使用其他行中的值将函数应用于 Pandas 数据帧行
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Apply function to pandas dataframe row using values in other rows
提问by lukewitmer
I have a situation where I have a dataframe row to perform calculations with, and I need to use values in following (potentially preceding) rows to do these calculations (essentially a perfect forecast based on the real data set). I get each row from an earlier df.applycall, so I could pass the whole df along to the downstream objects, but that seems less than ideal based on the complexity of objects in my analysis.
我有一种情况，我有一个数据框行来执行计算，我需要使用后续（可能在前面）行中的值来进行这些计算（基本上是基于真实数据集的完美预测）。我从较早的df.apply调用中获取每一行，因此我可以将整个 df 传递给下游对象，但根据我分析中对象的复杂性，这似乎不太理想。
I found one closely related question and answer [1], but the problem is actually fundamentally different in the sense that I do not need the whole df for my calcs, simply the following xnumber of rows (which might matter for large dfs).
我发现了一个密切相关的问题和答案 [1]，但问题实际上是根本不同的，因为我不需要整个 df 来计算我的计算，只需要以下x行数（这对于大型 dfs 可能很重要）。
So, for example:
因此，例如：
df = pd.DataFrame([100, 200, 300, 400, 500, 600, 700, 800, 900, 1000],
columns=['PRICE'])
horizon = 3
I need to access values in the following 3 (horizon) rows in my row-wise df.applycall. How can I get a naive forecast of the next 3 data points dynamically in my row-wise apply calcs? e.g. for row the first row, where the PRICEis 100, I need to use [200, 300, 400]as a forecast in my calcs.
我需要horizon在我的按行df.apply调用中访问以下 3 ( ) 行中的值。如何在我的按行应用计算中动态获得接下来 3 个数据点的天真预测？例如，对于第一行，其中PRICE是100，我需要[200, 300, 400]在我的计算中用作预测。
[1] apply a function to a pandas Dataframe whose returned value is based on other rows
回答by lukewitmer
By getting the row's index inside of the df.applycall using row.name[1], you can generate the 'forecast' data relative to which row you are currently on. This is effectively a preprocessing step to put the 'forecast' onto the relevant row, or it could be done as part of the initial df.applycall if the df is available downstream.
通过df.apply使用row.name[1]在调用中获取行的索引，您可以生成与当前所在行相关的“预测”数据。这实际上是将“预测”放在相关行上的预处理步骤，或者df.apply如果 df 在下游可用，则它可以作为初始调用的一部分完成。
df = pd.DataFrame([100, 200, 300, 400, 500, 600, 700, 800, 900, 1000], columns=['PRICE'])
horizon = 3
df['FORECAST'] = df.apply(lambda x: [df['PRICE'][x.name+1:x.name+horizon+1]], axis=1)
Results in this:
结果如下：
PRICE FORECAST
0 100 [200, 300, 400]
1 200 [300, 400, 500]
2 300 [400, 500, 600]
3 400 [500, 600, 700]
4 500 [600, 700, 800]
5 600 [700, 800, 900]
6 700 [800, 900, 1000]
7 800 [900, 1000]
8 900 [1000]
9 1000 []
Which can be used in your row-wise df.applycalcs.
可以在您的行式df.apply计算中使用。
EDIT:If you want to strip the index from the resulting 'Forecast':
编辑：如果您想从结果“预测”中删除索引：
df['FORECAST'] = df.apply(lambda x: [df['PRICE'][x.name+1:x.name+horizon+1].reset_index(drop=True)], axis=1)
回答by piRSquared
You may find this useful as well.
您可能会发现这也很有用。
keys = range(horizon + 1)
pd.concat([df.shift(-i) for i in keys], axis=1, keys=keys)
0 1 2 3
PRICE PRICE PRICE PRICE
0 100 200.0 300.0 400.0
1 200 300.0 400.0 500.0
2 300 400.0 500.0 600.0
3 400 500.0 600.0 700.0
4 500 600.0 700.0 800.0
5 600 700.0 800.0 900.0
6 700 800.0 900.0 1000.0
7 800 900.0 1000.0 NaN
8 900 1000.0 NaN NaN
9 1000 NaN NaN NaN
if you assign the concatto df_c
如果你分配concat给df_c
keys = range(horizon + 1)
df_c = pd.concat([df.shift(-i) for i in keys], axis=1, keys=keys)
df_c.apply(lambda x: pd.Series([x[0].values, x[1:].values]), axis=1)
0 1
0 [100.0] [200.0, 300.0, 400.0]
1 [200.0] [300.0, 400.0, 500.0]
2 [300.0] [400.0, 500.0, 600.0]
3 [400.0] [500.0, 600.0, 700.0]
4 [500.0] [600.0, 700.0, 800.0]
5 [600.0] [700.0, 800.0, 900.0]
6 [700.0] [800.0, 900.0, 1000.0]
7 [800.0] [900.0, 1000.0, nan]
8 [900.0] [1000.0, nan, nan]
9 [1000.0] [nan, nan, nan]
