How To Convert Vertical Pandas Table Of 2 Columns To Horizontal Table Based On Common Id Value In Python

df1 = pd.DataFrame({'foo': ['one', 'one', 'one', 'two', 'two', 'two'], 'bar': ['A', 'B', 'C', 'A', 'B', 'C']}) I would like to convert this to And the code I

Solution 1:

We can enumerate groups with groupby cumcount and use those as the pivot columns then add_prefix to the numerical values and reset_index to return the 'foo' values to the columns:

new_df = (
    df1.pivot_table(index='foo',
                    columns=df1.groupby('foo').cumcount() + 1,
                    values='bar', 
                    aggfunc='first')
        .add_prefix('val')
        .reset_index()
)

   foo val1 val2 val3
0  one    AB    C
1  two    AB    C

---

See how df1.groupby('foo').cumcount() + 1 makes the columns:

   foo  columns
0one1  # First instance of "one"
1one2  # Second instance of "one"
2one3  # Third instance of "one"
3  two        14  two        25  two        3

Code to generate the above DataFrame:

demo_df = pd.DataFrame({
    'foo': df1['foo'],
    'columns': df1.groupby('foo').cumcount() + 1
})

Solution 2:

Try:

df1.groupby([df1.groupby('foo').cumcount() + 1,
             'foo']).first()['bar'].unstack(0).add_prefix('val').reset_index()

Output:

   foo val1 val2 val3
0  one    AB    C
1  two    AB    C

Solution 3:

Another solution:

x = df1.pivot("foo", "bar", "bar")
x.columns = [f"var{i}"for i inrange(1, len(x.columns) + 1)]
x = x.reset_index()
print(x)

Prints:

   foo var1 var2 var3
0  one    A    B    C
1  two    A    B    C

Solution 4:

pd.pivot_table(df1,index='foo',columns=df1.groupby(df1.foo).cumcount()+1,
               values='bar',aggfunc=(lambda x:min(x)), dropna=True).add_prefix('val')

output:

    val1  val2  val 3
foo         
one   A     B     C
two   A     B     C

Solution 5:

Taking a cue from @HenryEcker's solution, a combination of groupby and pivot can get the result:

(df1.assign(counter = lambda df: df.groupby('foo').cumcount())
    .pivot('foo', 'counter', 'bar')
    .rename(columns = lambda col: f"val{col}"if col!=0else"val")
    .rename_axis(columns=None)
)
    val val1 val2
foo              
one   A    B    C
two   A    B    C