Create a Variable Name With "Paste" in R

Create a variable name with paste in R?

You can use assign (doc) to change the value of perf.a1:

> assign(paste("perf.a", "1", sep=""),5)
> perf.a1
[1] 5

How to use object name in paste?

you can use assign

j <- network(i)
assign(paste0(i,'network',sep = '_'), j)

Dynamically create variable names and values using paste in R

With tidyverse you could do the following. After selecting Q1 through Q5 (the columns needed to reproduce the new columns and omitting the already desired Q3 and Q4), you can put the data frame into long format (often preferable for future analyses). Row numbers rn are assigned to track rows, and all columns are made character so can be combined.

When in long format, the items sharing the same number (e.g., Q3a, Q3b, etc.) are grouped and then summarised, putting multiple values of the same group together separated by commas. Then, to get to your desired wide data frame, you can use pivot_wider.

library(tidyverse)

df %>%
  select(Q1:Q5) %>%
  mutate_all(as.character) %>%
  mutate(rn = row_number()) %>%
  pivot_longer(cols = -rn, 
               names_to = c("item", "subitem"), 
               names_pattern = "Q(\\d+)(\\w*)") %>%
  group_by(rn, item) %>%
  summarise(value = toString(value)) %>%
  pivot_wider(id_cols = rn, 
              names_from = item, 
              values_from = value, names_prefix = "Q") %>%
  ungroup %>%
  select(-rn)

Output

  Q1    Q2    Q3      Q4         Q5   
  <chr> <chr> <chr>   <chr>      <chr>
1 1     2     a, b, c a, b, c, d e    
2 2     3     a, b, c a, b, c, d f    
3 2     1     c, d, e c, d, e, f g  

R: Call a pasted variable name and use it as position argument

get is like assign, in that it lets you refer to variables by string instead of name.

r[get(paste0("p", x[1]))]=x[1]

But get is one of those "flags" of something that could be written in a much clearer and safer way.

Paste a string of variable names into function in R

Something like this?

a1 = 1:2
a2 = 3:4
b1 = 5:6
b2 = 7:8
expand.grid(do.call(rbind, mget(paste("a", 1:2, sep = ""))),
         do.call(rbind, mget(paste("b", 1:2, sep = ""))))
#   Var1 Var2
#1     1    5
#2     3    5
#3     2    5
#4     4    5
#5     1    7
#6     3    7
#7     2    7
#8     4    7
#9     1    6
#10    3    6
#11    2    6
#12    4    6
#13    1    8
#14    3    8
#15    2    8
#16    4    8

Using recode() with variable names generated through paste()

You simply could put a named vector v together with a list of your variables from mget into Map and subset it.

v <- c("Totally agree"=1, "Indifferent"=2, "Don't agree at all"=3)

Map(function(x, y) unname(y[x]), mget(ls(pattern="^item")), list(v))
# $item1
# [1] 3 1
# 
# $item2
# [1] 2 1

Or, suppose you have a data frame like this one,

head(dat1)
#   id         item1              item2         x
# 1  1 Totally agree      Totally agree 0.0356312
# 2  2 Totally agree      Totally agree 1.3149588
# 3  3 Totally agree        Indifferent 0.9781675
# 4  4 Totally agree        Indifferent 0.8817912
# 5  5   Indifferent        Indifferent 0.4822047
# 6  6   Indifferent Don't agree at all 0.9657529

then you may do this in a similar way. We may even simplify the code because we don't need the Map to return unnamed objects anymore.

v1 <- c("Totally agree"=1, "Indifferent"=2, "Don't agree at all"=3)

item_nm <- c("item1", "item2")
dat1[item_nm] <- Map(`[`, list(v1), dat2[item_nm])
dat1
#    id item1 item2          x
# 1   1     1     1  0.0356312
# 2   2     1     1  1.3149588
# 3   3     1     2  0.9781675
# 4   4     1     2  0.8817912
# 5   5     2     2  0.4822047
# 6   6     2     3  0.9657529
# 7   7     2     3 -0.8145709
# 8   8     1     1  0.2839578
# 9   9     3     1 -0.1616986
# 10 10     3     3  1.9355718

The second argument gets recycled for each Map iteration (i.e. list(v1, v1) would also work).

