Question

我创建了一个函数，该函数基于三个包含字符A到F的字符串来计算数字分数。通过确定每个字符串的最频繁字符来计算中间分数。最终分数由多数投票计算得出。如果所有三个中间分数都不相同，则最终数字分数将是所有中间分数的平均值。我知道该功能尚未优化-有几种方法可以更好地编写它。任何建议都值得欢迎！

我从以下StackOverflow问题中获得了部分代码... Character frequency in a string

fncLetterValue <- function(strLetter)
{
  intValue = 9
  if(strLetter=="A") intValue <- 1
  if(strLetter=="B") intValue <- 2
  if(strLetter=="C") intValue <- 3
  if(strLetter=="D") intValue <- 4
  if(strLetter=="E") intValue <- 5
  if(strLetter=="F") intValue <- 6
  return(intValue)
}

fncScore <- function(str1, str2, str3)
{
  # determine frequency of letters in str1
  A1 <- sum(charToRaw(str1)==charToRaw("A"))
  B1 <- sum(charToRaw(str1)==charToRaw("B"))
  C1 <- sum(charToRaw(str1)==charToRaw("C"))
  D1 <- sum(charToRaw(str1)==charToRaw("D"))
  E1 <- sum(charToRaw(str1)==charToRaw("E"))
  F1 <- sum(charToRaw(str1)==charToRaw("F"))
  Max1 <- max(A1,B1,C1,D1,E1,F1) # determine the maximum frequency
  Score1 <- 9
  if(A1==Max1) Score1 <- 1
  if(B1==Max1) Score1 <- 2
  if(C1==Max1) Score1 <- 3
  if(D1==Max1) Score1 <- 4
  if(E1==Max1) Score1 <- 5
  if(F1==Max1) Score1 <- 6

  # determine frequency of letters in str2
  A2 <- sum(charToRaw(str2)==charToRaw("A"))
  B2 <- sum(charToRaw(str2)==charToRaw("B"))
  C2 <- sum(charToRaw(str2)==charToRaw("C"))
  D2 <- sum(charToRaw(str2)==charToRaw("D"))
  E2 <- sum(charToRaw(str2)==charToRaw("E"))
  F2 <- sum(charToRaw(str2)==charToRaw("F"))
  Max2 <- max(A2,B2,C2,D2,E2,F2) # determine the maximum frequency
  Score2 <- 9
  if(A2==Max2) Score2 <- 1
  if(B2==Max2) Score2 <- 2
  if(C2==Max2) Score2 <- 3
  if(D2==Max2) Score2 <- 4
  if(E2==Max2) Score2 <- 5
  if(F2==Max2) Score2 <- 6

  # determine frequency of letters in str3
  A3 <- sum(charToRaw(str3)==charToRaw("A"))
  B3 <- sum(charToRaw(str3)==charToRaw("B"))
  C3 <- sum(charToRaw(str3)==charToRaw("C"))
  D3 <- sum(charToRaw(str3)==charToRaw("D"))
  E3 <- sum(charToRaw(str3)==charToRaw("E"))
  F3 <- sum(charToRaw(str3)==charToRaw("F"))
  Max3 <- max(A3,B3,C3,D3,E3,F3) # determine the maximum frequency
  Score3 <- 9
  if(A3==Max3) Score3 <- 1
  if(B3==Max3) Score3 <- 2
  if(C3==Max3) Score3 <- 3
  if(D3==Max3) Score3 <- 4
  if(E3==Max3) Score3 <- 5
  if(F3==Max3) Score3 <- 6

  # get final score by majority voting
  dblFinalScore <- 9
  if(Score1==Score2 | Score1==Score3) dblFinalScore <- Score1
  if(Score2==Score1 | Score2==Score3) dblFinalScore <- Score2
  if(Score3==Score1 | Score3==Score2) dblFinalScore <- Score3
  if(dblFinalScore==9) dblFinalScore <- mean(c(Score1,Score2,Score3))

  return(dblFinalScore)
}

# read csv
setwd("~/Downloads")
df <- read.csv("CompositeScore.csv", header = TRUE)

df$score <- fncScore(df$Vector1, df$Vector2, df$Vector3) #THIS LINE GIVES AN ERROR!

该功能确实可以在控制台中处理一组字符串... 例如fncScore（“ AAAABBBBBBBBB”，“ ABBCCCCCCCCCC”，“ FFFFFFFFFFF”）

[1] 3.666667

但是，相同的功能不适用于数据框。我收到以下错误： “ charToRaw（str1）中的错误：参数必须是长度为1的字符向量”

我希望在数据框中添加一个带有最终得分的新列。

我在做什么错了？

Answer 1

如果我正确理解了您的计分系统，我认为这可以满足您使用tidyverse ...

的需求

library(tidyverse)

df <- read_csv("CompositeScore.csv")

scores <- function(x){   #function to identify most common letter in each string
  map_int(x, ~which.max(str_count(., LETTERS[1:6])))
}

df <- df %>% 
  gather(key = Vector, value = value, -Observation) %>%  #change to 'long' format
  mutate(score = scores(value)) %>%                      #calculate scores
  group_by(Observation) %>%                              #group by Observation for next line
  mutate(score = ifelse(sum(score == median(score)) > 1, #if two the same
                        median(score),                   #then median
                        mean(score))) %>%                #otherwise mean
  spread(key = Vector, value = value)                    #back to wide format

head(df)

  Observation score Vector1      Vector2      Vector3     
1           1  3.33 CCEDDBEACBAD ADAABEEAEADD ACEFBAFDFDCB
2           2  3.33 ECBDEFACDAEA AFDEECDBEDFF EBEFCCEAEDFB
3           3  5    BDDDBBAFDFFF BBEEDEDBDCAE FFBADEEFCFFF
4           4  4    FDDFDEFBCBBA FECEEFDDCDAF FDFCDFEFBBCE
5           5  2    DBBEEDCBEECB CBFCAAFEBBCD FCFFBEBEEBDA
6           6  6    CBAEEEDBEBDF DCABCEAEDFFF CEFFFDBCADFC

复合评分功能在数据框架上失败，但仅适用于一组值

1 个答案: