How to Count the Frequency of a String for Each Row in R

How to count the frequency of a string for each row in R

df$count <- rowSums(df[-1] == "NC")
#    V1 V2 V3 V4 count
# 1 rs1 NC AB NC     2
# 2 rs2 AB NC AA     1
# 3 rs3 NC NC NC     3

We can use rowSums on the matrix that is created from this expression df[-1] == "NC".

Counting overall word frequency when each sentence is a separate row in a dataframe

You can just use table() on the unlisted strsplit() of your column

table(unlist(strsplit(df$Words, " ")))

# Luke     Luker       Sky Skywalker     Syker      Walk 
#    3         1         1         1         1         2 

and if you need it sorted

sort(table(unlist(strsplit(df$Words, " "))), decreasing = TRUE)

#     Luke      Walk     Luker       Sky Skywalker     Syker 
#        3         2         1         1         1         1 

where df$words is your column of interest.

R: Count the frequency of pairwise matching strings between all rows of a matrix

You are comparing wrong values:

apply(labels, 1, function(x) colMeans(x == t(labels)))

     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
 [1,]  1.0  0.0  0.0  0.0  0.0  0.4  0.0  0.0  1.0   0.0
 [2,]  0.0  1.0  0.0  0.0  0.2  0.4  0.0  0.0  0.0   1.0
 [3,]  0.0  0.0  1.0  0.0  0.0  0.0  0.0  0.2  0.0   0.0
 [4,]  0.0  0.0  0.0  1.0  0.0  0.0  0.2  0.6  0.0   0.0
 [5,]  0.0  0.2  0.0  0.0  1.0  0.0  0.0  0.0  0.0   0.2
 [6,]  0.4  0.4  0.0  0.0  0.0  1.0  0.0  0.0  0.4   0.4
 [7,]  0.0  0.0  0.0  0.2  0.0  0.0  1.0  0.0  0.0   0.0
 [8,]  0.0  0.0  0.2  0.6  0.0  0.0  0.0  1.0  0.0   0.0
 [9,]  1.0  0.0  0.0  0.0  0.0  0.4  0.0  0.0  1.0   0.0
[10,]  0.0  1.0  0.0  0.0  0.2  0.4  0.0  0.0  0.0   1.0

Count the frequency of strings in a dataframe R

You can use sapply() to go the counts and match every item in counts against the strings column in df using grepl() this will return a logical vector (TRUE if match, FALSE if non-match). You can sum this vector up to get the number of matches.

sapply(df, function(x) {
  sapply(counts, function(y) {
    sum(grepl(y, x))
  })
})

This will return:

    strings
pi        5
in        2
pie       2
ie        2

Frequency of each word in a set of strings

Pipes do the job.

df <- data.frame(column_x = c("hello world", "hello morning hello", 
                              "bye bye world"), stringsAsFactors = FALSE)
require(dplyr)
df$column_x %>%
  na.omit() %>%
  tolower() %>%
  strsplit(split = " ") %>% # or strsplit(split = "\\W") 
  unlist() %>%
  table() %>%
  sort(decreasing = TRUE)

Counting number of instances of a condition per row R

You can use rowSums.

df$no_calls <- rowSums(df == "nc")
df
#  rsID sample1 sample2 sample3 sample1304 no_calls
#1 abcd      aa      bb      nc         nc        2
#2 efgh      nc      nc      nc         nc        4
#3 ijkl      aa      ab      aa         nc        1

Or, as pointed out by MrFlick, to exclude the first column from the row sums, you can slightly modify the approach to

df$no_calls <- rowSums(df[-1] == "nc")

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Regarding the row names: They are not counted in rowSums and you can make a simple test to demonstrate it:

rownames(df)[1] <- "nc"  # name first row "nc"
rowSums(df == "nc")      # compute the row sums
#nc  2  3             
# 2  4  1        # still the same in first row

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