Question
Let's say I have this dataframe:
# mock data set
df.size = 10
cluster.id<- sample(c(1:5), df.size, replace = TRUE)
letters <- sample(LETTERS[1:5], df.size, replace = TRUE)
test.set <- data.frame(cluster.id, letters)
Will be something like:
cluster.id letters
<int> <fctr>
1 5 A
2 4 B
3 4 B
4 3 A
5 3 E
6 3 D
7 3 C
8 2 A
9 2 E
10 1 A
Now I want to group these per cluster.id and see what kind of letters I can find within a cluster, so for example cluster 3 contains the letters A,E,D,C. Then I want to get all unique pairwise combinations (but not combinations with itself so no A,A e.g.): A,E ; A,D, A,C etc. Then I want to update the pairwise distance for these combination in an adjacency matrix/data frame.
Idea
# group by cluster.id
# per group get all (unique) pairwise combinations for the letters (excluding pairwise combinations with itself, e.g. A,A)
# update adjacency for each pairwise combinations
What I tried
# empty adjacency df
possible <- LETTERS
adj.df <- data.frame(matrix(0, ncol = length(possible), nrow = length(possible)))
colnames(adj.df) <- rownames(adj.df) <- possible
# what I tried
update.adj <- function( data ) {
for (comb in combn(data$letters,2)) {
# stucked
}
}
test.set %>% group_by(cluster.id) %>% update.adj(.)
Probably there is an easy way to do this because I see adjacency matrices all the time, but I'm not able to figure it out.. Please let me know if it's not clear
Answer to comment
Answer to #Manuel Bickel:
For the data I gave as example (the table under "will be something like"):
This matrix will be A-->Z for the full dataset, keep that in mind.
A B C D E
A 0 0 1 1 2
B 0 0 0 0 0
C 1 0 0 1 1
D 1 0 1 0 1
E 2 0 1 1 0
I will explain what I did:
cluster.id letters
<int> <fctr>
1 5 A
2 4 B
3 4 B
4 3 A
5 3 E
6 3 D
7 3 C
8 2 A
9 2 E
10 1 A
Only the clusters containing more > 1 unique letter are relevant (because we don't want combinations with itself, e.g cluster 1 containing only letter B, so it would result in combination B,B and is therefore not relevant):
4 3 A
5 3 E
6 3 D
7 3 C
8 2 A
9 2 E
Now I look for each cluster what pairwise combinations I can make:
cluster 3:
A,E
A,D
A,C
E,D
E,C
D,C
Update these combination in the adjacency matrix:
A B C D E
A 0 0 1 1 1
B 0 0 0 0 0
C 1 0 0 1 1
D 1 0 1 0 1
E 2 0 1 1 0
Then go to the next cluster
cluster 2
A,E
Update the adjacency matrix again:
A B C D E
A 0 0 1 1 2 <-- note the 2 now
B 0 0 0 0 0
C 1 0 0 1 1
D 1 0 1 0 1
E 2 0 1 1 0
As reaction to the huge dataset
library(reshape2)
test.set <- read.table(text = "
cluster.id letters
1 5 A
2 4 B
3 4 B
4 3 A
5 3 E
6 3 D
7 3 C
8 2 A
9 2 E
10 1 A", header = T, stringsAsFactors = F)
x1 <- reshape2::dcast(test.set, cluster.id ~ letters)
x1
#cluster.id A B C D E
#1 1 1 0 0 0 0
#2 2 1 0 0 0 1
#3 3 1 0 1 1 1
#4 4 0 2 0 0 0
#5 5 1 0 0 0 0
x2 <- table(test.set)
x2
# letters
#cluster.id A B C D E
# 1 1 0 0 0 0
# 2 1 0 0 0 1
# 3 1 0 1 1 1
# 4 0 2 0 0 0
# 5 1 0 0 0 0
x1.c <- crossprod(x1)
#Error in crossprod(x, y) :
# requires numeric/complex matrix/vector arguments
x2.c <- crossprod(x2)
#works fine
Following above comment, here the code of Tyler Rinker used with your data. I hope this is what you want.
