I have got the following dataset:
ID s1 s2 s3
A 0.6 1 0.3
B 3 0.4 0.4
C 3 2 1
D 0 0.3 0.2
E 3 2 0.1
i would like to retain the rows which have the value >=0.5 at least two of the 3 samples
So, the new data frame would be:
ID s1 s2 s3
A 0.6 1 0.3
C 3 2 1
E 3 2 0.1
Thanks in advance
You can do
df[rowSums(df[-1] > 0.5) >= 2, ]
# ID s1 s2 s3
#1 A 0.6 1 0.3
#3 C 3.0 2 1.0
#5 E 3.0 2 0.1
We create a logical matrix df[-1] > 0.5 and check if at least two values per row are TRUE.
data
df <- read.table(text="ID s1 s2 s3
A 0.6 1 0.3
B 3 0.4 0.4
C 3 2 1
D 0 0.3 0.2
E 3 2 0.1", header = TRUE, stringsAsFactor = FALSE)
Related
I did some searches but could not find the best keywords to phrase my question so I think I will attempt to ask it here.
I am dealing with a data frame in R that have two variables represent the identity of the data points. In the following example, A and 1 represent the same individual, B and 2 are the same and so are C and 3 but they are being mixed in the original data.
ID1 ID2 Value
A 1 0.5
B 2 0.8
C C 0.7
A A 0.6
B 2 0.3
3 C 0.4
2 2 0.3
1 A 0.4
3 3 0.6
What I want to achieve is to unify the identity by using only one of the identifiers so it can be either:
ID1 ID2 Value ID
A 1 0.5 A
B 2 0.8 B
C C 0.7 C
A A 0.6 A
B 2 0.3 B
3 C 0.4 C
2 2 0.3 B
1 A 0.4 A
3 3 0.6 C
or:
ID1 ID2 Value ID
A 1 0.5 1
B 2 0.8 2
C C 0.7 3
A A 0.6 1
B 2 0.3 2
3 C 0.4 3
2 2 0.3 2
1 A 0.4 1
3 3 0.6 3
I can probably achieve it by using ifelse function but that means I have to write two ifelse statements for each condition and it does not seem efficient so I was wondering if there is a better way to do it. Here is the example data set.
df=data.frame(ID1=c("A","B","C","A","B","3","2","1","3"),
ID2=c("1","2","C","A","2","C","2","A","3"),
Value=c(0.5,0.8,0.7,0.6,0.3,0.4,0.3,0.4,0.6))
Thank you so much for the help!
Edit:
To clarify, the two identifiers I have in my real data are longer string of texts instead of just ABC and 123. Sorry I did not make it clear.
An option is to to detect the elements that are only digits, convert to integer, then get the corresponding LETTERS in case_when
library(dplyr)
library(stringr)
df %>%
mutate(ID = case_when(str_detect(ID1, '\\d+')~
LETTERS[as.integer(ID1)], TRUE ~ ID1))
# ID1 ID2 Value ID
#1 A 1 0.5 A
#2 B 2 0.8 B
#3 C C 0.7 C
#4 A A 0.6 A
#5 B 2 0.3 B
#6 3 C 0.4 C
#7 2 2 0.3 B
#8 1 A 0.4 A
#9 3 3 0.6 C
Or more compactly
df %>%
mutate(ID = coalesce(LETTERS[as.integer(ID1)], ID1))
If we have different sets of values, then create a key/value dataset and do a join
keyval <- data.frame(ID1 = c('1', '2', '3'), ID = c('A', 'B', 'C'))
left_join(df, keyval) %>% mutate(ID = coalesce(ID, ID1))
A base R option using replace
within(
df,
ID <- replace(
ID1,
!ID1 %in% LETTERS,
LETTERS[as.numeric(ID1[!ID1 %in% LETTERS])]
)
)
or ifelse
within(
df,
ID <- suppressWarnings(ifelse(ID1 %in% LETTERS,
ID1,
LETTERS[as.integer(ID1)]
))
)
which gives
ID1 ID2 Value ID
1 A 1 0.5 A
2 B 2 0.8 B
3 C C 0.7 C
4 A A 0.6 A
5 B 2 0.3 B
6 3 C 0.4 C
7 2 2 0.3 B
8 1 A 0.4 A
9 3 3 0.6 C
I have two data base, df and cf. I want to multiply each value of A in df by each coefficient in cf depending on the value of B and C in table df.
For example
row 2 in df A= 20 B= 4 and C= 2 so the correct coefficient is 0.3,
the result is 20*0.3 = 6
There is a simple way to do that in R!?
Thanks in advance!!
df
A B C
20 4 2
30 4 5
35 2 2
24 3 3
43 2 1
cf
C
B/C 1 2 3 4 5
1 0.2 0.3 0.5 0.6 0.7
2 0.1 0.5 0.3 0.3 0.4
3 0.9 0.1 0.6 0.6 0.8
4 0.7 0.3 0.7 0.4 0.6
One solution with apply:
#iterate over df's rows
apply(df, 1, function(x) {
x[1] * cf[x[2], x[3]]
})
#[1] 6.0 18.0 17.5 14.4 4.3
Try this vectorized:
df[,1] * cf[as.matrix(df[,2:3])]
#[1] 6.0 18.0 17.5 14.4 4.3
A solution using dplyr and a vectorised function:
df = read.table(text = "
A B C
20 4 2
30 4 5
35 2 2
24 3 3
43 2 1
", header=T, stringsAsFactors=F)
cf = read.table(text = "
0.2 0.3 0.5 0.6 0.7
0.1 0.5 0.3 0.3 0.4
0.9 0.1 0.6 0.6 0.8
0.7 0.3 0.7 0.4 0.6
")
library(dplyr)
# function to get the correct element of cf
# vectorised version
f = function(x,y) cf[x,y]
f = Vectorize(f)
df %>%
mutate(val = f(B,C),
result = val * A)
# A B C val result
# 1 20 4 2 0.3 6.0
# 2 30 4 5 0.6 18.0
# 3 35 2 2 0.5 17.5
# 4 24 3 3 0.6 14.4
# 5 43 2 1 0.1 4.3
The final dataset has both result and val in order to check which value from cf was used each time.
