R Create a new variable name by partially matching a string from other variable names

Question

I have multiple variable names that I need to combine into a single variable based on a common string of text. My sample data are:

structure(list(And = c(10L, NA, 10L), and = c(20L, 10L, 10L), 
andbc = c(1L, NA, NA), baNdc = c(4L, NA, 5L), ban = c(1L, 
NA, 1L)), .Names = c("And", "and", "andbc", "baNdc", "ban"), class = "data.frame", row.names = c(NA, -3L))

I would like to create a new variable x, the value of which would be a row sum of the values of the other variables that share the common text string "and" ignoring the case of any of the letters in that string.

I attempted creating the variable by specifying the permutations, which I'm hoping to avoid:

names1[, 1:5][is.na(names1[, 1:5])] <- 0
names1$x <- sum(names1[which(grepl("And|and|aNd", names(names1)))])

The result I get for values of x is a sum total of all values for the variables that meet the text string criteria:

structure(list(And = c(10, 0, 10), and = c(20L, 10L, 10L), andbc = c(1, 0, 0), baNdc = c(4, 0, 5), ban = c(1, 0, 1), x = c(70, 70, 70)), .Names = c("And", "and", "andbc", "baNdc", "ban", "x"), row.names = c(NA, -3L), class ="data.frame"

How can I obtain the row sums based on the text string criteria and avoid having to specify the permutations of upper or lower case?

67342343 · Accepted Answer · 2017-03-28 23:40:16Z

2

The following would do the trick

df <- structure(list(And = c(10L, NA, 10L), and = c(20L, 10L, 10L), 
           andbc = c(1L, NA, NA), baNdc = c(4L, NA, 5L), ban = c(1L, 
                                                                 NA, 1L)), .Names = c("And", "and", "andbc", "baNdc", "ban"), class = "data.frame", row.names = c(NA, -3L))

x <- rowSums(df[, grep("and", tolower(colnames(df)))], na.rm = TRUE)

edited Mar 28, 2017 at 23:40

answered Mar 28, 2017 at 23:35

67342343

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1 Comment

thelatemail Over a year ago

rowSums(df[grepl("and", names(df), ignore.case=TRUE)], na.rm=TRUE) as a base R translation.

jwells · Accepted Answer · 2017-03-28 23:50:33Z

1

colnames(names1) <- tolower(colnames(names1))

will rid you for the need for permutations

names1$x <- rowSums(names1[which(grepl('and', colnames(names1)))], na.rm = TRUE)

answered Mar 28, 2017 at 23:50

jwells

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1 Comment

Simon Jackson Over a year ago

which(grepl(...)) can simply be grep()

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