Consider these two df examples
df1=data.frame(names=c('a','b','c'),value=1:3)
df2=data.frame(names=c('c','a','b'),value=1:3)
so that
> df1
names value
1 a 1
2 b 2
3 c 3
> df2
names value
1 c 1
2 a 2
3 b 3
Now, I would like to sort the df1 to the same order as the names column in df2, to obtain
names value
c 3
a 1
b 2
How can I achieve this?
try
df1[match(df2$names,df1$names),]
> df1[match(df2$names,df1$names),]
names value
3 c 3
1 a 1
2 b 2
Related
I'd like to add rows to a dataframe based on a vector within the dataframe. Here are the dataframes (df2 is the one I'd like to add rows to; df1 is the one I'd like to take the rows from):
ID=c(1:5)
x=c(rep("a",3),rep("b",2))
y=c(rep(0,5))
df1=data.frame(ID,x,y)
df2=df1[2:4,1:2]
df2$y=c(5,2,3)
df1
ID x y
1 1 a 0
2 2 a 0
3 3 a 0
4 4 b 0
5 5 b 0
df2
ID x y
2 2 a 5
3 3 a 2
4 4 b 3
I'd like to add to df2 any rows that aren't in df1, based on the ID vector. so my output dataframe would look like this:
ID x y
1 b 0
5 b 0
2 a 5
3 a 2
4 b 3
Can anyone see a way of doing this neatly, please? I need to do it for a lot of dataframes, all with different numbers of rows. I've tried using merge or rbind but I haven't been able to work out how to do it based on the vector.
Thank you!
A solution with dplyr:
bind_rows(df2,anti_join(df1,df2,by="ID"))
# ID x y
#1 2 a 5
#2 3 a 2
#3 4 b 3
#4 1 a 0
#5 5 b 0
You can do the following:
missingIDs <- which(!df1$ID %in% df2$ID) #check which df1 ID's are not in df2, see function is.element()
df.toadd <- df1[missingIDs,] #define the data frame to add to df2
result <- rbind(df.toadd, df2) #use rbind to add it
result
ID x y
1 1 a 0
5 5 b 0
2 2 a 5
3 3 a 2
4 4 b 3
What about this one-liner?
rbind(df2, df1[!df1$ID %in% df2$ID,])
ID x y
2 2 a 5
3 3 a 2
4 4 b 3
1 1 a 0
5 5 b 0
Given the following df:
a=c('a','b','c')
b=c(1,2,5)
c=c(2,3,4)
d=c(2,1,6)
df=data.frame(a,b,c,d)
a b c d
1 a 1 2 2
2 b 2 3 1
3 c 5 4 6
I'd like to apply a function that normally takes a vector (and returns a vector) like cummax row by row to the columns in position b to d.
Then, I'd like to have the output back in the df, either as a vector in a new column of the df, or replacing the original data.
I'd like to avoid writing it as a for loop that would iterate every row, pull out the content of the cells into a vector, do its thing and put it back.
Is there a more efficient way? I've given the apply family functions a go, but I'm struggling to first get a good way to vectorise content of columns by row and get the right output.
the final output could look something like that (imagining I've applied a cummax() function).
a b c d
1 a 1 2 2
2 b 2 3 3
3 c 5 5 6
or
a b c d output
1 a 1 2 2 (1,2,2)
2 b 2 3 1 (2,3,3)
3 c 5 4 6 (5,5,6)
where output is a vector.
Seems this would just be a simple apply problem that you want to cbind to df:
> cbind(df, apply(df[ , 4:2] # work with columns in reverse order
, 1, # do it row-by-row
cummax) )
a b c d 1 2 3
d a 1 2 2 2 1 6
c b 2 3 1 2 3 6
b c 5 4 6 2 3 6
Ouch. Bitten by failing to notice that this would be returned in a column oriented matrix and need to transpose that result; Such a newbie mistake. But it does show the value of having a question with a reproducible dataset I suppose.
> cbind(df, t(apply(df[ , 4:2] , 1, cummax) ) )
a b c d d c b
1 a 1 2 2 2 2 2
2 b 2 3 1 1 3 3
3 c 5 4 6 6 6 6
To destructively assign the result to df you would just use:
df <- # .... that code.
