基于Pandas中带条件的行的新列

original_df = [100, 200, 300, 400]
weights_df = [10, 20, 30, 40]
conditions_df = [1, 2, 3, 4]

df = pd.DataFrame({'x':original_df, 'y':weights_df, 'z':conditions_df})
def div(x, y, z):
    if z > 2:
        return float(x/y)
    else:
        return float(-1*x/y)

df['new_feature'] = df.apply(lambda p: div(p['x'], p['y'], p['z']), axis=1)

这是解决问题的一种方法。如果您的条件_df包含“+”/“-”,则可以在中更改条件 def div(x, y, z) 照着

您可以使用 numpy.where 对于条件掩码:

#data from lisa answer
#df = pd.DataFrame({'x':original_df, 'y':weights_df, 'z':conditions_df})

df['new_feature'] = df['x'] / df['y'] *  np.where(df['z'] > 2, 1, -1)
print (df)
     x   y  z  new_feature
0  100  10  1        -10.0
1  200  20  2        -10.0
2  300  30  3         10.0
3  400  40  4         10.0

时间安排 :

#4k rows
df = pd.concat([df]*1000).reset_index(drop=True)

#lisa answer
In [95]: %timeit df['new_feature1'] = df.apply(lambda p: div(p['x'], p['y'], p['z']), axis=1)
10 loops, best of 3: 123 ms per loop

In [96]: %timeit df['new_feature2'] = df['x'] / df['y'] *  np.where(df['z'] > 2, 1, -1)
1000 loops, best of 3: 595 µs per loop