I have two data.frames:
word1=c("a","a","a","a","b","b","b")
word2=c("a","a","a","a","c","c","c")
values1 = c(1,2,3,4,5,6,7)
values2 = c(3,3,0,1,2,3,4)
df1 = data.frame(word1,values1)
df2 = data.frame(word2,values2)
df1:
word1 values1
1 a 1
2 a 2
3 a 3
4 a 4
5 b 5
6 b 6
7 b 7
df2:
word2 values2
1 a 3
2 a 3
3 a 0
4 a 1
5 c 2
6 c 3
7 c 4
I would like to split these dataframes by word*, and perform two sample t.tests in R.
For example, the word "a" is in both data.frames. What's the t.test between the data.frames for the word "a"? And do this for all the words that are in both data.frames.
The result is a data.frame(result):
word tvalues
1 a 0.4778035
Thanks
Find the words common to both dataframes, then loop over these words, subsetting both dataframes and performing the t.test on the subsets.
E.g.:
df1 <- data.frame(word=sample(letters[1:5], 30, replace=TRUE),
x=rnorm(30))
df2 <- data.frame(word=sample(letters[1:5], 30, replace=TRUE),
x=rnorm(30))
common_words <- sort(intersect(df1$word, df2$word))
setNames(lapply(common_words, function(w) {
t.test(subset(df1, word==w, x), subset(df2, word==w, x))
}), common_words)
This returns a list, where each element is the output of the t.test for one of the common words. setNames just names the list elements so you can see which words they correspond to.
Note I've created new example data here since your example data only have one word in common (a) and so don't really resemble your true problem.
If you just want a matrix of statistics, you can do something like:
t(sapply(common_words, function(w) {
test <- t.test(subset(df1, word==w, x), subset(df2, word==w, x))
c(test$statistic, test$parameter, p=test$p.value,
`2.5%`=test$conf.int[1], `97.5%`=test$conf.int[2])
}))
## t df p 2.5% 97.5%
## a 0.9141839 8.912307 0.38468553 -0.4808054 1.1313220
## b -0.2182582 7.589109 0.83298193 -1.1536056 0.9558315
## c -0.2927253 8.947689 0.77640684 -1.5340097 1.1827691
## d -2.7244728 12.389709 0.01800568 -2.5016301 -0.2826952
## e -0.3683153 7.872407 0.72234501 -1.9404345 1.4072499
Related
I have a vector of variable names and several matrices with single rows.
I want to create a new matrix. The new matrix is created by match/merge the row names of the matrices with single rows.
Example:
A vector of variable names
Complete_names <- c("D","C","A","B")
Several matrices with single rows
Matrix_1 <- matrix(c(1,2,3),3,1)
rownames(Matrix_1) <- c("D","C","B")
Matrix_2 <- matrix(c(4,5,6),3,1)
rownames(Matrix_1) <- c("A","B","C")
Desired output:
Desired_output <- matrix(c(1,2,NA,3,NA,6,4,5),4,2)
rownames(Desired_output) <- c("D","C","A","B")
[,1] [,2]
D 1 NA
C 2 6
A NA 4
B 3 5
I know there are several similar postings like this, but those previous answers do not work perfectly for this one.
The main job can be done with merge, returning a data frame:
merge(Matrix_1, Matrix_2, by = "row.names", all = TRUE)
# Row.names V1.x V1.y
# 1 A NA 4
# 2 B 3 5
# 3 C 2 6
# 4 D 1 NA
Depending on your purposes you may then further modify names or get rid of Row.names.
