Question

这个问题是counting specific words across multiple columns in R的修改版本，但是增加了给某些列赋予不同权重的复杂性。如何使某些列计为1，另一些列计为0.5？

可复制的示例：

df <- data.frame(id=c(1, 2, 3, 4, 5), staple_1=c("potato",       "potato","rice","fruit","coffee"), 
             staple2_half1=c("yams","beer","potato","rice","yams"), 
             staple2_half2=c("potato","rice","yams","rice","yams"), 
             staple_3=c("rice","peanuts","fruit","fruit","rice"))
potato<-c("potato")
yams<-c("yams")
staples<-c("potato","cassava","rice","yams")

给予：

id staple_1 staple2_half1 staple2_half2 staple_3
 1   potato          yams        potato     rice
 2   potato          beer          rice  peanuts
 3     rice        potato          yams    fruit
 4    fruit          rice          rice    fruit
 5   coffee          yams          yams     rice

现在，我想创建2个额外的列来汇总“马铃薯”和“薯类”的计数，但是通过修改以下代码，以便“半”列（staple2_half1和Staple2_half2）中的任何计数仅计为0.5，而不是1.

使用原始答案的错误结果：

df$staples <- apply(df, 1, function(x) sum(staples %in% x))
df$potato<- apply(df, 1, function(x) sum(potato %in% x))
df$yams<- apply(df, 1, function(x) sum(yams %in% x))

礼物：

  id staple_1 staple2_half1 staple2_half2 staple_3 staples potato yams
  1   potato          yams        potato     rice       3      1    1
  2   potato          beer          rice  peanuts       2      1    0
  3     rice        potato          yams    fruit       3      1    1
  4    fruit          rice          rice    fruit       1      0    0
  5   coffee          yams          yams     rice       2      0    1

基于加权计数的所需结果：

  id staple_1 staple2_half1 staple2_half2 staple_3 staples potato yams
  1   potato          yams        potato     rice       3     1.5  0.5
  2   potato          beer          rice  peanuts      1.5      1    0
  3     rice        potato          yams    fruit       2     0.5  0.5
  4    fruit          rice          rice    fruit       1      0    0
  5   coffee          yams          yams     rice       2      0    1

Answer 1

如果您在apply的列上%in% df[, -1]函数，则会得到一个真值和假值矩阵。然后，要进行加权和，可以将此矩阵乘以权重向量。

words <- data.frame(staples, potato, yams)
weights <- 1 - 0.5*grepl('half', names(df[, -1]))

df[names(words)] <- 
  lapply(words, function(x) apply(df[, -1], 2, `%in%`, x) %*% weights)


df

#   id staple_1 staple2_half1 staple2_half2 staple_3 staples potato yams
# 1  1   potato          yams        potato     rice     3.0    1.5  0.5
# 2  2   potato          beer          rice  peanuts     1.5    1.0  0.0
# 3  3     rice        potato          yams    fruit     2.0    0.5  0.5
# 4  4    fruit          rice          rice    fruit     1.0    0.0  0.0
# 5  5   coffee          yams          yams     rice     2.0    0.0  1.0

apply(df1[, -1], 2, ...的输出示例

apply(df[, -1], 2, `%in%`, potato)
#      staple_1 staple2_half1 staple2_half2 staple_3
# [1,]     TRUE         FALSE          TRUE    FALSE
# [2,]     TRUE         FALSE         FALSE    FALSE
# [3,]    FALSE          TRUE         FALSE    FALSE
# [4,]    FALSE         FALSE         FALSE    FALSE
# [5,]    FALSE         FALSE         FALSE    FALSE

apply(df[, -1], 2, `%in%`, potato) %*% weights
#      [,1]
# [1,]  1.5
# [2,]  1.0
# [3,]  0.5
# [4,]  0.0
# [5,]  0.0

Answer 2

很多方法可以做到这一点，但这是使用tidyverse的一种方法。通过“收集”数据，使订书钉全都放在一栏中，我认为更容易应用正确的权重。

library(tidyverse)

df <- data.frame(id=c(1, 2, 3, 4, 5), staple_1=c("potato",       "potato","rice","fruit","coffee"), 
                 staple2_half1=c("yams","beer","potato","rice","yams"), 
                 staple2_half2=c("potato","rice","yams","rice","yams"), 
                 staple_3=c("rice","peanuts","fruit","fruit","rice"))
potato<-c("potato")
yams<-c("yams")
staples<-c("potato","cassava","rice","yams")


freqs <- df %>% 
  mutate_if(is.factor, as.character) %>% # avoids a warning about converting types
  gather("column", "item", -id) %>%
  mutate(scalar = if_else(str_detect(column, "half"), 0.5, 1)) %>%
  group_by(id) %>%
  summarize(
    staples = sum(item %in% staples * scalar),
    potato = sum(item %in% potato * scalar),
    yams = sum(item %in% yams * scalar)
  )

left_join(df, freqs, by = "id")
#>   id staple_1 staple2_half1 staple2_half2 staple_3 staples potato yams
#> 1  1   potato          yams        potato     rice     3.0    1.5  0.5
#> 2  2   potato          beer          rice  peanuts     1.5    1.0  0.0
#> 3  3     rice        potato          yams    fruit     2.0    0.5  0.5
#> 4  4    fruit          rice          rice    fruit     1.0    0.0  0.0
#> 5  5   coffee          yams          yams     rice     2.0    0.0  1.0

^{由reprex package（v0.2.1）于2018-12-11创建}

如何计算跨列的单词计数的加权和？

2 个答案: