How to extract data based on number of duplicate rows?

Question

I have a dataset with mulitple rows. I want to create a new dataset based on number of duplicate rows for a column. For first dataset, i want a dataset of no duplicate rows, meaning only the row with one value. For second dataset, i want two duplicate rows and three duplicate rows,but only up to the second one. For third dataset, i want a dataset with just three duplicate rows. So as an example, i wrote codes to describe this situation. Let's Say i have a dataframe as such

x = {'column1': ['a','a','b','b','b','c','c','c','d'],
    'column2': [22000,25000,27000,350,0,3,5,4,312]
    }
df = pd.DataFrame(x, columns = ['column1', 'column2'])
print (df)

The first dataset should look like this:

x = {'column1': ['d'],
    'column2': [312]
    }
df = pd.DataFrame(x, columns = ['column1', 'column2'])
print (df)

Second dataset should look like this:

x = {'column1': ['a','a','b','b','c','c'],
    'column2': [22000,25000,27000,350,3,5]
    }
df = pd.DataFrame(x, columns = ['column1', 'column2'])
print (df)

Third dataset should look like this:

x = {'column1': ['b','b','b','c','c','c'],
    'column2': [27000,350,0,3,5,4]
    }
df = pd.DataFrame(x, columns = ['column1', 'column2'])
print (df)

How would i do this not manually?

Answer 1

First create counter Series by counts:

x = {'column1': ['a','a','b','b','b','c','c','c','d'],
    'column2': [22000,25000,27000,350,0,3,5,4,312]
    }
df = pd.DataFrame(x, columns = ['column1', 'column2'])
print (df)

s = df.groupby('column1')['column1'].transform('size')
#alternative
#s = df['column1'].map(df['column1'].value_counts())
print (s)
0    2
1    2
2    3
3    3
4    3
5    3
6    3
7    3
8    1
Name: column1, dtype: int64

And then on list comprehension create DataFrames with filter by GroupBy.head:

L = [df[s >= i].groupby(['column1',s]).head(i) if i > 1 else g for i, g in df.groupby(s)]
print (L[0])
print (L[1])
print (L[2])
  column1  column2
8       d      312
  column1  column2
0       a    22000
1       a    25000
2       b    27000
3       b      350
5       c        3
6       c        5
  column1  column2
2       b    27000
3       b      350
4       b        0
5       c        3
6       c        5
7       c        4

EDIT1: Because solution is more complicated custom function is created and for rows with 3 values is used GroupBy.nth:

def func(x, g):
    if x == 1:
        return g
    else:
        df1 = df[s >= x].groupby(['column1',s]).head(x)
        if x == 3:
            return (df1.groupby(['column1',s], group_keys=False)
                       .nth([0, -1])
                       .reset_index(level=1, drop=True)
                       .reset_index())

        else:
            return df1

L = [func(i, g) for i, g in df.groupby(s)]
print (L[0])
print (L[1])
print (L[2])
  column1  column2
8       d      312
  column1  column2
0       a    22000
1       a    25000
2       b    27000
3       b      350
5       c        3
6       c        5
  column1  column2
0       b    27000
1       b        0
2       c        3
3       c        4