
is doing the same as if x is not None else None
here, which will return None
in the case a list is empty. For a more detailed explanation on this you may check this post:
df = pd.DataFrame({'col1':[["Apple", "Banana"], ["Kiwi"], None, ["Apple"], ["Banana", "Kiwi"]]})
d = {1: ["Apple", "Banana"], 2: ["Kiwi"]}
d = {i:k for k, v in d.items() for i in v}
# {'Apple': 1, 'Banana': 1, 'Kiwi': 2}
out = [list(set(d[j] for j in i)) or None for i in df.col1.fillna('')]
# [[1], [2], None, [1], [1, 2]]
pd.DataFrame([out]).T
0
0 [1]
1 [2]
2 None
3 [1]
4 [1, 2]
Option 1
Rebuild the dictionary
m = {v: k for k, V in d.items() for v in V}
Rebuild
x = s.dropna()
v = [*map(m.get, np.concatenate(x.to_numpy()))]
i = x.index.repeat(x.str.len())
y = pd.Series(v, i)
y.groupby(level=0).unique().reindex(s.index)
0 [1]
1 [2]
2 NaN
3 [1]
4 [1, 2]
dtype: object
If you insist on None
rather than NaN
y.groupby(level=0).unique().reindex(s.index).mask(pd.isna, None)
0 [1]
1 [2]
2 None
3 [1]
4 [1, 2]
dtype: object
Setup
s = pd.Series([
['Apple', 'Banana'],
['Kiwi'],
None,
['Apple'],
['Banana', 'Kiwi']
])
d = {1: ['Apple', 'Banana'], 2: ['Kiwi']}