CODEWITHSUNDEEP
r
user11768059
user11768059

Reputation:

Repetition of a sequence in a column

I want to repeat again and again the given(below) sequence to fill approx 5000 rows in R.

Time data set:

8.00.00 AM
9.00.00 AM
10.00.00 AM
11.00.00 AM
12.00.00 PM
1.00.00 PM
2.00.00 PM
3.00.00 PM
4.00.00 PM
5.00.00 PM
6.00.00 PM
7.00.00 PM
8.00.00 PM
9.00.00 PM

Upvotes: 0

Views: 146

Answers (4)

r2evans
r2evans

Reputation: 160417

There could be several reasons why you see blanks. I'll focus on two possibles: NAs, and literal blanks.

srcvec <- c("8.00.00 AM", "9.00.00 AM", "10.00.00 AM", "11.00.00 AM", "12.00.00 PM", 
"1.00.00 PM", "2.00.00 PM", "3.00.00 PM", "4.00.00 PM", "5.00.00 PM", 
"6.00.00 PM", "7.00.00 PM", "8.00.00 PM", "9.00.00 PM", NA, ""
)
rep(srcvec, len=30)
#  [1] "8.00.00 AM"  "9.00.00 AM"  "10.00.00 AM" "11.00.00 AM" "12.00.00 PM" "1.00.00 PM" 
#  [7] "2.00.00 PM"  "3.00.00 PM"  "4.00.00 PM"  "5.00.00 PM"  "6.00.00 PM"  "7.00.00 PM" 
# [13] "8.00.00 PM"  "9.00.00 PM"  NA            ""            "8.00.00 AM"  "9.00.00 AM" 
# [19] "10.00.00 AM" "11.00.00 AM" "12.00.00 PM" "1.00.00 PM"  "2.00.00 PM"  "3.00.00 PM" 
# [25] "4.00.00 PM"  "5.00.00 PM"  "6.00.00 PM"  "7.00.00 PM"  "8.00.00 PM"  "9.00.00 PM"

To remove the NAs, we can simply use na.omit:

rep(na.omit(srcvec), len=30)
#  [1] "8.00.00 AM"  "9.00.00 AM"  "10.00.00 AM" "11.00.00 AM" "12.00.00 PM" "1.00.00 PM" 
#  [7] "2.00.00 PM"  "3.00.00 PM"  "4.00.00 PM"  "5.00.00 PM"  "6.00.00 PM"  "7.00.00 PM" 
# [13] "8.00.00 PM"  "9.00.00 PM"  ""            "8.00.00 AM"  "9.00.00 AM"  "10.00.00 AM"
# [19] "11.00.00 AM" "12.00.00 PM" "1.00.00 PM"  "2.00.00 PM"  "3.00.00 PM"  "4.00.00 PM" 
# [25] "5.00.00 PM"  "6.00.00 PM"  "7.00.00 PM"  "8.00.00 PM"  "9.00.00 PM"  ""

To remove blanks as well, we can filter on nzchar, which returns true when strings continue 1 or more characters:

rep(Filter(nzchar, na.omit(srcvec)), len=30)
#  [1] "8.00.00 AM"  "9.00.00 AM"  "10.00.00 AM" "11.00.00 AM" "12.00.00 PM" "1.00.00 PM" 
#  [7] "2.00.00 PM"  "3.00.00 PM"  "4.00.00 PM"  "5.00.00 PM"  "6.00.00 PM"  "7.00.00 PM" 
# [13] "8.00.00 PM"  "9.00.00 PM"  "8.00.00 AM"  "9.00.00 AM"  "10.00.00 AM" "11.00.00 AM"
# [19] "12.00.00 PM" "1.00.00 PM"  "2.00.00 PM"  "3.00.00 PM"  "4.00.00 PM"  "5.00.00 PM" 
# [25] "6.00.00 PM"  "7.00.00 PM"  "8.00.00 PM"  "9.00.00 PM"  "8.00.00 AM"  "9.00.00 AM"

If you have non-empty blanks (e.g., whitespace), you can use this:

srcvec <- c(srcvec, "   ")
rep(Filter(function(a) !is.na(a) & nzchar(gsub("\\s", "", a)), srcvec), len=30)
#  [1] "8.00.00 AM"  "9.00.00 AM"  "10.00.00 AM" "11.00.00 AM" "12.00.00 PM" "1.00.00 PM" 
#  [7] "2.00.00 PM"  "3.00.00 PM"  "4.00.00 PM"  "5.00.00 PM"  "6.00.00 PM"  "7.00.00 PM" 
# [13] "8.00.00 PM"  "9.00.00 PM"  "8.00.00 AM"  "9.00.00 AM"  "10.00.00 AM" "11.00.00 AM"
# [19] "12.00.00 PM" "1.00.00 PM"  "2.00.00 PM"  "3.00.00 PM"  "4.00.00 PM"  "5.00.00 PM" 
# [25] "6.00.00 PM"  "7.00.00 PM"  "8.00.00 PM"  "9.00.00 PM"  "8.00.00 AM"  "9.00.00 AM"

Upvotes: 1

yixi zhou
yixi zhou

Reputation: 297

First create a vector includes the data and then use matrix

x <- c("8.00.00 AM",
"9.00.00 AM",
"10.00.00 AM",
"11.00.00 AM",
"12.00.00 PM",
"1.00.00 PM",
"2.00.00 PM",
"3.00.00 PM",
"4.00.00 PM",
"5.00.00 PM",
"6.00.00 PM",
"7.00.00 PM",
"8.00.00 PM",
"9.00.00 PM")

output <- matrix(x,5000)

Upvotes: 0

zhang jing
zhang jing

Reputation: 181

You can use my onetree package from github yikeshu0611 to deal with problem.

devtools::install_github("yikeshu0611/onetree")
library(onetree)

df=read_Text("
time PAM
8.00.00 AM
9.00.00 AM
10.00.00 AM
11.00.00 AM
12.00.00 PM
1.00.00 PM
2.00.00 PM
3.00.00 PM
4.00.00 PM
5.00.00 PM
6.00.00 PM
7.00.00 PM
8.00.00 PM
9.00.00 PM
")
nrow(df)
358 ≈ 5000/14

df is your original data. Then we add the repeted number 358 to df.

df$n=358

Finally, we use flat_strech to strech data by n

df2=flat_strech(data=df,strech="n")
nrow(df2)
5012

Upvotes: 0

Omar Abd El-Naser
Omar Abd El-Naser

Reputation: 704

Time = paste(c(8:12,1:9),".00.00",sep = "")
PM_AM <- paste(Time,rep(c("AM","PM"),c(4,10)))
n_rep <- 5000 %/% length(PM_AM)
n_remain <- 5000 %% length(PM_AM)
x <- rep(PM_AM,n_rep)
x <- c(x,PM_AM[1:n_remain])

Upvotes: 1

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