This does the concatenation with commas (and as a result no longer needs to be transposed:
> cbind(df, output=apply(df[ , 4:2] , 1, function(x) paste( cummax(x), collapse=",") ) )
a b c d output
1 a 1 2 2 2,2,2
2 b 2 3 1 1,3,3
3 c 5 4 6 6,6,6
I have a data frame initial of the following format
> head(initial)
Strings
1 A,A,B,C
2 A,B,C
3 A,A,A,A,A,B
4 A,A,B,C
5 A,B,C
6 A,A,A,A,A,B
and the data frame I want is final
> head(final)
Strings A B C
1 A,A,B,C 2 1 1
2 A,B,C 1 1 1
3 A,A,A,A,A,B 5 1 0
4 A,A,B,C 2 1 1
5 A,B,C 1 1 1
6 A,A,A,A,A,B 5 1 0
to generate the data frames the following codes can be used to keep the number of rows high
initial<-data.frame(Strings=rep(c("A,A,B,C","A,B,C","A,A,A,A,A,B"),100))
final<-data.frame(Strings=rep(c("A,A,B,C","A,B,C","A,A,A,A,A,B"),100),A=rep(c(2,1,5),100),B=rep(c(1,1,1),100),C=rep(c(1,1,0),100))
What is the fastest way I can achieve this? Any help will be greatly appreciated
We can use base R methods for this task. We split the 'Strings' column (strsplit(...)), set the names of the output list with the sequence of rows, stack to convert to data.frame with key/value columns, get the frequency with table, convert to 'data.frame' and cbind with the original dataset.
cbind(df1, as.data.frame.matrix(
table(
stack(
setNames(
strsplit(as.character(df1$Strings),','), 1:nrow(df1))
)[2:1])))
# Strings A B C D
#1 A,B,C,D 1 1 1 1
#2 A,B,B,D,D,D 1 2 0 3
#3 A,A,A,A,B,C,D,D 4 1 1 2
or we can use mtabulate after splitting the column.
library(qdapTools)
cbind(df1, mtabulate(strsplit(as.character(df1$Strings), ',')))
# Strings A B C D
#1 A,B,C,D 1 1 1 1
#2 A,B,B,D,D,D 1 2 0 3
#3 A,A,A,A,B,C,D,D 4 1 1 2
Update
For the new dataset 'initial', the second method works. If we need to use the first method with the correct order, convert to factor class with levels specified as the unique elements of 'ind'.
df1 <- stack(setNames(strsplit(as.character(initial$Strings), ','),
seq_len(nrow(initial))))
df1$ind <- factor(df1$ind, levels=unique(df1$ind))
cbind(initial, as.data.frame.matrix(table(df1[2:1])))
I have a dataframe that looks something like this, where each row represents a samples, and has repeats of the the same strings
> df
V1 V2 V3 V4 V5
1 a a d d b
2 c a b d a
3 d b a a b
4 d d a b c
5 c a d c c
I want to be able to create a new dataframe, where ideally the headers would be the string variables in the previous dataframe (a, b, c, d) and the contents of each row would be the number of occurrences of each the respective variable from
the original dataframe. Using the example from above, this would look like
> df2
a b c d
1 2 1 0 2
2 2 1 1 1
3 2 1 0 1
4 1 1 1 2
5 1 0 3 1
In my actual dataset, there are hundreds of variables, and thousands of samples, so it'd be ideal if I could automatically pull out the names from the original dataframe, and alphabetize them into the headers for the new dataframe.
You may try
library(qdapTools)
mtabulate(as.data.frame(t(df)))
Or
mtabulate(split(as.matrix(df), row(df)))
Or using base R
Un1 <- sort(unique(unlist(df)))
t(apply(df ,1, function(x) table(factor(x, levels=Un1))))
You can stack the columns and then use table:
table(cbind(id = 1:nrow(mydf),
stack(lapply(mydf, as.character)))[c("id", "values")])
# values
# id a b c d
# 1 2 1 0 2
# 2 2 1 1 1
# 3 2 2 0 1
# 4 1 1 1 2
# 5 1 0 3 1
I have a longitudinal data in a long format. I want to create an ID variable based on the variable-column that identifies each observation of my data. How do I do that in R?
Example: I have this data
name year var1 var2
A 1 4 3
A 2 5 1
A 3 4 2
B 1 . .
B 2 4 3
B 3 5 1
I want to produce a new column called 'id' with a unique number for every name, such as:
name id year var1 var2
A 1 1 4 3
A 1 2 5 1
A 1 3 4 2
B 2 1 . .
B 2 2 4 3
B 2 3 5 1
Any help?
If your name column doesn't just contain single letters (or even if it does), you can use:
dat$id <- as.numeric(as.factor(dat$name))
or, more simply:
dat$id <- c(as.factor(dat$name))
where dat is your data.frame.
tc='
name year var1 var2
A 1 4 3
A 2 5 1
A 3 4 2
B 1 . .
B 2 4 3
B 3 5 1'
df <- read.table(text=tc, header=T)
df$ID <- match(df$name, LETTERS)
Although is not clear if name is a column or are the rownames of the data frame.
If is not a column then try rownames(df) instead of df$name