UPDATE: Following below comments, I added a solution using the package reshape2 in order to be able to handle larger amounts of data.
test.set <- read.table(text = "
cluster.id letters
1 5 A
2 4 B
3 4 B
4 3 A
5 3 E
6 3 D
7 3 C
8 2 A
9 2 E
10 1 A", header = T, stringsAsFactors = F)
x <- table(test.set)
x
letters
#cluster.id A B C D E
# 1 1 0 0 0 0
# 2 1 0 0 0 1
# 3 1 0 1 1 1
# 4 0 2 0 0 0
# 5 1 0 0 0 0
#base approach, based on answer by Tyler Rinker
x <- crossprod(x)
diag(x) <- 0 #this is to set matches such as AA, BB, etc. to zero
x
# letters
# letters
# A B C D E
# A 0 0 1 1 2
# B 0 0 0 0 0
# C 1 0 0 1 1
# D 1 0 1 0 1
# E 2 0 1 1 0
#reshape2 approach
x <- acast(test.set, cluster.id ~ letters)
x <- crossprod(x)
diag(x) <- 0
x
# A B C D E
# A 0 0 1 1 2
# B 0 0 0 0 0
# C 1 0 0 1 1
# D 1 0 1 0 1
# E 2 0 1 1 0
Related
I have a questionnaire data that look like below:
items no_stars1 no_stars2 no_stars3 average satisfied bad
1 A 1 0 0 0 0 1
2 B 0 1 0 1 0 0
3 C 0 0 1 0 1 0
4 D 0 1 0 0 1 0
5 E 0 0 1 1 0 0
6 F 0 0 1 0 1 0
7 G 1 0 0 0 0 1
Basically, the header columns (no. of stars rating and satisfactory) are ordinal ranking for each Items. I would like to summarize the no_stars(col 2:4) and satisfactory(col 5:7) into one column so that the output would look like this :
items no_stars satisfactory
1 A 1 1
2 B 2 2
3 C 3 3
4 D 2 3
5 E 3 2
6 F 3 3
7 G 1 1
$no_stars <- 1 is for no_stars1, 2 for no_stars2, 3 for no_stars3
$satisfactory <- 1 is for bad, 2 for average, 3 for good
I have tried the code below
df$no_stars2[df$no_stars2 == 1] <- 2
df$no_stars3[df$no_stars3 == 1] <- 3
df$average[df$average == 1] <- 2
df$satisfied[df$satisfied == 1] <- 3
no_stars <- df$no_stars1 + df$no_stars2 + df$no_stars3
satisfactory <- df$bad + df$average + df$satisfied
tidy_df <- data.frame(df$Items, no_stars, satisfactory)
tidy_df
Is there any function in R that can do the same thing? or
anyone got better and simpler solution ?
