[英]Pandas apply but only for rows where a condition is met
I would like to use Pandas df.apply
but only for certain rows我想使用 Pandas df.apply
但仅限于某些行
As an example, I want to do something like this, but my actual issue is a little more complicated:作为一个例子,我想做这样的事情,但我的实际问题有点复杂:
import pandas as pd
import math
z = pd.DataFrame({'a':[4.0,5.0,6.0,7.0,8.0],'b':[6.0,0,5.0,0,1.0]})
z.where(z['b'] != 0, z['a'] / z['b'].apply(lambda l: math.log(l)), 0)
What I want in this example is the value in 'a' divided by the log of the value in 'b' for each row, and for rows where 'b' is 0, I simply want to return 0.在这个例子中我想要的是'a'中的值除以每行'b'中的值的对数,对于'b'为0的行,我只想返回0。
The other answers are excellent, but I thought I'd add one other approach that can be faster in some circumstances – using broadcasting and masking to achieve the same result:其他答案非常好,但我想我会添加另一种在某些情况下可以更快的方法——使用广播和屏蔽来达到相同的结果:
import numpy as np
mask = (z['b'] != 0)
z_valid = z[mask]
z['c'] = 0
z.loc[mask, 'c'] = z_valid['a'] / np.log(z_valid['b'])
Especially with very large dataframes, this approach will generally be faster than solutions based on apply()
.特别是对于非常大的数据帧,这种方法通常比基于apply()
的解决方案更快。
You can just use an if statement in a lambda function.您可以在 lambda 函数中使用 if 语句。
z['c'] = z.apply(lambda row: 0 if row['b'] in (0,1) else row['a'] / math.log(row['b']), axis=1)
I also excluded 1, because log(1) is zero.我也排除了 1,因为 log(1) 为零。
Output:输出:
a b c
0 4 6 2.232443
1 5 0 0.000000
2 6 5 3.728010
3 7 0 0.000000
4 8 1 0.000000
Hope this helps.希望这可以帮助。 It is easy and readable它简单易读
df['c']=df['b'].apply(lambda x: 0 if x ==0 else math.log(x))
You can use a lambda with a conditional to return 0 if the input value is 0 and skip the whole where
clause:如果输入值为 0,则可以使用带有条件的 lambda 来返回 0 并跳过整个where
子句:
z['c'] = z.apply(lambda x: math.log(x.b) if x.b > 0 else 0, axis=1)
You also have to assign the results to a new column ( z['c']
).您还必须将结果分配给新列( z['c']
)。
Use np.where() which divides a
by the log of the value in b
if the condition is met and returns 0 otherwise:如果满足条件,则使用np.where()将a
除以b
中值的对数,否则返回 0:
import numpy as np
z['c'] = np.where(z['b'] != 0, z['a'] / np.log(z['b']), 0)
Output:输出:
a b c
0 4.0 6.0 2.232443
1 5.0 0.0 0.000000
2 6.0 5.0 3.728010
3 7.0 0.0 0.000000
4 8.0 1.0 inf
You can use the numpy
function where
:您可以使用numpy
函数where
:
np.where(z.b != 0, z.a / np.log(z.b), 0)
or pandas
methods mask
and where
:或pandas
方法mask
和where
:
z.b.mask(z.b != 0, other=z.a / np.log(z.b))
z.b.where(z.b == 0, other=z.a / np.log(z.b))
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