The answers offered by Julius Vainora and achimneyswallow work well, but just to exactly obtain the desired output I want:
temp <- merge(Matrix_1, Matrix_2, by = "row.names", all = TRUE)
temp$Row.names <- factor(temp$Row.names, levels=Complete_names)
temp <- temp[order(temp$Row.names),]
rownames(temp) <- temp[,1]
Desired_output <- as.matrix(temp[,-1])
V1.x V1.y
D 1 NA
C 2 6
A NA 4
B 3 5
I have a simple problem which can be solved in a dirty way, but I'm looking for a clean way using data.table
I have the following data.table with n columns belonging to m unequal groups. Here is an example of my data.table:
dframe <- as.data.frame(matrix(rnorm(60), ncol=30))
cletters <- rep(c("A","B","C"), times=c(10,14,6))
colnames(dframe) <- cletters
A A A A A A
1 -0.7431185 -0.06356047 -0.2247782 -0.15423889 -0.03894069 0.1165187
2 -1.5891905 -0.44468389 -0.1186977 0.02270782 -0.64950716 -0.6844163
A A A A B B B
1 -1.277307 1.8164195 -0.3957006 -0.6489105 0.3498384 -0.463272 0.8458673
2 -1.644389 0.6360258 0.5612634 0.3559574 1.9658743 1.858222 -1.4502839
B B B B B B B
1 0.3167216 -0.2919079 0.5146733 0.6628149 0.5481958 -0.01721261 -0.5986918
2 -0.8104386 1.2335948 -0.6837159 0.4735597 -0.4686109 0.02647807 0.6389771
B B B B C C
1 -1.2980799 0.3834073 -0.04559749 0.8715914 1.1619585 -1.26236232
2 -0.3551722 -0.6587208 0.44822253 -0.1943887 -0.4958392 0.09581703
C C C C
1 -0.1387091 -0.4638417 -2.3897681 0.6853864
2 0.1680119 -0.5990310 0.9779425 1.0819789
What I want to do is to take a random subset of the columns (of a sepcific size), keeping the same number of columns per group (if the chosen sample size is larger than the number of columns belonging to one group, take all of the columns of this group).
I have tried an updated version of the method mentioned in this question:
sample rows of subgroups from dataframe with dplyr
but I'm not able to map the column names to the by argument.
Can someone help me with this?
Here's another approach, IIUC:
idx <- split(seq_along(dframe), names(dframe))
keep <- unlist(Map(sample, idx, pmin(7, lengths(idx))))
dframe[, keep]
Explanation:
The first step splits the column indices according to the column names:
idx
# $A
# [1] 1 2 3 4 5 6 7 8 9 10
#
# $B
# [1] 11 12 13 14 15 16 17 18 19 20 21 22 23 24
#
# $C
# [1] 25 26 27 28 29 30
In the next step we use
pmin(7, lengths(idx))
#[1] 7 7 6
to determine the sample size in each group and apply this to each list element (group) in idx using Map. We then unlist the result to get a single vector of column indices.
Not sure if you want a solution with dplyr, but here's one with just lapply:
dframe <- as.data.frame(matrix(rnorm(60), ncol=30))
cletters <- rep(c("A","B","C"), times=c(10,14,6))
colnames(dframe) <- cletters
# Number of columns to sample per group
nc <- 8
res <- do.call(cbind,
lapply(unique(colnames(dframe)),
function(x){
dframe[,if(sum(colnames(dframe) == x) <= nc) which(colnames(dframe) == x) else sample(which(colnames(dframe) == x),nc,replace = F)]
}
))
It might look complicated, but it really just takes all columns per group if there's less than nc, and samples random nc columns if there are more than nc columns.
And to restore your original column-name scheme, gsub does the trick:
colnames(res) <- gsub('.[[:digit:]]','',colnames(res))
I have a very simply question about lapply. I am transitioning from STATA to R and I think there is some very basic concept that I am not getting about looping in R. But I have been reading about it all afternoon and can't figure out a reasonable way to do this very simple thing.
I have three data frames df1, df2, and df3 that all have the same column names, in the same order, etc.
I want to rename their columns all at once.
I put the data frames in a list:
dflist <- list(df1, df2, df3)
What I want the new names to be:
varlist <- c("newname1", "newname2", "newname3")
Write a function that replaces names with those in varlist, and lapply it over the data frames
ChangeNames <- function(x) {
names(x) <- varlist
return(x)
}
dflist <- lapply(dflist, ChangeNames)
So, as far as I understand, R has changed the names of the copies of the data frames that I put in the list, but not the original data frames themselves. I want the data frames themselves to be renamed, not the elements of the list (which are trapped in a list).
