Combining for loop print outputs in R

Question

I'm trying to generate a sentence in Markdown to handle dynamic data

###Generate some sample data

Type <- c("A","A","A","A","A","A","A","A","A",
      "B","B","B","B","B","B","B","B",
      "C","C","C","C","C","C","C",
      "ABC","ABC","ABC","ABC","ABC")
Type <- as.data.frame(Type)

###Set the tables and iterations

l <- length(unique(Type$Type))
t <- table(as.character(Type$Type))
pt <- prop.table((table(as.character(Type$Type))))

###Loop to print the first type in sentence

for(i in seq(from=1, to=1)) {
  typebegin <- print(paste0("Type ", 
           names(pt)[i], 
           " accounted for ", 
           t[i], 
           " (",round(pt[i]*100),"%),"))
}

Here's where the problem is:

###Loop to print all the types in the middle

for(i in seq(from=2, to=(l-1),by=1)) {
  typemid <- print(paste0("type ", 
           names(pt)[i], 
           " accounted for ", 
           t[i], 
           " (",round(pt[i]*100),"%),"))
}

I get an output from the function as:

[1] "type ABC accounted for 5 (17%),"

[1] "type B accounted for 8 (28%),"

I don't know how to concatenate these.

###Loop to end the sentence

for(i in seq(from=l, to=l)) {
  typeend <- print(paste0("type ", 
           names(pt)[i], 
           " accounted for ", 
           t[i], 
           " (",round(pt[i]*100),"%)."))
}

###Print the sentence

paste(typebegin, typemid, typeend)

[1] "Type C accounted for 7 (24%), type B accounted for 8 (28%), type C accounted for 7 (24%)."

pp <- table(Type); pp <- matrix(c(names(pp), pp, round(prop.table(pp) * 100)), ncol = 3); pr <- apply(pp, 1, function(x) sprintf('type %s accounted for %s (%s%%)', x[1], x[2], x[3])); sub('(.)', '\\U\\1', paste(pr, collapse = ', '), perl = TRUE) — rawr
– rawr, Commented Dec 16, 2016 at 18:30
pp <- table(Type); pp <- matrix(c(names(pp), pp, round(prop.table(pp) * 100)), ncol = 3); pr <- apply(pp, 1, function(x) sprintf('type %s accounted for %s (%s%%)', x[1], x[2], x[3])); pr <- paste0(sub('(.)', '\\U\\1', paste(pr, collapse = ', '), perl = TRUE),".") pr — SCDCE
– SCDCE, Commented Dec 27, 2016 at 13:25

Hack-R · Accepted Answer · 2016-12-16 16:31:03Z

1

a <- as.character()
for(i in 1:length(pt)) {
  if(i ==1){
    a <- c(a,   
           paste0("Type ", 
                  names(pt)[i], 
                  " accounted for ", 
                  t[i], 
                  " (",round(pt[i]*100),"%),"))
  }
  if(i < length(pt) & i > 1){
  a <- c(a,   
                          paste0("type ", 
                           names(pt)[i], 
                           " accounted for ", 
                           t[i], 
                           " (",round(pt[i]*100),"%),")
  )
  } else if (i == length(pt)){
    a <- c(a,   
           paste0("type ", 
                  names(pt)[i], 
                  " accounted for ", 
                  t[i], 
                  " (",round(pt[i]*100),"%).")
    )

  }
}

cat(a)

Type A accounted for 9 (31%), type ABC accounted for 5 (17%), type B accounted for 8 (28%), type C accounted for 7 (24%).

If you need to save the setence in an object do it this way:

a <- capture.output(cat(a))

edited Dec 16, 2016 at 16:31

answered Dec 16, 2016 at 16:25

Hack-R

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1 Comment

Hack-R Over a year ago

The if statements are just to handle the capitalization and punctuation of different parts of the sentence.

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