CODEWITHSUNDEEP
r
u31889
u31889

Reputation: 351

Subset a table by columns and rows using a named vector in R

Using the diamonds dataset (from the ggplot2 library) as an example, I am trying to subset this table by columns and rows based on a vector of named elements (the names of the vector should be used to subset by columns and the corresponding vector elements by rows).

library(ggplot2)
diamonds
# A tibble: 53,940 x 10
   carat cut       color clarity depth table price     x     y     z
   <dbl> <ord>     <ord> <ord>   <dbl> <dbl> <int> <dbl> <dbl> <dbl>
 1 0.23  Ideal     E     SI2      61.5    55   326  3.95  3.98  2.43
 2 0.21  Premium   E     SI1      59.8    61   326  3.89  3.84  2.31
 3 0.23  Good      E     VS1      56.9    65   327  4.05  4.07  2.31
 4 0.290 Premium   I     VS2      62.4    58   334  4.2   4.23  2.63
 5 0.31  Good      J     SI2      63.3    58   335  4.34  4.35  2.75
 6 0.24  Very Good J     VVS2     62.8    57   336  3.94  3.96  2.48
 7 0.24  Very Good I     VVS1     62.3    57   336  3.95  3.98  2.47
 8 0.26  Very Good H     SI1      61.9    55   337  4.07  4.11  2.53
 9 0.22  Fair      E     VS2      65.1    61   337  3.87  3.78  2.49
10 0.23  Very Good H     VS1      59.4    61   338  4     4.05  2.39
# … with 53,930 more rows

myVector <- c(cut="Ideal", cut="Good", color="E", color="J")
myVector
    cut     cut   color   color 
"Ideal"  "Good"     "E"     "J"

What I intend to do, would be something like follows but using myVector:

library(dplyr)
diamonds %>% subset(., (cut=="Ideal" | cut=="Good") & (color=="E" | color=="J")) %>%
select(cut, color)

Upvotes: 6

Views: 136

Answers (6)

r.user.05apr
r.user.05apr

Reputation: 5456

If you don't use the vector which has characters (and not expressions) as names, it gets a lot easier and maybe more readable:

library(ggplot2)
library(tidyverse)
library(rlang)

my_filter <- function(d, x, selection) {
  cmd <- map2(x, selection, ~quo(`%in%`(!!.x, !!.y))) # create filter expression
  d %>%
    filter(!!!cmd) %>% # filter
    select(!!!x) # select columns cut and color (in this case)
}    
diamonds %>%
  my_filter(x = vars(cut, color),
            sel = list(c("Ideal", "Good"), c("E", "J")))

# # A tibble: 6,039 x 2
# cut   color
# <ord> <ord>
#   1 Ideal E    
# 2 Good  E    
# 3 Good  J    
# 4 Good  J    
# 5 Ideal J    
# 6 Ideal J    
# 7 Good  J    
# 8 Good  J    
# 9 Good  E    
# 10 Ideal J    
# # ... with 6,029 more rows

Upvotes: 0

Rui Barradas
Rui Barradas

Reputation: 76663

Starting with the split idea of ThomasIsCoding, slightly changed, here is a base R solution based on having Reduce/Map created a logical index.

v <- split(unname(myVector), names(myVector))
i <- Reduce('&', Map(function(x, y){x %in% y}, diamonds[names(v)], v))
diamonds[i, ]
## A tibble: 6,039 x 10
#   carat cut   color clarity depth table price     x     y     z
#   <dbl> <ord> <ord> <ord>   <dbl> <dbl> <int> <dbl> <dbl> <dbl>
# 1  0.23 Ideal E     SI2      61.5    55   326  3.95  3.98  2.43
# 2  0.23 Good  E     VS1      56.9    65   327  4.05  4.07  2.31
# 3  0.31 Good  J     SI2      63.3    58   335  4.34  4.35  2.75
# 4  0.3  Good  J     SI1      64      55   339  4.25  4.28  2.73
# 5  0.23 Ideal J     VS1      62.8    56   340  3.93  3.9   2.46
# 6  0.31 Ideal J     SI2      62.2    54   344  4.35  4.37  2.71
# 7  0.3  Good  J     SI1      63.4    54   351  4.23  4.29  2.7 
# 8  0.3  Good  J     SI1      63.8    56   351  4.23  4.26  2.71
# 9  0.23 Good  E     VS1      64.1    59   402  3.83  3.85  2.46
#10  0.33 Ideal J     SI1      61.1    56   403  4.49  4.55  2.76
## ... with 6,029 more rows

Package dplyr

The code above can be written as a function and used in dplyr::filter.