Now, I can go
df1 <- as.data.frame(dflist[1])
df2 <- as.data.frame(dflist[2])
df2 <- as.data.frame(dflist[3])
But that seems weird. You need a loop to get back the elements of a loop?
Basically: once you've put some data frames in a list and run your function on them via lapply, how do you get them back out of the list, without starting back at square one?
If you just want to change the names, that isn't too hard in R. Bear in mind that the assignment operator, <-, can be applied in sequence. Hence:
names(df1) <- names(df2) <- names(df3) <- c("newname1", "newname2", "newname3")
I am not sure I understand correctly, do you want to rename the columns of the data frames or the components of the list that contain the data frames?
If it is the first, please always search before asking, the question has been asked here.
So what you can easily do in case you have even more data frames in the list is:
# Creating some sample data first
> dflist <- list(df1 = data.frame(a = 1:3, b = 2:4, c = 3:5),
+ df2 = data.frame(a = 4:6, b = 5:7, c = 6:8),
+ df3 = data.frame(a = 7:9, b = 8:10, c = 9:11))
# See how it looks like
> dflist
$df1
a b c
1 1 2 3
2 2 3 4
3 3 4 5
$df2
a b c
1 4 5 6
2 5 6 7
3 6 7 8
$df3
a b c
1 7 8 9
2 8 9 10
3 9 10 11
# And do the trick
> dflist <- lapply(dflist, setNames, nm = c("newname1", "newname2", "newname3"))
# See how it looks now
> dflist
$df1
newname1 newname2 newname3
1 1 2 3
2 2 3 4
3 3 4 5
$df2
newname1 newname2 newname3
1 4 5 6
2 5 6 7
3 6 7 8
$df3
newname1 newname2 newname3
1 7 8 9
2 8 9 10
3 9 10 11
So the names were changed from a, b and c to newname1, newname2and newname3 for each data frame in the list.
If it is the second, you can do this:
> names(dflist) <- c("newname1", "newname2", "newname3")
I've noticed that aggregate() appears to return its result ordered by the grouping column(s). Is this a guarantee? Can this be relied upon in surrounding logic?
A couple of examples:
set.seed(1); df <- data.frame(group=sample(letters[1:3],10,replace=T),value=1:10);
aggregate(value~group,df,sum);
## group value
## 1 a 16
## 2 b 22
## 3 c 17
And with two groups (notice the second group is ordered first, then the first group to break ties):
set.seed(1); df <- data.frame(group1=sample(letters[1:3],10,replace=T),group2=sample(letters[4:6],10,replace=T),value=1:10);
aggregate(value~group1+group2,df,sum);
## group1 group2 value
## 1 a d 1
## 2 b d 2
## 3 b e 9
## 4 c e 10
## 5 a f 15
## 6 b f 11
## 7 c f 7
Note: I'm asking because I just came up with an answer for Aggregating while merging two dataframes in R which, at least in its current form at the time of writing, depends on aggregate() returning its result ordered by the grouping column.
Yes, as long as you understand the natural ordering of factors to be by their integer keys. You can see this in the code:
y <- as.data.frame(by, stringsAsFactors = FALSE)
... # y becomes the "integerized" dataframe of index vectors
grp <- rank(do.call(paste, c(lapply(rev(y), ident), list(sep = "."))),
ties.method = "min")
y <- y[match(sort(unique(grp)), grp, 0L), , drop = FALSE]
...
I have an aggregation problem which I cannot figure out how to perform efficiently in R.
Say I have the following data:
group1 <- c("a","b","a","a","b","c","c","c","c",
"c","a","a","a","b","b","b","b")
group2 <- c(1,2,3,4,1,3,5,6,5,4,1,2,3,4,3,2,1)
value <- c("apple","pear","orange","apple",
"banana","durian","lemon","lime",
"raspberry","durian","peach","nectarine",
"banana","lemon","guava","blackberry","grape")
df <- data.frame(group1,group2,value)
I am interested in sampling from the data frame df such that I randomly pick only a single row from each combination of factors group1 and group2.
