CODEWITHSUNDEEP
pandasdataframevectorization
el_Rinaldo
el_Rinaldo

Reputation: 1018

Best way to flatten dataframe based on values on column

I have to process a whole dataframe with some hundered thousands rows, but I can simplify it as below:

df = pd.DataFrame([
('a', 1, 1),
('a', 0, 0),
('a', 0, 1),
('b', 0, 0),
('b', 1, 0),
('b', 0, 1),
('c', 1, 1),
('c', 1, 0),
('c', 1, 0)
], columns=['A', 'B', 'C'])

print (df)

   A  B  C
0  a  1  1
1  a  0  0
2  a  0  1
3  b  0  0
4  b  1  0
5  b  0  1
6  c  1  1
7  c  1  0
8  c  1  0

My goal it to flatten the columns "B" and "C" based on the label they have in the "A" column

   A  B_1  B_2  B_3  C_1  C_2  C_3
0  a    1    0    0    1    0    1
3  b    0    1    0    0    0    1
6  c    1    1    1    1    0    0

The code I wrote gives the result I want, but it is pretty slow as it uses a simple for loop on the unique labels. The solution I see is to write some vectorized function that optimize my code. Anyone has some idea? Below I append the code.

added_col = ['B_1', 'B_2', 'B_3', 'C_1', 'C_2', 'C_3']

new_df = df.drop(['B', 'C'], axis=1).copy()
new_df = new_df.iloc[[x for x in range(0, len(df), 3)], :]
new_df = pd.concat([new_df,pd.DataFrame(columns=added_col)], sort=False)

for e, elem in new_df['A'].iteritems():
    new_df.loc[e, added_col] = df[df['A'] == elem].loc[:,['B','C']].T.values.flatten()

Upvotes: 16

Views: 4458

Answers (4)

piRSquared
piRSquared

Reputation: 294358

In an effort to improve performance, I've used numba and numpy assignment

from numba import njit

@njit
def f(i, vals, n, m, k):

  out = np.empty((n, k, m), vals.dtype)
  out.fill(0)

  c = np.zeros(n, np.int64)

  for j in range(len(i)):
    x = i[j]
    out[x, :, c[x]] = vals[j]
    c[x] += 1

  return out.reshape(n, m * k)


d0 = df.drop('A', 1)
cols = [*d0]

i, r = pd.factorize(df.A)

n = len(r)
m = np.bincount(i).max()
k = len(cols)

vals = d0.values

pd.DataFrame(
    f(i, vals, n, m, k),
    pd.Index(r, name='A'),
    [f"{c}_{i}" for c in cols for i in range(1, m + 1)]
).reset_index()

   A  B_1  B_2  B_3  C_1  C_2  C_3
0  a    1    0    0    1    0    1
1  b    0    1    0    0    0    1
2  c    1    1    1    1    0    0

Upvotes: 5

BENY
BENY

Reputation: 323306

Modify your index by using %

df.index=df.index%3+1
df.set_index('A',append=True,inplace=True)
newdf=df.unstack(level=0)
newdf.columns=newdf.columns.map('{0[0]}_{0[1]}'.format)
newdf
Out[291]: 
   B_1  B_2  B_3  C_1  C_2  C_3
A                              
a    1    0    0    1    0    1
b    0    1    0    0    0    1
c    1    1    1    1    0    0

Upvotes: 2

rafaelc
rafaelc

Reputation: 59274

Another approach using groupby and ravel()

>>> df.groupby('A')[['B','C']].apply(lambda s: pd.Series(s.T.values.ravel(), 
                                                         index=[f'{x}_{i}' for x in s.columns for i in range(1, len(s)+1)]))

    B_1 B_2 B_3 C_1 C_2 C_3
A                       
a   1   0   0   1   0   1
b   0   1   0   0   0   1
c   1   1   1   1   0   0

Upvotes: 2

akuiper
akuiper

Reputation: 214987

Here is one way:

# create a row number by group
df['rn'] = df.groupby('A').cumcount() + 1

# pivot the table
new_df = df.set_index(['A', 'rn']).unstack()

# rename columns
new_df.columns = [x + '_' + str(y) for (x, y) in new_df.columns]

new_df.reset_index()
#   A  B_1  B_2  B_3  C_1  C_2  C_3
#0  a    1    0    0    1    0    1
#1  b    0    1    0    0    0    1
#2  c    1    1    1    1    0    0

Upvotes: 15

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