Split pandas dataframe based on groupby

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.

Answers:

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Method 1

gb = df.groupby('ZZ')    
[gb.get_group(x) for x in gb.groups]

Method 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')]

Method 3

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]

Method 4

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

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0