# Input:
# X - a data set to be filtered
# values - a named list
values_in <- function(X, values){
  v <- split(unname(values), names(values))
  i <- Reduce('&', Map(function(x, y){x %in% y}, X[names(v)], v))
  i
}

diamonds %>% filter( values_in(., myVector) )

The output is the same as above and, therefore, omited.

Upvotes: 3

u31889
u31889

Reputation: 351

Using both approaches proposed by @Roman (generating all combinations of vector element and joining) and @ThomaslsCoding (splitting the vector) seems to do the trick:

data.frame(split(myVector, names(myVector))) %>% 
expand.grid() %>% 
inner_join(diamonds[,unique(names(myVector))])

Upvotes: 2

jay.sf
jay.sf

Reputation: 73702

Similar idea to @ThomasIsCoding's, just in base R.

al <- split(myVector, names(myVector))

res <- with(diamonds, diamonds[eval(parse(text=paste(sapply(names(al), function(x) 
  paste0(x, " %in% ", "al[['", x, "']]")), collapse=" & "))), ])

unique(res$cut)
# [1] Ideal Good 
# Levels: Fair < Good < Very Good < Premium < Ideal
unique(res$color)
# [1] E J
# Levels: D < E < F < G < H < I < J

Upvotes: 1

ThomasIsCoding
ThomasIsCoding

Reputation: 102710

I am not sure if you want something like below

u <- split(myVector,names(myVector))
eval(str2expression(sprintf("diamonds %%>%% filter(%s)",paste0(sapply(names(u),function(x) paste0(x," %in% u$",x)),collapse = " & "))))

such that

> eval(str2expression(sprintf("diamonds %%>%% filter(%s)",paste0(sapply(names(u),function(x) paste0(x," %in% u$",x)),collapse = " & "))))
# A tibble: 6,039 x 10
   carat cut   color clarity depth table price     x     y     z
   <dbl> <ord> <ord> <ord>   <dbl> <dbl> <int> <dbl> <dbl> <dbl>
 1  0.23 Ideal E     SI2      61.5    55   326  3.95  3.98  2.43
 2  0.23 Good  E     VS1      56.9    65   327  4.05  4.07  2.31
 3  0.31 Good  J     SI2      63.3    58   335  4.34  4.35  2.75
 4  0.3  Good  J     SI1      64      55   339  4.25  4.28  2.73
 5  0.23 Ideal J     VS1      62.8    56   340  3.93  3.9   2.46
 6  0.31 Ideal J     SI2      62.2    54   344  4.35  4.37  2.71
 7  0.3  Good  J     SI1      63.4    54   351  4.23  4.29  2.7
 8  0.3  Good  J     SI1      63.8    56   351  4.23  4.26  2.71
 9  0.23 Good  E     VS1      64.1    59   402  3.83  3.85  2.46
10  0.33 Ideal J     SI1      61.1    56   403  4.49  4.55  2.76
# ... with 6,029 more rows

Upvotes: 3

Roman
Roman

Reputation: 17668

you can try

my_vec_cut = myVector[names(myVector) == "cut"]
my_vec_color = myVector[names(myVector) == "color"]

I splitted the vector in two since you filter for two columns using and and or

diamonds %>% 
  filter(.data[[unique(names(my_vec_cut))]] %in%  my_vec_cut & .data[[unique(names(my_vec_color))]] %in%  my_vec_color)

A general way would be a joining approach. First you build all required combinations from your vector, then you left join the data.

library(tidyverse)
tibble(a=names(myVector), b=myVector) %>%
    group_by(a) %>% 
    mutate(n=1:n()) %>% 
    pivot_wider(names_from = a, values_from=b) %>%
    select(-n) %>% 
    complete(cut, color) 
# A tibble: 4 x 2
  cut   color
  <chr> <chr>
1 Good  E    
2 Good  J    
3 Ideal E    
4 Ideal J    

# now left_joining:
tibble(a=names(myVector), b=myVector) %>%
  group_by(a) %>% 
  mutate(n=1:n()) %>% 
  pivot_wider(names_from = a, values_from=b) %>%
  select(-n) %>% 
  complete(cut, color) %>% 
  left_join(diamonds) 
  count(cut, color)

Upvotes: 1

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