I am currently trying to run multiple data using loop, and ended up with the results below. Each line corresponds to the output from one data that has been filtered.
I am using this code to get the results below.
output <- print(paste(data.final$Peptide, collapse = ','))
It was previously given the form of table with one of the column name "Peptide", so I am pasting the peptide into a string separated comma, as shown here :
[1]"LPPAYTNSF,LPFNDGVYF,WMESEFRVY,YSSANNCTF,SANNCTFEY,KIYSKHTPI,WTAGAAAYY,YLQPRTFLL,CVADYSVLY,YNSASFSTF,ASFSTFKCY,RLFRKSNLK,TSNQVAVLY"
[1]"LPPAYTNSF,LPFNDGVYF,WMESEFRVY,YSSANNCTF,SANNCTFEY,KIYSKHTPI,WTAGAAAYY,YLQPRTFLL,CVADYSVLY,YNSASFSTF,ASFSTFKCY,RLFRKSNLK,QSYGFQPTY,TSNQVAVLY"
[1]"LPPAYTNSF,LPFNDGVYF,WMESEFRVY,YSSANNCTF,SANNCTFEY,KIYSKHTPI,WTAGAAAYY,YLQPRTFLL,CVADYSVLY,YNSASFSTF,ASFSTFKCY,RLFRKSNLK,QSYGFQPTY,TSNQVAVLY"
[1]"LPPAYTNSF,LPFNDGVYF,WMESEFRVY,YSSANNCTF,SANNCTFEY,KIYSKHTPI,WTAGAAAYY,YLQPRTFLL,CVADYSVLY,YNSASFSTF,ASFSTFKCY,RLFRKSNLK,QSYGFQPTY,TSNQVAVLY"
[1]"LPPAYTNSF,LPFNDGVYF,WMESEFRVY,YSSANNCTF,SANNCTFEY,KIYSKHTPI,WTAGAAAYY,YLQPRTFLL,CVADYSVLY,YNSASFSTF,ASFSTFKCY,RLFRKSNLK,QSYGFQPTY,TSNQVAVLY"
[1]"LPSAYTNSF,SAYTNSFTR,LPFNDGVYF,WMESEFRVY,YSSANNCTF,SANNCTFEY,KIYSKHTPI,WTAGAAAYY,YLQPRTFLL,CVADYSVLY,YNSASFSTF,ASFSTFKCY,RLFRKSNLK,QSYGFQPTY,TSNQVAVLY"
[1]"LPSAYTNSF,SAYTNSFTR,LPFNDGVYF,WMESEFRVY,YSSANNCTF,SANNCTFEY,KIYSKHTPI,WTAGAAAYY,YLQPRTFLL,CVADYSVLY,YNSASFSTF,ASFSTFKCY,RLFRKSNLK,QSYGFQPTY,TSNQVAVLY"
[1]"LPPAYTNSF,LPFNDGVYF,WMESEFRVY,YSSANNCTF,SANNCTFEY,KIYSKHTPI,WTAGAAAYY,YLQPRTFLL,CVADYSVLY,YNSASFSTF,ASFSTFKCY,RLFRKSNLK,TSNQVAVLY"
I would like to find the number of duplicates from each comma-separated strings (eg. LPPAYTNSF) between the lines.
Is there anyways to do this?
If I understand you well, there is no reason to first collapse them to a string. Why not just get the counts from your data.frame? Any n > 1 has duplicates.
