Convert elements of list in pandas series using a dict

Question

I have the following Pandas dataframe:

1    ["Apple", "Banana"]
2    ["Kiwi"]
3    None
4    ["Apple"]
5    ["Banana", "Kiwi"]

and the following dict:

{1: ["Apple", "Banana"],
2: ["Kiwi"]}

I would now like to map all the entries in the lists in my dataframe using the dictionary. The result should be the following:

1    [1]
2    [2]
3    None
4    [1]
5    [1, 2]

How can this be done most efficiently?

Answer 1

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"]}

Answer 2

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]

Answer 3

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']}