CODEWITHSUNDEEP
rdplyrtidyverse
Henrique
Henrique

Reputation: 146

Mapping values based on a dictionary in R

I'm looking for an algorithm to map the values in each column based on a chosen dictionary.

Here's an example:

Artificial data:

df = data.frame(
  sex = c(0,0,0,1,0,1,0,1,1,1,0),
  icu = c(1,1,0,1,0,1,0,1,1,1,1),
  niv = c(0,1,0,1,0,1,0,0,0,1,0),
  mv = c(1,0,0,1,1,1,0,0,0,1,0),
  o2 = c(1,0,0,1,0,1,0,0,0,1,0)
)

I have two dictionaries. The goal is to create a new_column based on the values of each dictionary. In the end, sum row-wise the values of this set of columns and save them in a new column.

Dictionaries

dict1 <- list(
  sex = 0,
  icu = 2,
  niv = 1,
  mv = 3,
  o2 = 2
)

dict2 <- list(
  sex = 3,
  icu = 4,
  niv = 2,
  mv = 6,
  o2 = 1
)

I managed to operationalize this using the following algorithm. But it is not scalable to N variables.

Current solution:

set_index <- function(dataset, dict){
  temp = dataset
  temp$score = rep(0, times=nrow(temp))
  for(row in seq_len(nrow(temp))){
    for(c in names(dict2)){
      if(temp[row, c] == 1){
        temp[row, "score"] = temp[row, "score"] + dict2[[c]]
      }
    }
  }
  return(temp)
}

dataset <- set_index(dataset, dict2)

dataset$score <- tidyr::replace_na(data = dataset$score, 0)

I have a solution in Python, but I couldn't transport it to R.

import numpy as np
def ReplaceAndSumValues(dataset, dict):
    out = df.transform(lambda x, dct: np.where(x, dct[x.name][1], dct[x.name][0]), dct=d)
    return out.assign(sum=out.sum(axis=1))

Upvotes: 1

Views: 708

Answers (1)

Gregor Thomas
Gregor Thomas

Reputation: 145805

This is just matrix multiplication:

foo = function(df, dict) {
  df = df[names(dict)]
  as.matrix(df) %*% unlist(dict)
}

df$result1 = foo(df, dict1)
df$result2 = foo(df, dict2)
df
#    sex icu niv mv o2 result1 result2
# 1    0   1   0  1  1       7      11
# 2    0   1   1  0  0       3       6
# 3    0   0   0  0  0       0       0
# 4    1   1   1  1  1       8      16
# 5    0   0   0  1  0       3       6
# 6    1   1   1  1  1       8      16
# 7    0   0   0  0  0       0       0
# 8    1   1   0  0  0       2       7
# 9    1   1   0  0  0       2       7
# 10   1   1   1  1  1       8      16
# 11   0   1   0  0  0       2       4

Upvotes: 1

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