I have two data frames containing float values.

1. The first one is a 1 column data frame that contains positions.
2. The second one is a matrix of ncol equal to the number of IDs and nrows equal to the nrow of the first data frame.

The idea is to create a new data frame of the same size as the second one. It needs to contain an equation between each value of the 1st data frame and each value for each column of the second one. The idea is that it will iterate over each row for one column before passing to the next one.

The ecuation would be something like df1 * df2 / len(df1) 1 Example data:

df1 = pd.DataFrame([10,20,30,40,50,60], columns=['POS'])
df2 = pd.DataFrame({"ID1" : [0,2,4,6,8,10] , "ID2" :[1,3,5,7,9,11]})
final = pd.DataFrame({"ID1" : [0, 5.714285714, 17.14285714, 34.28571429, 57.14285714, 85.71428571] , "ID2" :[1.428571429, 8.571428571, 21.42857143, 40, 64.28571429, 94.28571429]})

I think the the nested loop would be something like this, but I still can't get theanswer right. What I'm missing?

final = pd.DataFrame([])
for i in list(range(0,len(df1))):
          for j in list(range(0,len(df2))):
                        final.append(df2.iloc[i,j] * df1[0][i] / len(df1) 1)

In R the answer is this:

for (i in 1:nrow(df1)){
  for (j in 1:ncol(df2)){
    final[i,j] <- (df2[i,j] * df1[i,1]) / nrow(df1) 1
  }
}

CodePudding user response：

In a Pandorable way, you can do it with pandas.DataFrame.squeeze with pandas.DataFrame.mul :

result = df2.mul(df1.squeeze(), axis=0).div(len(df1) 1)

Output :

print(result)

         ID1        ID2
0   0.000000   1.428571
1   5.714286   8.571429
2  17.142857  21.428571
3  34.285714  40.000000
4  57.142857  64.285714
5  85.714286  94.285714

CodePudding user response：

Explicitly for df1 * df2 / (len(df1) * df2) calculation:

pd.DataFrame((df1.values * df2.values) / (len(df1) * df2.values), columns=df2.columns)

         ID1        ID2
0        NaN   1.666667
1   3.333333   3.333333
2   5.000000   5.000000
3   6.666667   6.666667
4   8.333333   8.333333
5  10.000000  10.000000