data.final %>% count(Peptide)
solution based on your string
table(unlist(strsplit(v, ",")))
results
ASFSTFKCY CVADYSVLY KIYSKHTPI LPFNDGVYF LPPAYTNSF LPSAYTNSF QSYGFQPTY RLFRKSNLK SANNCTFEY SAYTNSFTR TSNQVAVLY WMESEFRVY
8 8 8 8 6 2 6 8 8 2 8 8
WTAGAAAYY YLQPRTFLL YNSASFSTF YSSANNCTF
8 8 8 8
data
v <- "LPPAYTNSF,LPFNDGVYF,WMESEFRVY,YSSANNCTF,SANNCTFEY,KIYSKHTPI,WTAGAAAYY,YLQPRTFLL,CVADYSVLY,YNSASFSTF,ASFSTFKCY,RLFRKSNLK,TSNQVAVLY,LPPAYTNSF,LPFNDGVYF,WMESEFRVY,YSSANNCTF,SANNCTFEY,KIYSKHTPI,WTAGAAAYY,YLQPRTFLL,CVADYSVLY,YNSASFSTF,ASFSTFKCY,RLFRKSNLK,QSYGFQPTY,TSNQVAVLY,LPPAYTNSF,LPFNDGVYF,WMESEFRVY,YSSANNCTF,SANNCTFEY,KIYSKHTPI,WTAGAAAYY,YLQPRTFLL,CVADYSVLY,YNSASFSTF,ASFSTFKCY,RLFRKSNLK,QSYGFQPTY,TSNQVAVLY,LPPAYTNSF,LPFNDGVYF,WMESEFRVY,YSSANNCTF,SANNCTFEY,KIYSKHTPI,WTAGAAAYY,YLQPRTFLL,CVADYSVLY,YNSASFSTF,ASFSTFKCY,RLFRKSNLK,QSYGFQPTY,TSNQVAVLY,LPPAYTNSF,LPFNDGVYF,WMESEFRVY,YSSANNCTF,SANNCTFEY,KIYSKHTPI,WTAGAAAYY,YLQPRTFLL,CVADYSVLY,YNSASFSTF,ASFSTFKCY,RLFRKSNLK,QSYGFQPTY,TSNQVAVLY,LPSAYTNSF,SAYTNSFTR,LPFNDGVYF,WMESEFRVY,YSSANNCTF,SANNCTFEY,KIYSKHTPI,WTAGAAAYY,YLQPRTFLL,CVADYSVLY,YNSASFSTF,ASFSTFKCY,RLFRKSNLK,QSYGFQPTY,TSNQVAVLY,LPSAYTNSF,SAYTNSFTR,LPFNDGVYF,WMESEFRVY,YSSANNCTF,SANNCTFEY,KIYSKHTPI,WTAGAAAYY,YLQPRTFLL,CVADYSVLY,YNSASFSTF,ASFSTFKCY,RLFRKSNLK,QSYGFQPTY,TSNQVAVLY,LPPAYTNSF,LPFNDGVYF,WMESEFRVY,YSSANNCTF,SANNCTFEY,KIYSKHTPI,WTAGAAAYY,YLQPRTFLL,CVADYSVLY,YNSASFSTF,ASFSTFKCY,RLFRKSNLK,TSNQVAVLY"
Not quite sure what your data are like so I've scribbled together some toy data to show how a regex and a non-regex tidyverse solution would work on your task of "find[ing] the number of duplicates from each comma-separated strings":
Data:
x <- c("a,c,x,a,f,s,w,s,b,n,x,q",
"A,B,B,X,B,Q")
A regex tidyverse solution:
library(tidyverse)
data.frame(x) %>%
# create new column with duplicated values:
mutate(dups = str_extract_all(x, "([A-Za-z])+(?=.*\\1)"))
x dups
1 a,c,x,a,f,s,w,s,b,n,x,q a, x, s
2 A,B,B,X,B,Q B, B
NB: the + in the regex pattern makes sure you can use this solution also for longer-than-one-character alphabetic comma-separated strings
Alternatively, a non-regex tidyverse solution:
library(tidyverse)
data.frame(x) %>%
# make column with unique ID for each string:
mutate(stringID = row_number()) %>%
# separate values into rows:
separate_rows(x) %>%
# for each combination of `stringID` and `x`...
group_by(stringID, x) %>%
# ...count the number of tokens:
summarise(N = n()) %>%
# show only the duplicated values:
filter(N > 1)
# A tibble: 4 × 3
# Groups: stringID [2]
stringID x N
<int> <chr> <int>
1 1 a 2
2 1 s 2
3 1 x 2
4 2 B 3
I have all column names that start with 'm'. Example: mIncome, mAge. I want to remove the prefix. So far, I have tried the following:
df %>%
rename_all(~stringr::str_replace_all(.,"m",""))
This removes all the column names that has the letter 'm'. I just need it removed from from the start. Any suggestions?
