Group by name and add up the columns count in r - r

Question

I have a dataset with 405 observations and 39 variables. But just two columns are important for further analysis.
I would like to group the first row with similar names together and add up their number from the second column.
Reproducible dataset looks like this:
df1 <- data.frame(name=c("Google Ads", "Google Doubleclick","Facebook Login",
"Facebook Ads","Twitter MoPub","Flurry","Amazon advertisment","Microsoft ","Ad4screen","imobi"),
value=c(10,20,30,40,50,60,70,80,90,100),unimportant=c(1,2,3,4,5,6,7,8,9,10))
Outcome should be in an new data.frame and look like this:
df2 <- data.frame (name=c("Google","Facebook","Twitter","Flurry","Amazon","Microsoft","Others"),
value=c(30,70,50,60,70,80,190))

Answer 1

A tidyverse way of doing it.
First store all valid_names in a vector say valid_names
Thereafter create a new column say all_names in df1 by -
first splitting all strings at space ' ' using str_split
thereafter use purrr::map_chr() to check if any of the split string matches with your valid_names and if yes, retrieve that string only otherwise get others
Thereafter group_by on this field. (I omitted one step of mutate first and then group_by and directly created the new field in group_by statement, that works)
Now summarise your important values as desired.
valid_names =c("Google","Facebook","Twitter","Flurry","Amazon","Microsoft")
valid_names
#> [1] "Google" "Facebook" "Twitter" "Flurry" "Amazon" "Microsoft"
df1 <- data.frame(name=c("Google Ads", "Google Doubleclick","Facebook Login",
"Facebook Ads","Twitter MoPub","Flurry","Amazon advertisment","Microsoft ","Ad4screen","imobi"),
value=c(10,20,30,40,50,60,70,80,90,100),unimportant=c(1,2,3,4,5,6,7,8,9,10))
df1
#> name value unimportant
#> 1 Google Ads 10 1
#> 2 Google Doubleclick 20 2
#> 3 Facebook Login 30 3
#> 4 Facebook Ads 40 4
#> 5 Twitter MoPub 50 5
#> 6 Flurry 60 6
#> 7 Amazon advertisment 70 7
#> 8 Microsoft 80 8
#> 9 Ad4screen 90 9
#> 10 imobi 100 10
library(tidyverse)
df1 %>% group_by(all_names = str_split(name, ' '),
all_names = map_chr(all_names, ~ ifelse(any(.x %in% valid_names),.x[.x %in% valid_names], 'others'))) %>%
summarise(value = sum(value), .groups = 'drop')
#> # A tibble: 7 x 2
#> all_names value
#> <chr> <dbl>
#> 1 Amazon 70
#> 2 Facebook 70
#> 3 Flurry 60
#> 4 Google 30
#> 5 Microsoft 80
#> 6 others 190
#> 7 Twitter 50
Created on 2021-06-22 by the reprex package (v2.0.0)

Answer 2

This works on the sample data using the adist function and with partial=TRUE to look at partial string matches. It requires defining the known groups though, rather than trying to find them. I think this leg work is worth doing though as it simplifies the problem a lot once the output is known
df1 <- data.frame(name=c("Google Ads", "Google Doubleclick","Facebook Login",
"Facbook Ads","Twitter MoPub","Flurry","Amazon advertisment","Microsoft ","Ad4screen","imobi"),
value=c(10,20,30,40,50,60,70,80,90,100),unimportant=c(1,2,3,4,5,6,7,8,9,10))
# types we want to map. known is the groupings
types <- unique(df1$name)
known <- c("Google","Facebook","Twitter","Flurry","Amazon","Microsoft")
# use distrance measures, and look for matches on partial strings eg
# ignore the Doubleclick part when matching on Google
distance <- adist(known, types, partial=TRUE)
# cap controls leniancy in matching e.g. Facbook and Facebook have a dist of 1
# whilst Facebook and Facebook is a perfect match with score of 0
# Raise to be more leniant
cap <- 1
# loop through the types
map_all <- sapply(seq_along(types), function(i){
# find minimum value, check if its below the cap. If so, assign to the closest
# group, else assign to others
v <- min(distance[,i])
if(v <= cap){
map_i <- known[which.min(distance[,i])]
}else{
map_i <- "Others"
}
map_i
})
# now merge in to df1, then sum out using your preferred method
df_map <- data.frame(name=types, group=map_all)
df_merged <- merge(df1, df_map, by="name")
df2 <- aggregate(value ~ group, sum, data=df_merged)
df2
group value
1 Amazon 70
2 Facebook 70
3 Flurry 60
4 Google 30
5 Microsoft 80
6 Others 190
7 Twitter 50