Thanks
Just use max.col and set preferences:
starsOrder<-c("no_stars1","no_stars2","no_stars3")
satOrder<-c("bad","average","satisfied")
data.frame(items=df$items,no_stars=max.col(df[,starsOrder]),
satisfactory=max.col(df[,satOrder]))
# items no_stars satisfactory
#1 A 1 1
#2 B 2 2
#3 C 3 3
#4 D 2 3
#5 E 3 2
#6 F 3 3
#7 G 1 1
Another tidyverse solution making use of factor to integer conversions to encode no_stars and satisfactory and spreading from wide to long twice:
library(tidyverse)
df %>%
gather(no_stars, v1, starts_with("no_stars")) %>%
mutate(no_stars = as.integer(factor(no_stars))) %>%
gather(satisfactory, v2, average, satisfied, bad) %>%
filter(v1 > 0 & v2 > 0) %>%
mutate(satisfactory = as.integer(factor(
satisfactory, levels = c("bad", "average", "satisfied")))) %>%
select(-v1, -v2) %>%
arrange(items)
# items no_stars satisfactory
#1 A 1 1
#2 B 2 2
#3 C 3 3
#4 D 2 3
#5 E 3 2
#6 F 3 3
#7 G 1 1
While there may be more elegant solutions, using dplyr::case_when() gives you the flexibility to code things however you want:
library(dplyr)
df %>%
dplyr::mutate(
no_stars = dplyr::case_when(
no_stars1 == 1 ~ 1,
no_stars2 == 1 ~ 2,
no_stars3 == 1 ~ 3)
, satisfactory = dplyr::case_when(
average == 1 ~ 2,
satisfied == 1 ~ 3,
bad == 1 ~ 1)
)
# items no_stars1 no_stars2 no_stars3 average satisfied bad no_stars satisfactory
# 1 A 1 0 0 0 0 1 1 1
# 2 B 0 1 0 1 0 0 2 2
# 3 C 0 0 1 0 1 0 3 3
# 4 D 0 1 0 0 1 0 2 3
# 5 E 0 0 1 1 0 0 3 2
# 6 F 0 0 1 0 1 0 3 3
# 7 G 1 0 0 0 0 1 1 1
dat%>%
replace(.==1,NA)%>%
replace_na(setNames(as.list(names(.)),names(.)))%>%
replace(.==0,NA)%>%
mutate(s=coalesce(!!!.[2:4]),
no_stars=as.numeric(factor(s,unique(s))),
t=coalesce(!!!.[5:7]),
satisfactory=as.numeric(factor(t,unique(t))))%>%
select(items,no_stars,satisfactory)
items no_stars satisfactory
1 A 1 1
2 B 2 2
3 C 3 3
4 D 2 3
5 E 3 2
6 F 3 3
7 G 1 1
using apply and match :
data.frame(
items = df1$items,
no_stars = apply(df1[2:4], 1, match, x=1),
satisfactory = apply(df1[c(7,5:6)], 1, match, x=1))
# items no_stars satisfactory
# 1 A 1 1
# 2 B 2 2
# 3 C 3 3
# 4 D 2 3
# 5 E 3 2
# 6 F 3 3
# 7 G 1 1
data
df1 <- read.table(header=TRUE,stringsAsFactors=FALSE,text="
items no_stars1 no_stars2 no_stars3 average satisfied bad
1 A 1 0 0 0 0 1
2 B 0 1 0 1 0 0
3 C 0 0 1 0 1 0
4 D 0 1 0 0 1 0
5 E 0 0 1 1 0 0
6 F 0 0 1 0 1 0
7 G 1 0 0 0 0 1")
I have a following table in R
df <- data.frame('a' = c(1,0,0,1,0),
'b' = c(1,0,0,1,0),
'c' = c(1,1,0,1,1))
df
a b c
1 1 1 1
2 0 0 1
3 0 0 0
4 1 1 1
4 0 0 1
What I want is to replace the row value with the column name whenever the row is equal to 1. The output would be this one:
a b c
1 a b c
2 0 0 c
3 0 0 0
4 a b c
4 0 0 c
How can I do this in R? Thanks.
I would use Map and replace:
df[] <- Map(function(n, x) replace(x, x == 1, n), names(df), df)
df
# a b c
# 1 a b c
# 2 0 0 c
# 3 0 0 0
# 4 a b c
# 5 0 0 c
We can use
df[] <- names(df)[(NA^!df) * col(df)]
df[is.na(df)] <- 0
df
# a b c
#1 a b c
#2 0 0 c
#3 0 0 0
#4 a b c
#4 0 0 c
You can try stack and unstack
a=stack(df)
a
values ind
1 1 a
2 0 a
3 0 a
4 1 a
5 0 a
6 1 b
7 0 b
8 0 b
9 1 b
10 0 b
11 1 c
12 1 c
13 0 c
14 1 c
15 1 c
a$values[a$values==1]=as.character(a$ind)[a$values==1]
unstack(a)
a b c
1 a b c
2 0 0 c
3 0 0 0
4 a b c
5 0 0 c
We can try iterating over the names of the data frame, and then handling each column, for a base R option:
df <- data.frame(a=c(1,0,0,1,0), b=c(1,0,0,1,0), c=c(1,1,0,1,1))
df <- data.frame(sapply(names(df), function(x) {
y <- df[[x]]
y[y == 1] <- x
return(y)
}))
df
a b c
1 a b c
2 0 0 c
3 0 0 0
4 a b c
5 0 0 c
Demo
You can do it with ifelse, but you have to do some intermediate transposing to account for R's column-major order processing.
