Using dplyr to filter rows which contain partial string of column

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Assuming I have a data frame like

term     cnt
apple     10
apples     5
a apple on 3
blue pears 3
pears      1

How could I filter all partial found strings within this column, e.g. getting as a result

term     cnt
apple     10
pears      1

without indicating to which terms I want to filter (apple|pears), but through a self-referencing manner (i.e. it does check each term against the whole column and removes terms that are a partial match). The number of tokens is not limited, nor the consistency of strings (i.e. "mapples" would get matched by "apple"). This would result in an inverted generalized dplyr-based version of

d[grep("^apple$|^pears$", d$term), ]

Additionally, it would be interesting use this departialisation to get a cumulated sum, e.g.

term     cnt
apple     18
pears      4

I couldn't get it to work with contains() or grep().

Thanks

3

There are 3 answers

2
amrrs On BEST ANSWER

Hopefully the complete answer. Not very idiomatic (as Pythonista's call) but someone can suggest improvement to this:

> ssss <- data.frame(c('apple','red apple','apples','pears','blue pears'),c(15,3,10,4,3))
> 
> names(ssss) <- c('Fruit','Count')
> 
> ssss
       Fruit Count
1      apple    15
2  red apple     3
3     apples    10
4      pears     4
5 blue pears     3
> 
> root_list <- as.vector(ssss$Fruit[unlist(lapply(ssss$Fruit,function(x){length(grep(x,ssss$Fruit))>1}))])
> 
> 
> ssss %>% filter(ssss$Fruit %in% root_list)
  Fruit Count
1 apple    15
2 pears     4
> 
> data <- data.frame(lapply(root_list, function(x){y <- stringr::str_extract(ssss$Fruit,x); ifelse(is.na(y),'',y)}))
> 
> cols <- colnames(data)
> 
> #data$x <- do.call(paste0, c(data[cols]))
> #for (co in cols) data[co] <- NULL
> 
> ssss$Fruit <- do.call(paste0, c(data[cols]))
> 
> ssss %>% group_by(Fruit) %>% summarise(val = sum(Count))
# A tibble: 2 x 2
  Fruit   val
  <chr> <dbl>
1 apple    28
2 pears     7
> 
1
Aramis7d On

you can try using tidyverse something like

1. define a list of the words as:

     k <- dft %>% 
          select(term) %>% 
          unlist() %>% 
          unique()

2. operate on the data as:

    dft %>%
      separate(term, c('t1', 't2')) %>%
      rowwise() %>%
      mutate( g = sum(t1 %in% k)) %>%
      filter( g > 0) %>%
      select(t1, cnt)

which gives:

      t1   cnt
   <chr> <int>
1  apple    10
2 apples     5
3  pears     1

this still doesn't handle apple and apples though. Will keep trying on that.

2
tushaR On

Try this:

df=data.frame(term=c('apple','apples','a apple on','blue pears','pears'),cnt=c(10,5,3,3,1))

matches = sapply(df$term,function(t,terms){grepl(pattern = t,x = terms)},df$term)

sapply(1:ncol(matches),function(t,mat){
  tempmat = mat[,t]&mat[,-t]
  indices=unlist(apply(tempmat,MARGIN = 2,which))
  df$term[indices]<<-df$term[t]
 },matches)

df%>%group_by(term)%>%summarize(cnt=sum(cnt))

 # A tibble: 2 x 2
 #  term   cnt
 #  <chr> <dbl>
 #1 apple    18
 #2 pears     4