如何根据该行中的条件为另一个数据框中的每一行选择一个数据框？

Question

I have the following 2 dataframes:

df1
   x  p  s
0  2  1  1
1  4  2  1
2  6  1  3
3  8  2  4

df2
     ts   1   2
0  1000  45  44
1  1001  46  46
2  1002  47  46
3  1003  48  48
4  1004  49  48
5  1005  50  50
6  1006  51  50
7  1007  52  52
8  1008  53  52

I would like to create a 3rd data frame with the same number of rows as df1 using values in df2 but based on the column values in df1. For example, for the first row of df1, I want to get every 'p' row from the 's' column up until the 'x' index in df2. I know how to do that using df.apply() as shown below but it is too slow of an operation for the program I am writing.

def foo(row):
    return str(df2[row['p']].iloc[0:row['x']+1:row['s']].to_list())

df3 = df1.apply(lambda x: foo(x), axis=1)
df3
0            [45, 46, 47]
1    [44, 46, 46, 48, 48]
2            [45, 48, 51]
3            [44, 48, 52]

Answer 1

I'm not sure how large the datasets are, but try the following

# We need to do "CROSS JOIN" so we add a dummy key to both datasets to allow this
df1["temp_key"] = 0
df2["temp_key"] = 0

# Next we need to shift the index into the DataFrame and call it row_number
df2 = df2.reset_index().rename(columns={"index":"row_number"})

# Now we perform the "CROSS JOIN"
df = df1.merge(df2, on="temp_key").drop(columns=["temp_key"])

df1 should now have 7 columns: ["x", "p", "s", "ts", "1", "2", "row_number"]

# We can now apply the 'x' logic
df = df[df["row_number"] <= df["x"]]

# And then the 's' logic
df = df[df["row_number"].mod(df["p"]) == 0]

# Next we chose the appropriate column based on the p value
df["value"] = df["1"]
df.iloc[df["p"] == 2, "value"] = df["2"]

# Finally we can group the DataFrame by the 'x' value and create the lists
# Note: I've made the assumption that x is unique in df1
df = df.groupby(["x"])["value"].apply(list).reset_index()

This should return a DataFrame with two columns: ["x", "value"] with x corresponding to the x value in df1 and value being the list of values similar to df3 in your example.