Convert unstructured csv file to a data frame

Question

I am learning R for text mining. I have a TV program schedule in form of CSV. The programs usually start at 06:00 AM and goes on until 05:00 AM the next day which is called

Answer 1

An alternative solution with data.table:

library(data.table)
library(zoo)
library(splitstackshape)

txt <- textConnection("Sunday|\n 01-Nov-15|\n 6|Tom\n some information about the program|\n 23.3|Jerry\n some information about the program|\n 5|Avatar\n some information about the program|\nMonday|\n 02-Nov-15|\n 6|Tom\n some information about the program|\n 23.3|Jerry\n some information about the program|\n 5|Avatar\n some information about the program|")
tv <- readLines(txt)
DT <- data.table(tv)[, tv := gsub('[|]$', '', tv)]

wd <- levels(weekdays(1:7, abbreviate = FALSE))

DT <- DT[, temp := tv %chin% wd
         ][, day := tv[temp], by = 1:nrow(tvDT)
           ][, day := na.locf(day)
             ][, temp := NULL
               ][, idx := rleid(day)
                 ][, date := tv[2], by = idx
                   ][, .SD[-c(1,2)], by = idx]

DT <- cSplit(DT, sep="|", "tv", "long")[, lbl := rep(c("Time","Program","Info")), by = idx]
DT <- dcast(DT, idx + day + date + rowid(lbl) ~ lbl, value.var = "tv")[, lbl := NULL]

DT <- DT[, datetime := as.POSIXct(paste(as.character(date), sprintf("%01.2f",as.numeric(as.character(Time)))), format = "%d-%b-%y %H.%M")
   ][, datetime := datetime + (+(datetime < shift(datetime, fill=datetime[1]) & datetime < 6) * 24 * 60 * 60)
     ][, .(datetime, Program, Info)]

The result:

> DT
              datetime Program                               Info
1: 2015-11-01 06:00:00     Tom some information about the program
2: 2015-11-01 23:30:00   Jerry some information about the program
3: 2015-11-02 05:00:00  Avatar some information about the program
4: 2015-11-02 06:00:00     Tom some information about the program
5: 2015-11-02 23:30:00   Jerry some information about the program
6: 2015-11-03 05:00:00  Avatar some information about the program

---

Explanation:

1: read data, convert to a data.table & remove trailing |:

txt <- textConnection("Sunday|\n 01-Nov-15|\n 6|Tom\n some information about the program|\n 23.3|Jerry\n some information about the program|\n 5|Avatar\n some information about the program|\nMonday|\n 02-Nov-15|\n 6|Tom\n some information about the program|\n 23.3|Jerry\n some information about the program|\n 5|Avatar\n some information about the program|")
tv <- readLines(txt)
DT <- data.table(tv)[, tv := gsub('[|]$', '', tv)]

2: extract the weekdays into a new column

wd <- levels(weekdays(1:7, abbreviate = FALSE)) # a vector with the full weekdays
DT[, temp := tv %chin% wd
   ][, day := tv[temp], by = 1:nrow(tvDT)
     ][, day := na.locf(day)
       ][, temp := NULL]

3: create an index per day & create a column with the dates

DT[, idx := rleid(day)][, date := tv[2], by = idx]

4: remove unnecessary lines

DT <- DT[, .SD[-c(1,2)], by = idx]

5: split the time and the program-name into separate rows & create a label column

DT <- cSplit(DT, sep="|", "tv", "long")[, lbl := rep(c("Time","Program","Info")), by = idx]

6: reshape into wide format using the 'rowid' function from the development version of data.table

DT <- dcast(DT, idx + day + date + rowid(idx2) ~ idx2, value.var = "tv")[, idx2 := NULL]

7: create a dattime column & set the late night time to the next day

DT[, datetime := as.POSIXct(paste(as.character(date), sprintf("%01.2f",as.numeric(as.character(Time)))), format = "%d-%b-%y %H.%M")
   ][, datetime := datetime + (+(datetime < shift(datetime, fill=datetime[1]) & datetime < 6) * 24 * 60 * 60)]

8: keep the needed columns

DT <- DT[, .(datetime, Program, Info)]