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VistasCreate a new column and assign values to the first row by each group in Python Pandas
I have an existing pandas dataframe below:
id time c d
0 1 1 2 3
1 1 3 1 6
2 2 2 3 2
3 2 3 8 6
I also have values stored in a list such as:
list = [0.4, 0.6]
I want to create a new column in the existing dateframe and assign each list element in the first row for each group (id) such as:
id time c d new_col
0 1 1 2 3 0.4
1 1 3 1 6
2 2 2 3 2 0.6
3 2 3 8 6
3 Respuestas
Responde la pregunta0
Group the dataframe by id and use cumcount to create a sequential counter, then use boolean indexing with loc to assign the list values where the value of the counter is 0
lst = [0.4, 0.6]
df.loc[df.groupby('id').cumcount().eq(0), 'new_col'] = lst
id time c d new_col
0 1 1 2 3 0.4
1 1 3 1 6 NaN
2 2 2 3 2 0.6
3 2 3 8 6 NaN
0
Use Series.map with dictionary created by enumerate and set values only first per groups in Series.mask with Series.duplicated:
L=[0.4, 0.6]
df["new_col"] = df['id'].sub(1).map(dict(enumerate(L))).mask(df['id'].duplicated())
print (df)
id time c d new_col
0 1 1 2 3 0.4
1 1 3 1 6 NaN
2 2 2 3 2 0.6
3 2 3 8 6 NaN
L=[0.4, 0.6]
df["new_col"] = df['id'].sub(1).map(dict(enumerate(L))).mask(df['id'].duplicated(),'')
print (df)
id time c d new_col
0 1 1 2 3 0.4
1 1 3 1 6
2 2 2 3 2 0.6
3 2 3 8 6
If possible any groups in id, e.g. 10, 20 use GroupBy.ngroup:
L=[0.4, 0.6]
df["new_col"] = (df.groupby('id').ngroup().map(dict(enumerate(L)))
.mask(df['id'].duplicated(),''))
print (df)
id time c d new_col
0 10 1 2 3 0.4
1 10 3 1 6
2 20 2 3 2 0.6
3 20 3 8 6
0
Does this help solve your problem? (I can't see any benefit to only putting that value in the first occurrence of the group and not the other values)
mapping = {1: 0.4, 2: 0.6}
df["new_col"] = df['id'].map(mapping)
Result:
Note: If all of your ids are sequential integers starting at 1 and all of the values in your list are also in that correct order you could convert it to a mapping dict using:
mapping = {i+1: value for i, value in enumerate(your_list)}