在 pandas 数据帧中向量化条件分配 [英] vectorize conditional assignment in pandas dataframe

问题描述

如果我的数据框df具有列x，并且想基于x的值创建列y，请在伪代码中使用此列:

 if df['x'] <-2 then df['y'] = 1 
 else if df['x'] > 2 then df['y']= -1 
 else df['y'] = 0

我将如何实现?我认为np.where是执行此操作的最佳方法，但不确定如何正确编码.

解决方案

一种简单的方法是先分配默认值，然后执行2次loc调用:

In [66]:

df = pd.DataFrame({'x':[0,-3,5,-1,1]})
df
Out[66]:
   x
0  0
1 -3
2  5
3 -1
4  1

In [69]:

df['y'] = 0
df.loc[df['x'] < -2, 'y'] = 1
df.loc[df['x'] > 2, 'y'] = -1
df
Out[69]:
   x  y
0  0  0
1 -3  1
2  5 -1
3 -1  0
4  1  0

如果您想使用np.where，则可以使用嵌套的np.where:

In [77]:

df['y'] = np.where(df['x'] < -2 , 1, np.where(df['x'] > 2, -1, 0))
df
Out[77]:
   x  y
0  0  0
1 -3  1
2  5 -1
3 -1  0
4  1  0

因此，在这里我们将第一个条件定义为x小于-2，返回1，然后有另一个np.where测试另一个条件，其中x大于2并返回-1，否则返回0

时间

In [79]:

%timeit df['y'] = np.where(df['x'] < -2 , 1, np.where(df['x'] > 2, -1, 0))

1000 loops, best of 3: 1.79 ms per loop

In [81]:

%%timeit
df['y'] = 0
df.loc[df['x'] < -2, 'y'] = 1
df.loc[df['x'] > 2, 'y'] = -1

100 loops, best of 3: 3.27 ms per loop

因此，对于此示例数据集，np.where方法的速度是其两倍快

If I have a dataframe df with column x and want to create column y based on values of x using this in pseudo code:

 if df['x'] <-2 then df['y'] = 1 
 else if df['x'] > 2 then df['y']= -1 
 else df['y'] = 0

How would I achieve this? I assume np.where is the best way to do this but not sure how to code it correctly.

解决方案

One simple method would be to assign the default value first and then perform 2 loc calls:

In [66]:

df = pd.DataFrame({'x':[0,-3,5,-1,1]})
df
Out[66]:
   x
0  0
1 -3
2  5
3 -1
4  1

In [69]:

df['y'] = 0
df.loc[df['x'] < -2, 'y'] = 1
df.loc[df['x'] > 2, 'y'] = -1
df
Out[69]:
   x  y
0  0  0
1 -3  1
2  5 -1
3 -1  0
4  1  0

If you wanted to use np.where then you could do it with a nested np.where:

In [77]:

df['y'] = np.where(df['x'] < -2 , 1, np.where(df['x'] > 2, -1, 0))
df
Out[77]:
   x  y
0  0  0
1 -3  1
2  5 -1
3 -1  0
4  1  0

So here we define the first condition as where x is less than -2, return 1, then we have another np.where which tests the other condition where x is greater than 2 and returns -1, otherwise return 0

timings

In [79]:

%timeit df['y'] = np.where(df['x'] < -2 , 1, np.where(df['x'] > 2, -1, 0))

1000 loops, best of 3: 1.79 ms per loop

In [81]:

%%timeit
df['y'] = 0
df.loc[df['x'] < -2, 'y'] = 1
df.loc[df['x'] > 2, 'y'] = -1

100 loops, best of 3: 3.27 ms per loop

So for this sample dataset the np.where method is twice as fast

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