Here is a sample data set:
sample1 <- data.frame(Names=letters[1:10], Values=sample(seq(0.1,1,0.1)))
When I'm reordering the data set, I'm losing the row names order
sample1[order(sample1$Values), ]
Names Values
7 g 0.1
4 d 0.2
3 c 0.3
9 i 0.4
10 j 0.5
5 e 0.6
8 h 0.7
6 f 0.8
1 a 0.9
2 b 1.0
Desired output:
Names Values
1 g 0.1
2 d 0.2
3 c 0.3
4 i 0.4
5 j 0.5
6 e 0.6
7 h 0.7
8 f 0.8
9 a 0.9
10 b 1.0
Try
rownames(Ordersample2) <- 1:10
or more generally
rownames(Ordersample2) <- NULL
I had a dplyr usecase:
df %>% as.data.frame(row.names = 1:nrow(.))
I have a data.frame that looks like this: (my real dataframe is bigger):
df <- data.frame(A=c("a","b","c","d","e","f","g","h","i"),
B=c("1","1","1","2","2","2","3","3","3"),
C=c(0.1,0.2,0.4,0.1,0.5,0.7,0.1,0.2,0.5))
> df
A B C
1 a 1 0.1
2 b 1 0.2
3 c 1 0.4
4 d 2 0.1
5 e 2 0.5
6 f 2 0.7
7 g 3 0.1
8 h 3 0.2
9 i 3 0.5
I want to add several n-columns (something similar to permutations) where the column D would be a random value from df$C but this value should only be picked from those rows with the dame value of df$B, an example of the desired output would be:
df <- data.frame(A=c("a","b","c","d","e","f","g","h","i"),
B=c("1","1","1","2","2","2","3","3","3"),
C=c(0.1,0.2,0.4,0.1,0.5,0.7,0.1,0.2,0.5),
D=c(0.2,0.2,0.1,0.5,0.7,0.1,0.5,0.5,0.2))
> df
A B C D
1 a 1 0.1 0.2
2 b 1 0.2 0.2
3 c 1 0.4 0.1
4 d 2 0.1 0.5
5 e 2 0.5 0.7
6 f 2 0.7 0.1
7 g 3 0.1 0.5
8 h 3 0.2 0.5
9 i 3 0.5 0.2
I've tried with plyr package but my approach does not work properly:
ddply(df, levels(.(B)), transform, D=sample(C))
I also have thought about splitting the dataframe based on df$B and then using a function to add the column in each dataframe using lapply however I have no clue how select for the levels of df$B,
Many thanks
No need for plyr, ave will do the trick.
transform(df, D=ave(C, B, FUN=function(b) sample(b, replace=TRUE)))
I have this data frame:
df <- data.frame(A=c("a","b","c","d","e","f","g","h","i"),
B=c("1","1","1","2","2","2","3","3","3"),
C=c(0.1,0.2,0.4,0.1,0.5,0.7,0.1,0.2,0.5))
> df
A B C
1 a 1 0.1
2 b 1 0.2
3 c 1 0.4
4 d 2 0.1
5 e 2 0.5
6 f 2 0.7
7 g 3 0.1
8 h 3 0.2
9 i 3 0.5
I would like to add 1000 further columns and fill this columns with the values generated by :
transform(df, D=ave(C, B, FUN=function(b) sample(b, replace=TRUE)))
I've tried with a for loop but it does not work:
for (i in 4:1000){
df[, 4:1000] <- NA
df[,i] = transform(df, D=ave(C, B, FUN=function(b) sample(b, replace=TRUE)))
}
For efficiency reasons, I suggest running sample only once for each group. This can be achieved with this:
sample2 <- function(x, size)
{
if(length(x)==1) rep(x, size) else sample(x, size, replace=TRUE)
}
new_df <- do.call(rbind, by(df, df$B,
function(d) cbind(d, matrix(sample2(d$C, length(d$C)*1000),
ncol=1000))))
Notes:
I've created sample2 in case there is a group with only one C value. Check ?sample to see what I mean.
The names of the columns will be numbers, from 1 to 1000. This can be changed as in the answer by #agstudy.
The row names are also changed. "Fixing" them is similar, just use row.names instead of col.names.
Using replicate for example:
cbind(df,replicate(1000,ave(df$C, df$B,
FUN=function(b) sample(b, replace=TRUE))))
To add 4 columns for example:
cbind(df,replicate(4,ave(df$C, df$B,
FUN=function(b) sample(b, replace=TRUE))))
A B C 1 2 3 4
1 a 1 0.1 0.2 0.2 0.1 0.2
2 b 1 0.2 0.4 0.2 0.4 0.4
3 c 1 0.4 0.1 0.1 0.1 0.1
4 d 2 0.1 0.1 0.5 0.5 0.1
5 e 2 0.5 0.7 0.1 0.5 0.1
6 f 2 0.7 0.1 0.7 0.7 0.7
7 g 3 0.1 0.2 0.5 0.2 0.2
8 h 3 0.2 0.2 0.1 0.2 0.1
9 i 3 0.5 0.5 0.5 0.1 0.5
Maybe you need to rename columns by something like :
gsub('([0-9]+)','D\\1',colnames(res))
1] "A" "B" "C" "D1" "D2" "D3" "D4"