And more generally, for each column you want to recode numerically, list one vector more in the second argument of Map.

head(dat2)
#   id         item1  item2         x
# 1  1 Totally agree Always 0.0356312
# 2  2 Totally agree Always 1.3149588
# 3  3 Totally agree   Both 0.9781675
# 4  4 Totally agree   Both 0.8817912
# 5  5   Indifferent   Both 0.4822047
# 6  6   Indifferent  Never 0.9657529

v2 <- c("Always"=1, "Both"=2, "Never"=3)

dat2[item_nm] <- Map(`[`, list(v1, v2), dat2[item_nm])
dat2
#    id item1 item2          x
# 1   1     1     1  0.0356312
# 2   2     1     1  1.3149588
# 3   3     1     2  0.9781675
# 4   4     1     2  0.8817912
# 5   5     2     2  0.4822047
# 6   6     2     3  0.9657529
# 7   7     2     3 -0.8145709
# 8   8     1     1  0.2839578
# 9   9     3     1 -0.1616986
# 10 10     3     3  1.9355718

---

Data:

dat1 <- structure(list(id = 1:10, item1 = c("Totally agree", "Totally agree", 
"Totally agree", "Totally agree", "Indifferent", "Indifferent", 
"Indifferent", "Totally agree", "Don't agree at all", "Don't agree at all"
), item2 = c("Totally agree", "Totally agree", "Indifferent", 
"Indifferent", "Indifferent", "Don't agree at all", "Don't agree at all", 
"Totally agree", "Totally agree", "Don't agree at all"), x = c(0.0356311982051355, 
1.31495884897891, 0.978167526364279, 0.881791226863203, 0.482204688262918, 
0.965752878105794, -0.814570938270238, 0.283957806364306, -0.161698647607024, 
1.93557176599585)), class = "data.frame", row.names = c(NA, -10L
))

dat2 <- structure(list(id = 1:10, item1 = c("Totally agree", "Totally agree", 
"Totally agree", "Totally agree", "Indifferent", "Indifferent", 
"Indifferent", "Totally agree", "Don't agree at all", "Don't agree at all"
), item2 = c("Always", "Always", "Both", "Both", "Both", "Never", 
"Never", "Always", "Always", "Never"), x = c(0.0356311982051355, 
1.31495884897891, 0.978167526364279, 0.881791226863203, 0.482204688262918, 
0.965752878105794, -0.814570938270238, 0.283957806364306, -0.161698647607024, 
1.93557176599585)), class = "data.frame", row.names = c(NA, -10L
))

How to get the value of a variable name generated using paste in R

First consider fortune(312):

> library(fortunes)
> fortune(312)

The problem here is that the $ notation is a magical shortcut 
and like any other magic if used incorrectly is likely to do 
the programmatic equivalent of turning yourself into a toad.    
-- Greg Snow (in response to a user that wanted to
   access a column whose name is stored in y via x$y
   rather than x[[y]])
   R-help (February 2012)

Now look back at where you learned to subset using $, if that source did not teach you that $ is a magical shortcut (or other phrasing) and that sometimes a shortcut is the longest path between 2 points, so use the regular indexing instead of the shortcut, then that person deserves to be hit up side the head with a wet fish. If they did teach that, then you should apologize to them for not paying attention (and now pay attention).

Also, for assigning to a vector, look at using sapply instead of the for loop. You do not tell us what t contains or what hurst should be, but here is some sample code that shows an approach without get:

> t <- lapply(1:15, function(i)list(Hs=i*10))
> names(t) <- paste0('V', 1:15)
> t$Other <- "don't grab me"
> 
> hurst <- sapply( paste0("V", 1:15), 
+                  function(v) t[[c(v,"Hs")]])
> hurst
 V1  V2  V3  V4  V5  V6  V7  V8  V9 V10 V11 V12 V13 V14 
 10  20  30  40  50  60  70  80  90 100 110 120 130 140 
V15 
150 
> 

How may I refer to values of variable name with paste() function

I would use mget and ls.

mget(ls(pattern="cut"))

# $aado2cut
# [1]  20 180 360 460
# 
# $agecut
# [1] 35 55 70 80
# 
# $albumincut
# [1] 1.2 1.8 2.2 2.8 4.4

The result is a list containing all your variables verifying a certain pattern.

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