I want to split the following dataframe based on column ZZ
df =
N0_YLDF ZZ MAT
0 6.286333 2 11.669069
1 6.317000 6 11.669069
2 6.324889 6 11.516454
3 6.320667 5 11.516454
4 6.325556 5 11.516454
5 6.359000 6 11.516454
6 6.359000 6 11.516454
7 6.361111 7 11.516454
8 6.360778 7 11.516454
9 6.361111 6 11.516454
As output, I want a new DataFrame with the N0_YLDF column split into 4, one new column for each unique value of ZZ. How do I go about this? I can do groupby, but do not know what to do with the grouped object.
gb = df.groupby('ZZ')
[gb.get_group(x) for x in gb.groups]
2
There is another alternative as the groupby returns a generator we can simply use a list-comprehension to retrieve the 2nd value (the frame).
dfs = [x for _, x in df.groupby('ZZ')]
2
In R there is a dataframe method called split. This is for all the R users out there:
def split(df, group):
gb = df.groupby(group)
return [gb.get_group(x) for x in gb.groups]
5
Store them in a dict, which allows you access to the group DataFrames based on the group keys.
d = dict(tuple(df.groupby('ZZ')))
d[6]
# N0_YLDF ZZ MAT
#1 6.317000 6 11.669069
#2 6.324889 6 11.516454
#5 6.359000 6 11.516454
#6 6.359000 6 11.516454
#9 6.361111 6 11.516454
If you need only a subset of the DataFrame, in this case just the 'NO_YLDF' Series, you can modify the dict comprehension.
d = dict((idx, gp['N0_YLDF']) for idx, gp in df.groupby('ZZ'))
d[6]
#1 6.317000
#2 6.324889
#5 6.359000
#6 6.359000
#9 6.361111
#Name: N0_YLDF, dtype: float64
0
You can iterate over unique values and get groups using loc or query:
[df.loc[df['ZZ'] == i] for i in df['ZZ'].unique()]
or
[df.query('ZZ == @i') for i in df['ZZ'].unique()]
Adding to user qwwqwwq answer:
gb = df.groupby('ZZ')
df_six = gb.get_group("6") #to create another dataframe with ZZ = 6
df_one = gb.get_group("7") #to create another dataframe with ZZ = 7