How to simplify handling with nested ifelse() structures in base R?

Question

I\'m facing nested ifelse() structures:

df1$var <- ifelse(x < a, u, ifelse(x < b, v, ifelse(x < c, w, ...)))

whereby t

Answer 1

You could use case_when from the dplyr library:

df1$y <- case_when(
    x == 1 ~ "s",
    x == 2 ~ "t",
    x == 3 ~ "u",
    x == 4 ~ "v",
    TRUE ~ "w"
)

Note that the final case above (TRUE) is the blanket else condition which will catch all cases not matching any earlier conditions.

Answer 2

I would create an empty vector then fill each condition one by one.

df1$y = rep(NA,nrow(df1))
df1$y[x < a] = u(x[x < a])
df1$y[x > a & x < b] = v(x[x > a & x < b])
df1$y[x > b & x < c] = w(x[x > b & x < c])
...

I have found this to be the clearest way of setting the values, I find it much easier to see at a glance what is happening when you have more conditions than nested ifelse statements.

df1$y = ifelse(x < a, u(x), ifelse(x < b, v(x), ifelse(x < c, w(x), ...)) )

An improvement on this would be to predefine the condition elements, that way each group is only calculated once:

aEls = which(x < a)
bEls = which(x > a & x < b)
cEls = which(x > b & x < c)
...

y = rep(NA,nrow(df1))
y[aEls] = u(x[aEls])
y[bEls] = v(x[bEls])
y[cEls] = w(x[cEls])
...

df$y = y

Answer 3

In base R, if there are multiple elements to be replaced, create a key/value dataset and do a merge

keyval <- data.frame(x = c(1, 2, 3, 4), y = c("s", "t", "u", "v"), stringsAsFactors = FALSE)
new <- merge(df1, keyval, by = 'x', all.x = TRUE)[['y']]
new[is.na(new)] <- "w"
df1$x <- new

data

set.seed(24)
df1 <- data.frame(x = rbinom(100, 5, .5))

Answer 4

Since you insist on base R, here are two possibilities:

Define a mapping data.frame:

# Define mapping
map <- cbind.data.frame(
    x = c(1, 2, 3, 4, NA),
    y = c("s", "t", "u", "v", "w"));

Method 1: match entries from map to df1.

# match entries
df1$y <- map[match(df1$x, map$x), 2];
df1$y[is.na(df1$y2)] <- "w";

Method 2: Loop through all mappings, and replace using direct indexing:

# for loop
df1$y <- factor("w", levels = map$y);
for (i in 1:nrow(map)) df1$y[df1$x == map$x[i]] <- map$y[i];

Output:

tail(df1);
#    x y
#95  4 v
#96  1 s
#97  4 v
#98  2 t
#99  4 v
#100 1 s

Note, the second method will also work for inequalities.

---

Sample data

set.seed(2017);
df1 <- data.frame(x = rbinom(100, 5, .5))