根据列上的值展平数据框的最佳方法

Question

我必须使用一些数千行来处理整个数据帧,但我可以简化如下:

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

我的目标是根据它们在"A"列中的标签来展平"B"和"C"列

   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

我写的代码给出了我想要的结果,但它很慢,因为它在唯一标签上使用了一个简单的for循环.我看到的解决方案是编写一些优化我的代码的矢量化函数.有人有想法吗？下面我附上代码.

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()

Answer 1

这是一种方式:

# 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

Answer 2

为了提高性能,我使用了numba和numpy赋值

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