As you can see, the results of table(df$group1,df$group2)
1 2 3 4 5 6
a 2 1 2 1 0 0
b 2 2 1 1 0 0
c 0 0 1 1 2 1
shows that some combinations are seen more than once, while others are never seen. For those that are seen more than once (e.g., group1="a" and group2=3), I want to randomly pick only one of the corresponding rows and return a new data frame that has only that subset of rows. That way, each possible combination of the grouping factors is represented by only a single row in the data frame.
One important aspect here is that my actual data sets can contain anywhere from 500,000 rows to >2,000,000 rows, so it is important to be mindful of performance.
I am relatively new at R, so I have been having trouble figuring out how to generate this structure correctly. One attempt looked like this (using the plyr package):
choice <- function(x,label) {
cbind(x[sample(1:nrow(x),1),],data.frame(state=label))
}
df <- ddply(df[,c("group1","group2","value")],
.(group1,group2),
pick_junc,
label="test")
Note that in this case, I am also adding an extra column to the data frame called "label" which is specified as an extra argument to the ddply function. However, I killed this after about 20 min.
In other cases, I have tried using aggregate or by or tapply, but I never know exactly what the specified function is getting, what it should return, or what to do with the result (especially for by).
I am trying to switch from python to R for exploratory data analysis, but this type of aggregation is crucial for me. In python, I can perform these operations very rapidly, but it is inconvenient as I have to generate a separate script/data structure for each different type of aggregation I want to perform.
I want to love R, so please help! Thanks!
Uri
Here is the plyr solution
set.seed(1234)
ddply(df, .(group1, group2), summarize,
value = value[sample(length(value), 1)])
This gives us
group1 group2 value
1 a 1 apple
2 a 2 nectarine
3 a 3 banana
4 a 4 apple
5 b 1 grape
6 b 2 blackberry
7 b 3 guava
8 b 4 lemon
9 c 3 durian
10 c 4 durian
11 c 5 raspberry
12 c 6 lime
EDIT. With a data frame that big, you are better off using data.table
library(data.table)
dt = data.table(df)
dt[,list(value = value[sample(length(value), 1)]),'group1, group2']
EDIT 2: Performance Comparison: Data Table is ~ 15 X faster
group1 = sample(letters, 1000000, replace = T)
group2 = sample(LETTERS, 1000000, replace = T)
value = runif(1000000, 0, 1)
df = data.frame(group1, group2, value)
dt = data.table(df)
f1_dtab = function() {
dt[,list(value = value[sample(length(value), 1)]),'group1, group2']
}
f2_plyr = function() {ddply(df, .(group1, group2), summarize, value =
value[sample(length(value), 1)])
}
f3_by = function() {do.call(rbind,by(df,list(grp1 = df$group1,grp2 = df$group2),
FUN = function(x){x[sample(nrow(x),1),]}))
}
library(rbenchmark)
benchmark(f1_dtab(), f2_plyr(), f3_by(), replications = 10)
test replications elapsed relative
f1_dtab() 10 4.764 1.00000
f2_plyr() 10 68.261 14.32851
f3_by() 10 67.369 14.14127
One more way:
with(df, tapply(value, list( group1, group2), length))
1 2 3 4 5 6
a 2 1 2 1 NA NA
b 2 2 1 1 NA NA
c NA NA 1 1 2 1
# Now use tapply to sample withing groups
# `resample` fn is from the sample help page:
# Avoids an error with sample when only one value in a group.
resample <- function(x, ...) x[sample.int(length(x), ...)]
#Create a row index
df$idx <- 1:NROW(df)
rowidxs <- with(df, unique( c( # the `c` function will make a matrix into a vector
tapply(idx, list( group1, group2),
function (x) resample(x, 1) ))))
rowidxs
# [1] 1 5 NA 12 16 NA 3 15 6 4 14 10 NA NA 7 NA NA 8
df[rowidxs[!is.na(rowidxs)] , ]