如何使用pandas-python递归构造一列数据框? [英] How to constuct a column of data frame recursively with pandas-python?

问题描述

给出这样一个数据框df:

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

我希望在 df 中添加一列 diff，它的每一行都等于，比如说，该行中的 val减去前一行的diff，乘以0.4，再加上前一天的diff:

I wish to add a column diff to df, and each row of it equals to, let's say, the val in that row minus the diff in the previous row and multiply 0.4 and then add diff in the previous day:

diff = (val - diff_previousDay) * 0.4 + diff_previousDay

并且第一行的diff等于该行的val * 4.也就是说，预期的 df 应该是:

And the diff in the first row equals to val * 4 in that row. That is, the expected df should be:

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

我已经尝试过:

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) 

但是得到了这样的错误:

But got such as error:

TypeError: ("不支持的操作数类型 -: 'float' 和 'NoneType'", 'occurred at index 1')

TypeError: ("unsupported operand type(s) for -: 'float' and 'NoneType'", 'occurred at index 1')

你知道如何解决这个问题吗?提前谢谢你！

Do you know how to solve this problem? Thank you in advance!

推荐答案

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

计算的迭代性质，其中输入取决于先前步骤的结果，使矢量化变得复杂.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.

The iterative nature of the calculation where the inputs depend on results of previous steps complicates vectorization. You could perhaps use apply with a function that does the same calculation as the loop, but behind the scenes this would also be a loop.

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