熊猫:如何更快地申请数据框？

Question

假设在我计算列该只大熊猫例如C乘以A与B和float使用,如果一定条件满足apply一个lambda功能:

import pandas as pd
df = pd.DataFrame({'A':[1,2,3,4,5,6,7,8,9],'B':[9,8,7,6,5,4,3,2,1]})

df['C'] = df.apply(lambda x: x.A if x.B > 5 else 0.1*x.A*x.B, axis=1)

预期结果将是:

   A  B    C
0  1  9  1.0
1  2  8  2.0
2  3  7  3.0
3  4  6  4.0
4  5  5  2.5
5  6  4  2.4
6  7  3  2.1
7  8  2  1.6
8  9  1  0.9

问题是这段代码很慢,我需要在大约5600万行的数据帧上执行此操作.

%timeit上述lambda操作的结果是:

1000 loops, best of 3: 1.63 ms per loop

从我的大型数据帧上的计算时间和内存使用情况开始,我假设此操作在进行计算时使用中间序列.

我尝试以不同的方式制定它,包括使用临时列,但我提出的每个替代解决方案都更慢.

有没有办法以不同的更快的方式获得我需要的结果,例如通过使用numpy？

Answer 1

为了提高性能,您可能最好使用NumPy阵列并使用np.where-

a = df.values # Assuming you have two columns A and B
df['C'] = np.where(a[:,1]>5,a[:,0],0.1*a[:,0]*a[:,1])

运行时测试

def numpy_based(df):
    a = df.values # Assuming you have two columns A and B
    df['C'] = np.where(a[:,1]>5,a[:,0],0.1*a[:,0]*a[:,1])

计时 -

In [271]: df = pd.DataFrame(np.random.randint(0,9,(10000,2)),columns=[['A','B']])

In [272]: %timeit numpy_based(df)
1000 loops, best of 3: 380 µs per loop

In [273]: df = pd.DataFrame(np.random.randint(0,9,(10000,2)),columns=[['A','B']])

In [274]: %timeit df['C'] = df.A.where(df.B.gt(5), df[['A', 'B']].prod(1).mul(.1))
100 loops, best of 3: 3.39 ms per loop

In [275]: df = pd.DataFrame(np.random.randint(0,9,(10000,2)),columns=[['A','B']])

In [276]: %timeit df['C'] = np.where(df['B'] > 5, df['A'], 0.1 * df['A'] * df['B'])
1000 loops, best of 3: 1.12 ms per loop

In [277]: df = pd.DataFrame(np.random.randint(0,9,(10000,2)),columns=[['A','B']])

In [278]: %timeit df['C'] = np.where(df.B > 5, df.A, df.A.mul(df.B).mul(.1))
1000 loops, best of 3: 1.19 ms per loop

仔细看看

让我们仔细看看NumPy的数字运算能力,并与大熊猫进行比较 -

# Extract out as array (its a view, so not really expensive
#   .. as compared to the later computations themselves)

In [291]: a = df.values 

In [296]: %timeit df.values
10000 loops, best of 3: 107 µs per loop

案例#1:使用NumPy数组并使用numpy.where:

In [292]: %timeit np.where(a[:,1]>5,a[:,0],0.1*a[:,0]*a[:,1])
10000 loops, best of 3: 86.5 µs per loop

再次,分配到一个新列:df['C']也不会非常昂贵 -

In [300]: %timeit df['C'] = np.where(a[:,1]>5,a[:,0],0.1*a[:,0]*a[:,1])
1000 loops, best of 3: 323 µs per loop

案例#2:使用pandas数据帧并使用其.where方法(无NumPy)

In [293]: %timeit df.A.where(df.B.gt(5), df[['A', 'B']].prod(1).mul(.1))
100 loops, best of 3: 3.4 ms per loop

案例#3:使用pandas数据帧(没有NumPy数组),但使用numpy.where-

In [294]: %timeit np.where(df['B'] > 5, df['A'], 0.1 * df['A'] * df['B'])
1000 loops, best of 3: 764 µs per loop

案例#4:再次使用pandas数据帧(没有NumPy数组),但使用numpy.where-

In [295]: %timeit np.where(df.B > 5, df.A, df.A.mul(df.B).mul(.1))
1000 loops, best of 3: 830 µs per loop

Answer 2

纯粹pandas
使用pd.Series.where

df['C'] = df.A.where(df.B.gt(5), df[['A', 'B']].prod(1).mul(.1))

   A  B    C
0  1  9  1.0
1  2  8  2.0
2  3  7  3.0
3  4  6  4.0
4  5  5  2.5
5  6  4  2.4
6  7  3  2.1
7  8  2  1.6
8  9  1  0.9