R: getting list of matching data frame values [duplicate] - r

This question already has answers here:
Collapse / concatenate / aggregate a column to a single comma separated string within each group
(6 answers)
Closed 7 years ago.
Two data sets:
people <- read.table(text="
pid
1
2
3
4
", header=TRUE)
comps <- read.table(text="
pid comp rank
1 1 0
1 3 1
1 2 2
2 4 0
2 1 1
2 3 2
3 1 0
3 2 1
3 4 2
", header=TRUE)
Trying to get a data frame of each unique pid with a list of their comparisons, like:
pid comps
1 1,3,2
2 4,1,3
3 1,2,4
Can't quite get there..

You can do this with aggregate:
aggregate(comp~pid, comps, paste, collapse=",")
# pid comp
# 1 1 1,3,2
# 2 2 4,1,3
# 3 3 1,2,4

Related

Receive the total sum score of every number [duplicate]

This question already has answers here:
Count number of occurences for each unique value
(14 answers)
Closed 2 years ago.
Using as input data frame:
df1 <- data.frame(num = c(1,1,1,2,2,2,3))
How is it possible to receive the sum of every number excited in the num column?
Example output:
num frequency
1 3
2 3
3 1
Using table and coerce it to a data frame.
as.data.frame(table(df1$num))
# Var1 Freq
# 1 1 3
# 2 2 3
# 3 3 1
or
with(df1, data.frame(num=unique(num), freq=tabulate(num)))
# num freq
# 1 1 3
# 2 2 3
# 3 3 1

R, dplyr: Is there a way to add order of groups when there are multiple rows per group without creating a new data frame? [duplicate]

This question already has answers here:
How to create a consecutive group number
(13 answers)
Closed 2 years ago.
I have data from an experiment that has multiple rows per item (each row has the reading time for one word of a sentence of n words), and multiple items per subject. Items can be varying numbers of rows. Items were presented in a random order, and their order in the data as initially read in reflects the sequence they saw the items in. What I'd like to do is add a column that contains the order in which the subject saw that item (i.e., 1 for the first item, 2 for the second, etc.).
Here's an example of some input data that has the relevant properties:
d <- data.frame(Subject = c(1,1,1,1,1,2,2,2,2,2),
Item = c(2,2,2,1,1,1,1,2,2,2))
Subject Item
1 2
1 2
1 2
1 1
1 1
2 1
2 1
2 2
2 2
2 2
And here's the output I want:
Subject Item order
1 2 1
1 2 1
1 2 1
1 1 2
1 1 2
2 1 1
2 1 1
2 2 2
2 2 2
2 2 2
I know I can do this by setting up a temp data frame that filters d to unique combinations of Subject and Item, adding order to that as something like 1:n() or row_number(), and then using a join function to put it back together with the main data frame. What I'd like to know is whether there's a way to do this without having to create a new data frame just to store the order---can this be done inside dplyr's mutate somehow if I group by Subject and Item, for instance?
Here's one way:
d %>%
group_by(Subject) %>%
mutate(order = match(Item, unique(Item))) %>%
ungroup()
# # A tibble: 10 x 3
# Subject Item order
# <dbl> <dbl> <int>
# 1 1 2 1
# 2 1 2 1
# 3 1 2 1
# 4 1 1 2
# 5 1 1 2
# 6 2 1 1
# 7 2 1 1
# 8 2 2 2
# 9 2 2 2
# 10 2 2 2
Here is a base R option
transform(d,
order = ave(Item, Subject, FUN = function(x) as.integer(factor(x, levels = unique(x))))
)
or
transform(d,
order = ave(Item, Subject, FUN = function(x) match(x, unique(x)))
)
both giving
Subject Item order
1 1 2 1
2 1 2 1
3 1 2 1
4 1 1 2
5 1 1 2
6 2 1 1
7 2 1 1
8 2 2 2
9 2 2 2
10 2 2 2

Select max or equal value from several columns in a data frame

I'm trying to select the column with the highest value for each row in a data.frame. So for instance, the data is set up as such.
> df <- data.frame(one = c(0:6), two = c(6:0))
> df
one two
1 0 6
2 1 5
3 2 4
4 3 3
5 4 2
6 5 1
7 6 0
Then I'd like to set another column based on those rows. The data frame would look like this.
> df
one two rank
1 0 6 2
2 1 5 2
3 2 4 2
4 3 3 3
5 4 2 1
6 5 1 1
7 6 0 1
I imagine there is some sort of way that I can use plyr or sapply here but it's eluding me at the moment.
There might be a more efficient solution, but
ranks <- apply(df, 1, which.max)
ranks[which(df[, 1] == df[, 2])] <- 3
edit: properly spaced!

Conditionally dropping duplicates from a data.frame

Im am trying to figure out how to subset my dataset according to the repeated value of the variable s, taking also into account the id associated to the row.
Suppose my dataset is:
dat <- read.table(text = "
id s
1 2
1 2
1 1
1 3
1 3
1 3
2 3
2 3
3 2
3 2",
header=TRUE)
What I would like to do is, for each id, to keep only the first row for which s = 3. The result with dat would be:
id s
1 2
1 2
1 1
1 3
2 3
3 2
3 2
I have tried to use both duplicated() and which() for using subset() in a second moment, but I am not going anywhere. The main problem is that it is not sufficient to isolate the first row of the s = 3 "blocks", because in some cases (as here between id = 1 and id = 2) the 3's overlap between one id and another.. Which strategy would you adopt?
Like this:
subset(dat, s != 3 | s == 3 & !duplicated(dat))
# id s
# 1 1 2
# 2 1 2
# 3 1 1
# 4 1 3
# 7 2 3
# 9 3 2
# 10 3 2
Note that subset can be dangerous to work with (see Why is `[` better than `subset`?), so the longer but safer version would be:
dat[dat$s != 3 | dat$s == 3 & !duplicated(dat), ]

