Convert a TIME8. to a Character Without First Converting to Numeric Format

» development » Convert a TIME8. to a Character Without First Converting to Numeric Format

By : حسین نوراحمدی

Date : November 28 2020, 09:01 AM

To fix the issue you can do I am trying to convert a TIME8. value to a character variable in the same original format the TIME8. was in. For example, if the value is 12:30:00, I want a character value that also appears as 12:30:00. , Just use the format TIME8. instead of 8. as shown below,

code :

data _null_;
  val='12:30:00'T;
  num=put(val,time8.);
  put _all_;
run;

1266  data test;
1267    val='12:30:00'T;
1268    num=put(val,time8.);
1269    put _all_;
1270  run;

val=45000 num=12:30:00 _ERROR_=0 _N_=1
NOTE: The data set WORK.TEST has 1 observations and 2 variables.
NOTE: DATA statement used (Total process time):
      real time           0.01 seconds
      cpu time            0.01 seconds

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converting multiple columns from character to numeric format in r

By : itkhanz

Date : March 29 2020, 07:55 AM

With these it helps What is the most efficient way to convert multiple columns in a data frame from character to numeric format? , You could try

code :

DF <- data.frame("a" = as.character(0:5),
                 "b" = paste(0:5, ".1", sep = ""),
                 "c" = letters[1:6],
                 stringsAsFactors = FALSE)

# Check columns classes
sapply(DF, class)

#           a           b           c 
# "character" "character" "character" 

cols.num <- c("a","b")
DF[cols.num] <- sapply(DF[cols.num],as.numeric)
sapply(DF, class)

#          a           b           c 
#  "numeric"   "numeric" "character"

In R: efficiently convert one format (character vector) to another format (numeric matrix)

By : Patrick Fokeu

Date : March 29 2020, 07:55 AM

like below fixes the issue Update
To avoid read.csv just use strsplit and the non-exported splitstackshape:::numMat functions:

code :

M <- strsplit(L, "\\s+|:")
cbind(mol = as.numeric(sapply(M, `[`, 1)),
    splitstackshape:::numMat(lapply(M, `[`, -1), fill=0))

L <- c("1  1:1 2:1 3:1 5:1 6:1 8:1",
       "5  1:1 2:1 4:1",
       "9  1:1 2:1 7:1 10:1")
M <- replicate(10000, L)

fun1 <- function() {
  spl <- lapply(strsplit(M,"\\s+|:.? |:.$"),as.numeric)
  vals <- lapply(spl,"[",-1)

  data.frame(
    mol=sapply(spl,"[",1),
    t(sapply(vals, function(x) {
      out <- rep(0,max(unlist(vals)))
      out[x] <- 1
      out} ))
  )
} 

system.time(out_late <- fun1())
#    user  system elapsed 
#   98.36    1.28  100.06
head(out_late)
#   mol X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
# 1   1  1  1  1  0  1  1  0  1  0   0
# 2   5  1  1  0  1  0  0  0  0  0   0
# 3   9  1  1  0  0  0  0  1  0  0   1
# 4   1  1  1  1  0  1  1  0  1  0   0
# 5   5  1  1  0  1  0  0  0  0  0   0
# 6   9  1  1  0  0  0  0  1  0  0   1

library(splitstackshape)
fun2 <- function() {
  M <- strsplit(M, "\\s+|:")
  cbind(mol = as.numeric(sapply(M, `[`, 1)),
        splitstackshape:::numMat(lapply(M, `[`, -1), fill=0))
}

system.time(out_ananda <- fun2())
#    user  system elapsed 
#    0.67    0.00    0.68
head(out_ananda)
#      mol 1 2 3 4 5 6 7 8 9 10
# [1,]   1 1 1 1 0 1 1 0 1 0  0
# [2,]   5 1 1 0 1 0 0 0 0 0  0
# [3,]   9 1 1 0 0 0 0 1 0 0  1
# [4,]   1 1 1 1 0 1 1 0 1 0  0
# [5,]   5 1 1 0 1 0 0 0 0 0  0
# [6,]   9 1 1 0 0 0 0 1 0 0  1

fun3 <- function() {
  t(sapply(strsplit(M, "\\s+"), function(l) {
    mol <- as.numeric(l[1])
    names(mol) <- 'mol'
    val <- numeric(10)
    names(val) <- 1:10
    for (x in strsplit(l[-1], ":"))
      val[x[1]] <- as.numeric(x[2])
    c(mol, val)
  }))
}

system.time(out_matthew <- fun3())
#    user  system elapsed 
#    2.33    0.00    2.34
head(out_matthew)
#      mol 1 2 3 4 5 6 7 8 9 10
# [1,]   1 1 1 1 0 1 1 0 1 0  0
# [2,]   5 1 1 0 1 0 0 0 0 0  0
# [3,]   9 1 1 0 0 0 0 1 0 0  1
# [4,]   1 1 1 1 0 1 1 0 1 0  0
# [5,]   5 1 1 0 1 0 0 0 0 0  0
# [6,]   9 1 1 0 0 0 0 1 0 0  1

Converting factor variables of character format into numeric

By : Salzchneider

Date : March 29 2020, 07:55 AM

hope this fix your issue Judging by your desired output, it doesn't look like you want to convert the factors to the numbers contained in their strings. Instead you want the internal representation of the factors.
Try this:

code :

df[,xxx] <- lapply(df[,factorvars], as.numeric)
#                               x       y  z x_Num y_Num
# 1    10: Already Delinquent 90+  00:Bad  2     1     1
# 2  11: Credit History <6 Months  01:Ind  2     2     2
# 3       12: Current Balance = 0 02:Good  1     3     3
# 4           13: Balance (2-6)=0    <NA>  1     4    NA
# 5                  20: 1+ x 90+  00:Bad  0     5     1
# 6                30: 3+ x 60-89  01:Ind  0     6     2
# 7                 31: 2 x 60-89 02:Good  0     7     3
# 8                 32: 1 x 60-89    <NA>  0     8    NA
# 9                40: 3+ x 30-59  00:Bad  2     9     1
# 10                41: 2 x 30-59  01:Ind  0    10     2
# 11                42: 1 x 30-59 02:Good  0    11     3
# 12 50: Insufficient Performance    <NA>  1    12    NA
# 13                60: 3+ x 1-29  00:Bad  1    13     1
# 14                 61: 2 x 1-29  01:Ind -1    14     2
# 15                 62: 1 x 1-29 02:Good -1    15     3
# 16         70: Never delinquent    <NA> -1    16    NA

is.na(df$y) <- df$y == "NA"
df$y <- droplevels(df$y)

Converting numeric/character column to date format in R

By : WilburMcMullen

Date : March 29 2020, 07:55 AM

To fix this issue Provided that you can start from the non-numeric version of the column (so, maintaining the 0), a possible solution is this:

code :

x <- "0825"
x <- paste(x, "2017", sep ="")
y<- as.Date(as.character(x), "%m%d%Y")
wd <- weekdays(y)

R tidyverse : Dataset structure format and error when converting from character to numeric

By : user3119326

Date : March 29 2020, 07:55 AM

wish helps you Every output value in case_when needs to have the same type. You're trying to create a numeric vector, so you can't have TRUE ~ . in case_when: . is character. You can just have the fallback case be NA:

code :

db %>% 
    mutate_at(vars(x1), list(~case_when(
        . == "1 - Nunca" ~ 1,
        . == "2 - Raramente" ~ 2,
        . == "3 - As vezes" ~ 3,
        . == "4 - Quase sempre" ~ 4,
        . == "5 - Sempre" ~ 5,
        TRUE ~ as.numeric(NA))))