How do you create a new column containing percentage data calculated from other columns? - r

Please excuse the very novice question, but I'm trying to create a new column in a data frame that contains percentages based on other columns. For example, the data I'm working with is similar to the following, where the That column is a binary factor (i.e. presence or absence of "that"), the Verb column is the individual verb (i.e. verbs that may or may not be following by "that"), and the Freq column indicates the frequency of each individual verb.
That Verb Freq
1 That believe 3
2 NoThat think 4
3 That say 3
4 That believe 3
5 That think 4
6 NoThat say 3
7 NoThat believe 3
8 NoThat think 4
9 That say 3
10 NoThat think 4
What I want is to add another column that provides the overall rate of "that" expression (coded as "That") for each of the different verbs. Something like the following:
That Verb Freq Perc.That
1 That believe 3 33.3
2 NoThat think 4 25.0
3 That say 3 33.3
4 That believe 3 33.3
5 That think 4 25.0
6 NoThat say 3 33.3
7 NoThat believe 3 33.3
8 NoThat think 4 25.0
9 That say 3 33.3
10 NoThat think 4 25.0
It may be that I've missed a similar question elsewhere. If so, my apologize. Nevertheless, thanks in advance for any help.

You want to use the ddply function in the plyr library:
#install.packages('plyr')
library(plyr)
dat # your data frame
ddply(dat, .(verb), transform, perc.that = freq/sum(freq))
# that verb freq perc.that
#1 That believe 3 0.3333333
#2 That believe 3 0.3333333
#3 NoThat believe 3 0.3333333
#4 That say 3 0.3333333
#...

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I'm currently reading "Practical Statistics for Data Scientists" and following along in R as they demonstrate some code. There is one chunk of code I'm particularly struggling to follow the logic of and was hoping someone could help. The code in question is creating a dataframe with 1000 rows where each observation is the mean of 5 randomly drawn income values from the dataframe loans_income. However, I'm getting confused about the logic of the code as it is fairly complicated with a tapply() function and nested rep() statements.
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For reference, the original dataset loans_income simply has a single column of 50,000 income values.
You have 50,000 loans_income in a single vector. Let's break your code down:
tapply(sample(loans_income$income,1000*5),
rep(1:1000,rep(5,1000)),
FUN = mean)
I will replace 1000 with 10 and income with random numbers, so it's easier to explain. I also set set.seed(1) so the result can be reproduced.
sample(loans_income$income,1000*5)
We 50 random incomes from your vector without replacement. They are (temporarily) put into a vector of length 50, so the output looks like this:
> sample(runif(50000),10*5)
[1] 0.73283101 0.60329970 0.29871173 0.12637654 0.48434952 0.01058067 0.32337850
[8] 0.46873561 0.72334215 0.88515494 0.44036341 0.81386225 0.38118213 0.80978822
[15] 0.38291273 0.79795343 0.23622492 0.21318431 0.59325586 0.78340477 0.25623138
[22] 0.64621658 0.80041393 0.68511759 0.21880083 0.77455662 0.05307712 0.60320912
[29] 0.13191926 0.20816298 0.71600799 0.70328349 0.44408218 0.32696205 0.67845445
[36] 0.64438336 0.13241312 0.86589561 0.01109727 0.52627095 0.39207860 0.54643661
[43] 0.57137320 0.52743012 0.96631114 0.47151170 0.84099503 0.16511902 0.07546454
[50] 0.85970500
rep(1:1000,rep(5,1000))
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> rep(1:10,rep(5,10))
[1] 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4 5 5 5 5 5 6 6 6
[29] 6 6 7 7 7 7 7 8 8 8 8 8 9 9 9 9 9 10 10 10 10 10
Those indices "group" the samples from step 1. So basically this vector tells R that the first 5 entries of your "sample vector" belong together (index 1), the next 5 entries belong together (index 2) and so on.
FUN = mean
Just apply the mean-function on the data.
tapply
So tapply takes the sampled data (sample-part) and groups them by the second argument (the rep()-part) and applies the mean-function on each group.
If you are familiar with data.frames and the dplyr package, take a look at this (only the first 10 rows are displayed):
set.seed(1)
df <- data.frame(income=sample(runif(5000),10*5), index=rep(1:10,rep(5,10)))
income index
1 0.42585569 1
2 0.16931091 1
3 0.48127444 1
4 0.68357403 1
5 0.99374923 1
6 0.53227877 2
7 0.07109499 2
8 0.20754511 2
9 0.35839481 2
10 0.95615917 2
I attached the an index to the random numbers (your income). Now we calculate the mean per group:
df %>%
group_by(index) %>%
summarise(mean=mean(income))
which gives us
# A tibble: 10 x 2
index mean
<int> <dbl>
1 1 0.551
2 2 0.425
3 3 0.827
4 4 0.391
5 5 0.590
6 6 0.373
7 7 0.514
8 8 0.451
9 9 0.566
10 10 0.435
Compare it to
set.seed(1)
tapply(sample(runif(5000),10*5),
rep(1:10,rep(5,10)),
mean)
which yields basically the same result:
1 2 3 4 5 6 7 8 9
0.5507529 0.4250946 0.8273149 0.3905850 0.5902823 0.3730092 0.5143829 0.4512932 0.5658460
10
0.4352546

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