问题描述
考虑下面的熊猫示例，在该示例中，如果使用apply和lambda函数满足特定条件，则将A与B和float相乘来计算列C:
Consider this pandas example where I'm calculating column C by multiplying A with B and a float if a certain condition is fulfilled using apply with a lambda function:
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
问题在于此代码运行缓慢，我需要在具有约5600万行的数据帧上执行此操作.
The problem is that this code is slow and I need to do this operation on a dataframe with around 56 million rows.
上述lambda操作的%timeit结果是:
The %timeit-result of the above lambda operation is:
1000 loops, best of 3: 1.63 ms per loop
从计算时间开始，以及在大型数据帧上执行此操作时的内存使用情况，我假定此操作在进行计算时会使用中间级数.
Going from the calculation time and also the memory usage when doing this on my large dataframe I presume this operation uses intermediary series while doing the calculations.
我试图用不同的方式来表达它，包括使用临时列，但是我想出的每一个替代解决方案都更慢.
I tried to formulate it in different ways including using temporary columns, but every alternative solution I came up with is even slower.
是否有一种方法可以以其他更快的方式获得所需的结果，例如通过使用numpy?
Is there a way to get the result I need in a different and faster way, e.g. by using numpy?
推荐答案
为了提高性能，您最好使用NumPy数组并使用np.where-
For performance, you might be better off working with NumPy array and using 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])
时间-
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
近距离观察
让我们仔细看看NumPy的数字处理能力，并与熊猫进行比较-
Let's take a closer look at NumPy's number crunching capability and compare with pandas into the mix -
# 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:
Case #1 : Work with NumPy array and use 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
同样，将其分配到新列中:df['C']也不会很昂贵-
Again, assigning into a new column : df['C'] would not be very expensive either -
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:处理pandas数据框并使用其.where方法(无NumPy)
Case #2 : Work with pandas dataframe and use its .where method (no 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数据框(无NumPy数组)，但使用numpy.where-
Case #3 : Work with pandas dataframe (no NumPy array), but use 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数据框(不使用NumPy数组)，但使用numpy.where-
Case #4 : Work with pandas dataframe again (no NumPy array), but use 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
这篇关于 pandas :如何更快地在数据框上应用?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！