Calculating the occurrences of numbers in the subsets of a data.frame

I have a data frame in R which is similar to the follows. Actually my real ’df’ dataframe is much bigger than this one here but I really do not want to confuse anybody so that is why I try to simplify things as much as possible.
So here’s the data frame.
id <-c(1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3)
a <-c(3,1,3,3,1,3,3,3,3,1,3,2,1,2,1,3,3,2,1,1,1,3,1,3,3,3,2,1,1,3)
b <-c(3,2,1,1,1,1,1,1,1,1,1,2,1,3,2,1,1,1,2,1,3,1,2,2,1,3,3,2,3,2)
c <-c(1,3,2,3,2,1,2,3,3,2,2,3,1,2,3,3,3,1,1,2,3,3,1,2,2,3,2,2,3,2)
d <-c(3,3,3,1,3,2,2,1,2,3,2,2,2,1,3,1,2,2,3,2,3,2,3,2,1,1,1,1,1,2)
e <-c(2,3,1,2,1,2,3,3,1,1,2,1,1,3,3,2,1,1,3,3,2,2,3,3,3,2,3,2,1,3)
df <-data.frame(id,a,b,c,d,e)
df
Basically what I would like to do is to get the occurrences of numbers for each column (a,b,c,d,e) and for each id group (1,2,3) (for this latter grouping see my column ’id’).
So, for column ’a’ and for id number ’1’ (for the latter see column ’id’) the code would be something like this:
as.numeric(table(df[1:10,2]))
##The results are:
[1] 3 7
Just to briefly explain my results: in column ’a’ (and regarding only those records which have number ’1’ in column ’id’) we can say that number '1' occured 3 times and number '3' occured 7 times.
Again, just to show you another example. For column ’a’ and for id number ’2’ (for the latter grouping see again column ’id’):
as.numeric(table(df[11:20,2]))
##After running the codes the results are:
[1] 4 3 3
Let me explain a little again: in column ’a’ and regarding only those observations which have number ’2’ in column ’id’) we can say that number '1' occured 4 times, number '2' occured 3 times and number '3' occured 3 times.
So this is what I would like to do. Calculating the occurrences of numbers for each custom-defined subsets (and then collecting these values into a data frame). I know it is not a difficult task but the PROBLEM is that I’m gonna have to change the input ’df’ dataframe on a regular basis and hence both the overall number of rows and columns might change over time…
What I have done so far is that I have separated the ’df’ dataframe by columns, like this:
for (z in (2:ncol(df))) assign(paste("df",z,sep="."),df[,z])
So df.2 will refer to df$a, df.3 will equal df$b, df.4 will equal df$c etc. But I’m really stuck now and I don’t know how to move forward…
Is there a proper, ”automatic” way to solve this problem?
How about -
> library(reshape)
> dftab <- table(melt(df,'id'))
> dftab
, , value = 1
variable
id a b c d e
1 3 8 2 2 4
2 4 6 3 2 4
3 4 2 1 5 1
, , value = 2
variable
id a b c d e
1 0 1 4 3 3
2 3 3 3 6 2
3 1 4 5 3 4
, , value = 3
variable
id a b c d e
1 7 1 4 5 3
2 3 1 4 2 4
3 5 4 4 2 5
So to get the number of '3's in column 'a' and group '1'
you could just do
> dftab[3,'a',1]
[1] 4
A combination of tapply and apply can create the data you want:
tapply(df$id,df$id,function(x) apply(df[id==x,-1],2,table))
However, when a grouping doesn't have all the elements in it, as in 1a, the result will be a list for that id group rather than a nice table (matrix).
$`1`
$`1`$a
1 3
3 7
$`1`$b
1 2 3
8 1 1
$`1`$c
1 2 3
2 4 4
$`1`$d
1 2 3
2 3 5
$`1`$e
1 2 3
4 3 3
$`2`
a b c d e
1 4 6 3 2 4
2 3 3 3 6 2
3 3 1 4 2 4
$`3`
a b c d e
1 4 2 1 5 1
2 1 4 5 3 4
3 5 4 4 2 5
I'm sure someone will have a more elegant solution than this, but you can cobble it together with a simple function and dlply from the plyr package.
ColTables <- function(df) {
counts <- list()
for(a in names(df)[names(df) != "id"]) {
counts[[a]] <- table(df[a])
}
return(counts)
}
results <- dlply(df, "id", ColTables)
This gets you back a list - the first "layer" of the list will be the id variable; the second the table results for each column for that id variable. For example:
> results[['2']]['a']
$a
1 2 3
4 3 3
For id variable = 2, column = a, per your above example.
A way to do it is using the aggregate function, but you have to add a column to your dataframe
> df$freq <- 0
> aggregate(freq~a+id,df,length)
a id freq
1 1 1 3
2 3 1 7
3 1 2 4
4 2 2 3
5 3 2 3
6 1 3 4
7 2 3 1
8 3 3 5
Of course you can write a function to do it, so it's easier to do it frequently, and you don't have to add a column to your actual data frame
> frequency <- function(df,groups) {
+ relevant <- df[,groups]
+ relevant$freq <- 0
+ aggregate(freq~.,relevant,length)
+ }
> frequency(df,c("b","id"))
b id freq
1 1 1 8
2 2 1 1
3 3 1 1
4 1 2 6
5 2 2 3
6 3 2 1
7 1 3 2
8 2 3 4
9 3 3 4
You didn't say how you'd like the data. The by function might give you the output you like.
by(df, df$id, function(x) lapply(x[,-1], table))

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