data.frame(t(ifelse(t(df)==1,names(df),0)))
a b c
1 a b c
2 0 0 c
3 0 0 0
4 a b c
5 0 0 c
I have a dataframe in which there are multiple columns (more than 30) that is saved in a list. I would like to apply the same criteria for all those columns without writing each code for each columns. I have example below to help understand my problem better
A<-c("A","B","C","D","E","F","G","H","I")
B<-c(0,0,0,1,2,3,0,0,0)
C<-c(0,1,0,0,1,2,0,0,0)
D<-c(0,0,0,0,1,1,0,1,0)
E<-c(0,0,0,0,0,0,0,1,0)
data<-data.frame(A,B,C,D,E)
Let say I have the above df as an example and I have saved the list of cols as below
list <- c("B","C","D","E")
I would like to use those cols with the same criteria as below
setDT(data)[B>=1 | C>=1 | D>=1 | E>=1]
And get the following result
A B C D E
1: B 0 1 0 0
2: D 1 0 0 0
3: E 2 1 1 0
4: F 3 2 1 0
5: H 0 0 1 1
However, is there a way to get the above answer without writing each individual column criteria (e.g. B>=1 | C>=1 ....) since I have more than 30 cols in the actual data. Thanks a lot
For your specific example of checking if at least one value in a row is at least 1, you could use rowSums
data[rowSums(data[,-1]) > 0, ]
# A B C D E
# 2 B 0 1 0 0
# 4 D 1 0 0 0
# 5 E 2 1 1 0
# 6 F 3 2 1 0
# 8 H 0 0 1 1
If you have other criteria in mind, you might as well consider using any within apply
ind <- apply(data[,-1], 1, function(x) {any(x >= 1)})
data[ind,]
# A B C D E
# 2 B 0 1 0 0
# 4 D 1 0 0 0
# 5 E 2 1 1 0
# 6 F 3 2 1 0
# 8 H 0 0 1 1
dplyr::filter_at will do just that.
library(dplyr)
data %>% filter_at(vars(-A),any_vars(.>=1))
# A B C D E
# 1 B 0 1 0 0
# 2 D 1 0 0 0
# 3 E 2 1 1 0
# 4 F 3 2 1 0
# 5 H 0 0 1 1
You could always use Reduce, this is nice because you can put any type of logic you want into the function:
A simple method might be:
data[Reduce("|", as.data.frame(data[,list] >= 1)),]
# A B C D E
#2 B 0 1 0 0
#4 D 1 0 0 0
#5 E 2 1 1 0
#6 F 3 2 1 0
#8 H 0 0 1 1
A little explanation: Reduce successively applies the same function to each element of x. In this case the "|" operator is applied to each of the logical columns of the data.frame.
If you wanted to do more complicated logic checks you could do that with your own anonymous function.
Please check this using applyin R.
B<-c(0,0,0,1,2,3,0,0,0)
C<-c(0,1,0,0,1,2,0,0,0)
D<-c(0,0,0,0,1,1,0,1,0)
ef=data.frame(B,C,D)
con=apply(ef,2,function(x) x>1 )
I have a list of IDs, each having multiple events. The data looks like an event log, i.e. one event per ID per row. For example:
n.ID=4
n.events=5
set.seed(1234)
df <- setNames(melt(replicate(n.ID, sort(sample(letters[c(1:10)], n.events))))[c(2:3)], c("ID", "Event"))
df
> df
ID Event
1 1 b
2 1 e
3 1 f
4 1 h
5 1 i
6 2 a
7 2 b
8 2 d
9 2 e
10 2 g
11 3 b
12 3 c
13 3 e
14 3 g
15 3 j
16 4 b
17 4 c
18 4 g
19 4 i
20 4 j
I want to select those IDs, that meet a list of criteria, that either use AND or OR.
