如何用pandas-python递归地构造一列数据帧？

Question

给出这样一个数据框df:

id_      val     
11111    12
12003    22
88763    19
43721    77
...

我想添加一列diff到df,并且它的每一行等于,我们说了,val在该行减去diff上一行和乘0.4,然后加入diff前一天:

diff = (val - diff_previousDay) * 0.4 + diff_previousDay

并且diff第一行中的等于val * 4该行.也就是说,预期df应该是:

id_      val     diff   
11111    12      4.8
12003    22      11.68
88763    19      14.608
43721    77      ...

我试过了:

mul = 0.4
df['diff'] = df.apply(lambda row: (row['val'] - df.loc[row.name, 'diff']) * mul + df.loc[row.name, 'diff'] if int(row.name) > 0 else row['val'] * mul, axis=1) 

但得到如错误:

TypeError :("不支持的操作数类型 - :'float'和'NoneType'",'发生在索引1')

你知道如何解决这个问题吗？先感谢您!

Answer 1

您可以使用：

df.loc[0, 'diff'] = df.loc[0, 'val'] * 0.4

for i in range(1, len(df)):
    df.loc[i, 'diff'] = (df.loc[i, 'val'] - df.loc[i-1, 'diff']) * 0.4  + df.loc[i-1, 'diff']

print (df)
     id_  val     diff
0  11111   12   4.8000
1  12003   22  11.6800
2  88763   19  14.6080
3  43721   77  39.5648

输入依赖于先前步骤的结果的计算的迭代性质使矢量化复杂化。您也许可以将 apply 与执行与循环相同的计算的函数一起使用，但在幕后这也将是一个循环。

Answer 2

递归函数不容易矢量化。但是，您可以使用numba. 这应该比常规循环更可取。

from numba import jit

@jit(nopython=True)
def foo(val):
    diff = np.zeros(val.shape)
    diff[0] = val[0] * 0.4
    for i in range(1, diff.shape[0]):
        diff[i] = (val[i] - diff[i-1]) * 0.4 + diff[i-1]
    return diff

df['diff'] = foo(df['val'].values)

print(df)

     id_  val     diff
0  11111   12   4.8000
1  12003   22  11.6800
2  88763   19  14.6080
3  43721   77  39.5648