Create Columns from Factors and Count

Create columns from factors and count

You only need to make some slight modification to your code. You should use .(Name) instead of c("Name"):

ddply(df1, .(Name), summarise,
      Score_1 = sum(Score == 1),
      Score_2 = sum(Score == 2),
      Score_3 = sum(Score == 3))

gives:

  Name Score_1 Score_2 Score_3
1  Ben       1       1       0
2 John       1       1       1

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Other possibilities include:

1. table(df1) as @alexis_laz mentioned in the comments, this gives:

> table(df1)
       Score
Name   1 2 3
  Ben  1 1 0
  John 1 1 1

2. The dcast function of the reshape2 package (or data.table which has the same dcast function):

library(reshape2) # or library(data.table)
dcast(df1, Name ~ paste0("Score_", Score), fun.aggregate = length) 

gives:

  Name Score_1 Score_2 Score_3
1  Ben       1       1       0
2 John       1       1       1

Count occurrences of factors across multiple columns in grouped dataframe

You can stack col1 & col2 together, count the number of each combination, and then transform the table to a wide form.

library(dplyr)
library(tidyr)

df %>%
  pivot_longer(col1:col2) %>%
  count(grp, name, value) %>%
  pivot_wider(grp, names_from = c(name, value), names_sort = TRUE,
              values_from = n, values_fill = 0)

# A tibble: 3 x 6
  grp   col1_A col1_B col2_B col2_C col2_D
  <chr>  <int>  <int>  <int>  <int>  <int>
1 a          1      2      2      0      1
2 b          2      0      0      2      0
3 c          1      2      0      2      1

A base solution (Thank @GKi to refine the code):

table(cbind(df["grp"], col=do.call(paste0, stack(df[-1])[2:1])))

   col
grp col1A col1B col2B col2C col2D
  a     1     2     2     0     1
  b     2     0     0     2     0
  c     1     2     0     2     1

R Creating new columns based on factors in another column

Try this

df %>% group_by(category) %>% mutate(id = row_number()) %>% ungroup() %>% pivot_wider(id_cols = id, names_from = category, values_from = cond) %>% select(-id) 

Counting factor instance of several columns of factors

We can use rowSums on a logical matrix

DF$CountYes <- rowSums(DF == 'Yes')
DF$CountYes
#[1] 3 2 0

Counting the number of factor variables in a data frame

A few problems here:

Your subscript is out of bounds problem is because df[1:5, ] is rows 1:5, whereas columns would be df[ ,1:5]. It appears that you only have 3 rows, not 5.

The second error no applicable method for 'as.quoted' applied to
an object of class "function" is referring to the as.factor, which is a function. It is saying that a function doesn't belong within the function count. You can check exactly what count wants by running ?count in the console

A third problem that I see is that R will not automatically think that integers are factors. You will have to specify this with numbers. If you read in words, they are often automatically set as factors.

Here is a reproducible example:

> df<-data.frame("var1"=rnorm(3),"var2"=c(1:3),"var3"=rnorm(3),"var4"=c(3,1,2),"var5"=rnorm(3))
> str(df)

'data.frame':   3 obs. of  5 variables:
 $ var1: num  0.716 1.43 -0.726
 $ var2: int  1 2 3
 $ var3: num  0.238 -0.658 0.492
 $ var4: num  3 1 2
 $ var5: num  1.71 1.54 1.05

Here I used the structure str() function to check what type of data I have. Note, var1 is read in as an integer when I generated it as c(1:3), whereas specifying c(3,1,2) was read in as numeric in var4

Here, I will tell R I want two of the columns to be factors, and I will make another column of words, which will automatically become factors.

> df<-data.frame("var1"=rnorm(3),"var2"=as.factor(c(1:3)),"var3"=rnorm(3),"var4"=as.factor(c(3,1,2))
+                ,"var5"=rnorm(3), "var6"=c("Green","Red","Blue"))
> str(df)
'data.frame':   3 obs. of  6 variables:
 $ var1: num  -1.18 1.26 -0.53
 $ var2: Factor w/ 3 levels "1","2","3": 1 2 3
 $ var3: num  1.38 -0.401 -0.924
 $ var4: Factor w/ 3 levels "1","2","3": 3 1 2
 $ var5: num  1.688 0.547 0.727
 $ var6: Factor w/ 3 levels "Blue","Green",..: 2 3 1

You can then as which are factors:

> sapply(df, is.factor)
 var1  var2  var3  var4  var5  var6 
FALSE  TRUE FALSE  TRUE FALSE  TRUE 

And if you wanted a number for how many are factors something like this would get you there:

> length(which(sapply(df, is.factor)==TRUE))
[1] 3

You have something similar: length(which(vec==as.factor)), but one problem with this is you are asking which things in the vec object are the same as a function as.factor, which doesn't make sense. So it is giving you the error Error in vec == as.factor :
comparison (1) is possible only for atomic and list types

as.factor is for setting things as factor (as I have shown above), but is.factor is for asking if something is a factor, which will return a logical (TRUE vs FALSE) - also shown above.

Calculate factors across columns in R

Use levels= when creating the factors. DF has character columns whereas DF2 has factor columns all having the same levels, levs.

# test data frame
DF <- as.data.frame(matrix(letters,, 2), stringsAsFactors = FALSE) 

DF2 <- DF
levs <- sort(unique(unlist(DF)))
DF2[] <- lapply(DF2, factor, levels = levs)

This could be written as a one-liner like this:

DF2 <- replace(DF, TRUE, lapply(DF, factor, levels = sort(unique(unlist(DF)))))

Creating a table of counts by some factor level in R

library(reshape2)
dcast(data = all, formula = food~day, fun.aggregate = length, value.var = "food")
   food 1 2 3
1 pizza 2 0 2
2 snake 1 1 1
3  taco 1 2 2

OR

table(all$food, all$day)

        1 2 3
  pizza 2 0 2
  snake 1 1 1
  taco  1 2 2

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