r 生成具有基于条件的值的列

Question

I have a dataset with 2 columns like this below

   w     p
   0.5   0.5267
   0.5   0.5239
   1.0   0.5267
   1.0   0.5267
   1.0   0.5267
   0.5   0.3870
   0.5   0.3566
   1.0   0.4914
   1.0   0.4914  
   0.125 0.5267 
   0.125 0.5239 
   0.125 0.3870 
   0.125 0.3844 
   0.125 0.4942 
   0.125 0.4914 
   0.125 0.3566 
   0.125 0.3540 

I am trying to create a third column based on this criteria below

Step1 : Start with Row 1 and check the value in Column w. 
        Row 1 column w is not 1  
Step2 : if the value in column w is not 1, then read the next value in column w. 
        Read the next column w value (Row 2)
Step3 : repeat step 2 until the sum of values from column w is 1.
        Column w row1 and row2 , 0.5 + 0.5 = 1
Step4 : Then read the corresponding values in column p.
        0.5267,  0.5239
Step5 : Multiply the values in column p with corresponding values in column w.
        0.5267*0.5 , 0.5239*0.5
Step6 : Add the values from Step 5
        0.5267*0.5 +  0.5239*0.5 
Step6 : Divide the values in column p with sum from step5.
        0.5267/(0.5267*0.5 +  0.5239*0.5) 
        0.5239/(0.5267*0.5 +  0.5239*0.5) 

The expected output is follows

   w     p        Result
   0.5   0.5267   0.5267/(0.5267*0.5 +  0.5239*0.5) 
   0.5   0.5239   0.5239/(0.5267*0.5 +  0.5239*0.5)
   1.0   0.5267   1
   1.0   0.5267   1
   1.0   0.5267   1 
   0.5   0.3870   0.3870/(0.3870*0.5 +  0.3566*0.5)
   0.5   0.3566   0.3566/(0.3870*0.5 +  0.3566*0.5)
   1.0   0.4914   1
   1.0   0.4914   1
   0.125 0.5267   0.5267/(0.5267*0.125 + 0.5239*0.125 + 0.3870*0.125 + 0.3844*0.125 + 0.4942*0.125 + 0.4914*0.125 + 0.3566*0.125 + 0.3540*0.125)
   0.125 0.5239   0.5239/(0.5267*0.125 + 0.5239*0.125 + 0.3870*0.125 + 0.3844*0.125 + 0.4942*0.125 + 0.4914*0.125 + 0.3566*0.125 + 0.3540*0.125)
   0.125 0.3870   0.3870/(0.5267*0.125 + 0.5239*0.125 + 0.3870*0.125 + 0.3844*0.125 + 0.4942*0.125 + 0.4914*0.125 + 0.3566*0.125 + 0.3540*0.125)
   0.125 0.3844   0.3844/(0.5267*0.125 + 0.5239*0.125 + 0.3870*0.125 + 0.3844*0.125 + 0.4942*0.125 + 0.4914*0.125 + 0.3566*0.125 + 0.3540*0.125)
   0.125 0.4942   0.4942/(0.5267*0.125 + 0.5239*0.125 + 0.3870*0.125 + 0.3844*0.125 + 0.4942*0.125 + 0.4914*0.125 + 0.3566*0.125 + 0.3540*0.125)
   0.125 0.4914   0.4914/(0.5267*0.125 + 0.5239*0.125 + 0.3870*0.125 + 0.3844*0.125 + 0.4942*0.125 + 0.4914*0.125 + 0.3566*0.125 + 0.3540*0.125)
   0.125 0.3566   0.3566/(0.5267*0.125 + 0.5239*0.125 + 0.3870*0.125 + 0.3844*0.125 + 0.4942*0.125 + 0.4914*0.125 + 0.3566*0.125 + 0.3540*0.125)
   0.125 0.3540   0.3540/(0.5267*0.125 + 0.5239*0.125 + 0.3870*0.125 + 0.3844*0.125 + 0.4942*0.125 + 0.4914*0.125 + 0.3566*0.125 + 0.3540*0.125)

I could do this using for loops ans ifelse statements, wondering if there is a more elegant way to accomplish this. Thanks.

Answer 1

We can create a group using cumulative sum of w values and calculate the result .

library(dplyr)

df %>%
  group_by(gr = ceiling(cumsum(w))) %>%
  mutate(result = p/sum(w * p)) %>%
  ungroup() %>%
  select(-gr)


# A tibble: 17 x 3
#       w     p result
#   <dbl> <dbl>  <dbl>
# 1 0.5   0.527  1.00 
# 2 0.5   0.524  0.997
# 3 1     0.527  1    
# 4 1     0.527  1    
# 5 1     0.527  1    
# 6 0.5   0.387  1.04 
# 7 0.5   0.357  0.959
# 8 1     0.491  1    
# 9 1     0.491  1    
#10 0.125 0.527  1.20 
#11 0.125 0.524  1.19 
#12 0.125 0.387  0.880
#13 0.125 0.384  0.874
#14 0.125 0.494  1.12 
#15 0.125 0.491  1.12 
#16 0.125 0.357  0.811
#17 0.125 0.354  0.805

---

This can be done in data.table as:

library(data.table)
setDT(df)[, result := p/sum(w * p), ceiling(cumsum(w))]