For example:
those IDs that have events "b" AND "c" AND "g" --> results in ID 3 & 4
those IDs that have events "a" OR "h" --> results in ID 1 & 2
The criteria vectors can be any length.
EDIT:
I am aware of %in% and "|", however,
keep.if <- c("b", "c", "g") # This list can be of any length
subset(df, Event %in% keep.if)
ID Event
1 1 b
7 2 b
10 2 g
11 3 b
12 3 c
14 3 g
16 4 b
17 4 c
18 4 g
I only want those that have 3 rows in the results, so i can do a table on this results, and select those IDs where the Freq == length(keep.if)... but I guess there should be an easier, less messy method...
I guess for the OR version I can just take:
unique(subset(df, Event %in% keep.if)$ID)
I would create a table then use tidyr::spread to create a contigency table type object. Then I would use data.table for easier sub setting and logical operations:
library(tidyr)
df.table<-as.data.frame(table(df)) %>% spread(Event, Freq)
df.table
ID a b c d e f g h i j
1 0 1 0 0 1 1 0 1 1 0
2 1 1 0 1 1 0 1 0 0 0
3 0 1 1 0 1 0 1 0 0 1
4 0 1 1 0 0 0 1 0 1 1
library(data.table)
##easier to subset with
df.table<-data.table(df.table)
df.table[b & c & g]
ID a b c d e f g h i j
3 0 1 1 0 1 0 1 0 0 1
4 0 1 1 0 0 0 1 0 1 1
df.table[a | h]
ID a b c d e f g h i j
1 0 1 0 0 1 1 0 1 1 0
2 1 1 0 1 1 0 1 0 0 0
Those are the 2 examples you gave in the question. You should be able to do just about any operation you want. Also, if you only want to know which IDs satisfy your logic (and not their entire contingency table) then:
df.table[b & c & g]$ID
[1] 3 4
I have a data frame with 5*n columns, where n is the number of categories listed in a vector. I want to break the data frame into chunks of 5 columns (eg. category 1 is columns 1:5, category 2 is columns 6:10) and then assign the category names from the vector to the chunks.
eg.
*original data frame* *vector of category names*
X a b c d e a b c d e a b c d e 1 apples
1 1 0 0 0 1 0 1 0 1 0 0 0 1 1 0 2 oranges
2 0 1 0 1 0 0 0 1 0 1 1 0 0 0 1 3 bananas
Will become
*apples* *oranges* *bananas*
X a b c d e X a b c d e X a b c d e
1 1 0 0 0 1 1 0 1 0 1 0 1 0 0 1 1 0
2 0 1 0 1 0 2 0 0 1 0 1 2 1 0 0 0 1
I can find a whole lot of information about splitting data.frames by rows, which is much more common to do, but I can't find anything about splitting a data frame into n chunks by columns. Thanks!
You could split your original_data_frame by column indices similarely:
df <- read.table(header=T, check.names = F, text="
X a b c d e a b c d e a b c d e
1 1 0 0 0 1 0 1 0 1 0 0 0 1 1 0
2 0 1 0 1 0 0 0 1 0 1 1 0 0 0 1")
n <- 5 # fixed chunksize (a-e)
lst <- lapply(split(2:ncol(df), rep(seq(ncol(df[-1])/n), each=n)), function(x) df[, x])
names(lst) <- c("apples", "oranges", "bananas")
# lst
# $apples
# a b c d e
# 1 1 0 0 0 1
# 2 0 1 0 1 0
#
# $oranges
# a b c d e
# 1 0 1 0 1 0
# 2 0 0 1 0 1
#
# $bananas
# a b c d e
# 1 0 0 1 1 0
# 2 1 0 0 0 1
I don't know if this is elegant, but it came to my mind, first.