Error in charToDate(x) : character string is not in a standard unambiguous format

Question

I am writing because I have nowhere else to go to get an answer. I am trying to shrink my existing table a bit. It is of the next form:

Živilec;     Proizvodnja;            Kariera d.o.o.;  18.11.2014 hh.mm.ss;  Ljubljana
Živilec;     Prehrambena industrija; Kariera d.o.o.;  18.11.2014 hh.mm.ss;  Ljubljana
Vodja;       Strojništvo;            Adecco;          18.11.2014 hh.mm.ss;  Maribor
Vodja;       Tehnične storitve;      Adecco;          18.11.2014 hh.mm.ss;  Maribor
Vodja;       Elektrotehnika;         Adecco;          18.11.2014 hh.mm.ss;  Celje

, the dates are actually inserted as 18.11.2014 8:35:59 but I dont need the time, just the date. And what I wish to get to is this:

Živilec;  Proizvodnja,Preh. industrija; Kariera d.o.o.; 18.11.2014;          Ljubljana
Vodja;    Stroj.,Teh. stor., Elektro.;  Adecco;         18.11.2014;          Maribor, Celje

I have tryed getting this with the help of this R-code:

matrik<-matrix(0,600,30)
for (i in 1:dim(a)[1]){
  if (is.element(a[i,3],matrik[,15])==TRUE & is.element(a[i,1],matrik[,1])==TRUE){
    katero<-which(a[i,1]==matrik[,1])
    kdo<-which(a[i,15]==matrik[,15])
    kje<-min(intersect(kdo,katero))
    if (kje!=0){
      prosto<-min(which(matrik[kje,2:14]==0))
      matrik[kje,prosto]<-as.character(a[i,2])
      prosti<-min(which(matrik[kje,17:30]==0))
      matrik[kje,prosti]<-as.character(a[i,5])
    }
    if (kje==0){
      povrsti<-min(which(matrik[,1]==0))
      matrik[povrsti,1]<-as.character(a[i,1])
      prosto<-min(which(matrik[povrsti,2:14]==0))+1
      matrik[povrsti,prosto]<-as.character(a[i,2])
      matrik[povrsti,15]<-as.character(a[i,3])
      matrik[povrsti,16]<-as.character(a[i,4])
      prosti<-min(which(matrik[povrsti,17:30]==0))+1
      matrik[povrsti,prosti]<-as.character(a[i,5])      
    }
  }
  else {
    povrsti<-min(which(matrik[,1]==0))
    matrik[povrsti,1]<-as.character(a[i,1])
    prosto<-min(which(matrik[povrsti,2:14]==0))+1
    matrik[povrsti,prosto]<-as.character(a[i,2])
    matrik[povrsti,15]<-as.character(a[i,3])
    matrik[povrsti,16]<-as.character(a[i,4])
    prosti<-min(which(matrik[povrsti,17:30]==0))+16
    matrik[povrsti,prosti]<-as.character(a[i,5])  
  }

}

Basically I make a new matrix in which I will store the values, because i cannot store the categories like teh. storitve, strojništvo, elektro in one cell and just 2 values in another cell in the same column I decided to look at the maximum value of all the categories and make that many cells. If this problem is solvable otherwise please let me know aswell if you could. So anyways after making a zero matrix, I check if the first element (so "Živilec") and the third element (so "Kariera d.o.o.) are the same, if that is true I would like to just add values to the second and fifth(last) column. If not I see that I must add a new row to the existing matrix with all the values from the table. As I run this code I get the error:

Error in charToDate(x) : 
  character string is not in a standard unambiguous format

What to do? Any solutions?

Thank you for your time.

Your intended result is a very bad format for data analysis and your code is too complicated and unreadable because your space key doesn't seem to work. Anyway, I can't find where you transform the datetimes in your code, but you should use something like as.Date(as.POSIXct(<your datetimes>, format = "%d.%m.%Y %H:%M:%S", tz = <yourtimezone>)). — Roland

Miha Trošt Miha Trošt · Accepted Answer · 2014-11-28T14:47:02

In order to parse the dates, you can do it like below:

library(lubridate)

x <- c("18.11.2014 8:35:59")
as.Date(dmy_hms(x))

Otherwise, you should give the community some sample data...use

dput(your_data)

people will show you the way in no time.

UPDATE

Here is a solution:

Load some useful libraries...

library(stringr)
library(dplyr)

your data...

toy_data <-
structure(list(V1 = structure(c(2L, 2L, 1L, 1L, 1L), .Label = c("Vodja", 
"Živilec"), class = "factor"), V2 = structure(c(5L, 4L, 2L, 3L, 
1L), .Label = c("       Elektrotehnika", "       Strojništvo", 
"       Tehnične storitve", "     Prehrambena industrija", "     Proizvodnja"
), class = "factor"), V3 = structure(c(2L, 5L, 1L, 4L, 3L), .Label = c("            Adecco", 
"            Kariera d.o.o.", "         Adecco", "      Adecco", 
" Kariera d.o.o."), class = "factor"), V4 = structure(c(2L, 2L, 
1L, 1L, 1L), .Label = c("          18.11.2014", "  18.11.2014"
), class = "factor"), V5 = structure(c(2L, 2L, 3L, 3L, 1L), .Label = c("  Celje", 
"  Ljubljana", "  Maribor"), class = "factor")), .Names = c("V1", 
"V2", "V3", "V4", "V5"), class = "data.frame", row.names = c(NA, 
-5L))

an useful function...

my_str_c <- function(x){str_c(unique(x), collapse = ";")}

a code for your desired output...

toy_data %>%
  mutate_each(funs(str_trim)) %>%
  group_by(V1) %>%
  summarise_each(funs(my_str_c))

Error in charToDate(x) : character string is not in a standard unambiguous format

1 Answers