How do I filter a range of numbers in R? [duplicate]

Answer 1

The above answer is probably more user friendly but here are a couple more...

Edit for down vote about index:

Mydata[Mydata$x >= 3 & Mydata$x <= 7, ]

  x  y
3 3 45
4 4 54
5 5 65
6 6 78
7 7 97

Which can be extended to return other columns, e.g., if you just wanted y:

Mydata[Mydata$x >= 3 & Mydata$x <= 7, 'y']

[1] 45 54 65 78 97

It can also return more than one column,e.g.:

Mydata <- data.frame(x = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10),
                     y = c(20, 30, 45, 54, 65, 78, 97, 102, 123, 156),
                     z = c(5, 4, 3, 2, 1, 0, -1, -2, -3, -4))   

Mydata[Mydata$x >= 3 & Mydata$x <= 7, c('y','z')]

   y  z
3 45  3
4 54  2
5 65  1
6 78  0
7 97 -1

Answer 2

Piggybacking on @Anna 's answer, I just ran a few of the options to see which was faster on a larger dataset for a problem have at work. I used the setup from here (Faster way to subset on rows of a data frame in R?) I checked it on a 1 billion row (16gb) dataset. Looks like data.table edged out dplyr by a little bit. I am just starting to use data.table though so I may have not used the most efficient code. Oh, also, I narrowed it down to these 4 based on times from a 100million row dataset. See below:

set.seed(42)  
# 1 billion rows
df <- data.frame(age=sample(1:65,1e9,replace=TRUE),x=rnorm(1e9),y=rpois(1e9,25))



microbenchmark(df1 <- df %>% filter(age >= 5 & age <= 25),
               df2 <- df %>% filter(dplyr::between(df$age, 5, 25)),
               times=10)


Unit: seconds
                                      expr      min     lq   mean median     uq    max  neval
df %>% filter(age >= 5 & age <= 25)          15.327 15.796 16.526 16.601 17.086 17.996    10
df %>% filter(dplyr::between(df$age, 5, 25)) 14.214 14.752 15.413 15.487 16.121 16.447    10


DT <- as.data.table(df)
microbenchmark(dt1 <- DT[age %inrange% c(5, 25)],
               dt2 <- DT[age %between% c(5, 25)],
               times = 10)


Unit: seconds
                              expr    min     lq   mean median     uq    max neval
 dt1 <- DT[age %inrange% c(5, 25)] 15.122 16.042 17.180 16.969 17.310 22.138    10
 dt2 <- DT[age %between% c(5, 25)] 10.212 11.121 11.675 11.436 12.132 13.913    10

Answer 3

Use %in%

library(dplyr)
new_frame<- Mydata%>% filter(x %in% (3:7))
new_frame
#   x  y
# 1 3 45
# 2 4 54
# 3 5 65
# 4 6 78
# 5 7 97

Answer 4

Base R solution:

df <- Mydata[Mydata$x >= 3 & Mydata$x <= 7, ]

df
  x  y
3 3 45
4 4 54
5 5 65
6 6 78
7 7 97

Answer 5

Plenty of good dplyr solutions such as filtering in or hard-coding the upper and lower bounds already present in some of the answers:

MydataTable%>% filter(between(x, 3, 70))
Mydata %>% filter(x %in% 3:7)
Mydata %>% filter(x>=3&x<=7)

You could also work with data.table, which is very fast for large data sets. inrange and between work identically for this purpose

library(data.table)
MydataTable <- data.table(x = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10),
                          y = c(20, 30, 45, 54, 65, 78, 97, 102, 123, 156))
MydataTable[x %inrange% c(3,7)]
MydataTable[x %between% c(3,7)]

A benefit of this method (besides the speed of data.table) is that you only need to specify the min and max range - you are not creating an array to subset the filter.

A time comparison of these methods:

> df <- data.frame(x = sample(1:10, 10000000, replace = T),
+                      y = sample(1:10, 10000000, replace = T))
> system.time({ df %>% filter(between(x, 3, 7)) })
   user  system elapsed 
   0.18    0.05    0.14 
> system.time({ df %>% filter(x %in% 3:7) })
       user  system elapsed 
       0.19    0.06    0.29 
> system.time({ df %>% filter(x>=3&x<=7)  })
   user  system elapsed 
   0.17    0.09    0.26 

> dt <- data.table(df)
> system.time( {dt[x %inrange% c(3,7)] })
   user  system elapsed 
   0.13    0.07    0.21 
> system.time( {dt[x %between% c(3,7)] })
   user  system elapsed 
   0.18    0.05    0.13

Answer 6

And the good old base::subset:

subset(Mydata, x >= 3 & x <= 7)
subset(Mydata, x %in% 3:7)