You can use sub in base R to remove "m" from the beginning of the column names.
names(df) <- sub('^m', '', names(df))
We need to specify the location. The ^ matches the start of the string (or here the column name). So, if we use ^m, it will only match 'm' at the beginning or start of the string and not elsewhere.
library(dplyr)
library(stringr)
df %>%
rename_all(~stringr::str_replace(.,"^m",""))
# ba Mbgeg gmba cfor
#1 1 2 4 6
#2 2 3 5 7
#3 3 4 6 8
Also, if the case should be ignored, wrap with regex and specify ignore_case = TRUE
df %>%
rename_all(~ stringr::str_replace(., regex("^m", ignore_case = TRUE), ""))
# ba bgeg gmba cfor
#1 1 2 4 6
#2 2 3 5 7
#3 3 4 6 8
Another option is word boundary (\\bm), but this could match the beginning of words where there are multi word column names
NOTE: str_replace_all is used when we want to replace multiple occurrence of the pattern. Here, we just need to replace the first instance and for that str_replace is enough.
data
df <- data.frame(mba = 1:3, Mbgeg = 2:4, gmba = 4:6, cfor = 6:8)
Another way you can try
library(tidyverse)
df <- data.frame(mma = 1:2, mbapbe = 1:2)
df2 <- df %>%
rename_at(vars(c("mma", "mbapbe")) ,function(x) gsub("^m", "", x))
# ma bapbe
# 1 1 1
# 2 2 2
I'm trying to detect how many words in a vector have a repeating letter and count the number of times that it is repeated in other words also, adding it to a data frame each time the repeated letters are encountered.
For example: x = c("google", "blood", "street")
the data frame will appear as
letter n
1 oo 2
2 ee 1
You can match repeating letters using regex and match using stringr::str_match_all():
library(stringr)
as.data.frame(table(unlist(sapply(str_match_all(x, regex("([A-Za-z]{1})\\1")), `[`, , 1))))
Var1 Freq
1 ee 1
2 oo 2
One option in base R is to convert to raw, use rle to get the run-length-encoding, subset only the elements having lengths greater than 1, reconvert to character and get the frequency count with table
stack(table(sapply(x, function(y) rawToChar(with(rle(charToRaw(y)),
rep(values[lengths > 1], lengths[lengths > 1]))))))[2:1]
# ind values
#1 ee 1
#2 oo 2
Or with str_extract (assuming there is only a single repeated substring)
library(stringr)
stack(table(str_extract(x, "(\\w)\\1")))[2:1]
# ind values
#1 ee 1
#2 oo 2
Or using dplyr
library(dplyr)
library(tidyr)
str_extract_all(x, "(\\w)\\1") %>%
tibble(letter = .) %>%
unnest(c(letter)) %>%
count(letter)
Another base R solution using regmatches + table
dfout <- as.data.frame(table(unlist(regmatches(x,gregexpr("(\\w)\\1+",x)))))
which gives
> dfout
Var1 Freq
1 ee 1
2 oo 2
I have a dataframe returned from a function that looks like this:
df <- data.frame(data = c(1,2,3,4,5,6,7,8))
rownames(df) <- c('firsta','firstb','firstc','firstd','seconda','secondb','secondc','secondd')
firsta 1
seconda 5
firstb 2
secondb 6
my goal is to turn it into this:
df_goal <- data.frame(first = c(1,2,3,4), second = c(5,6,7,8))
rownames(df_goal) <- c('a','b','c','d')
first second
a 1 5
b 2 6
Basically the problem is that there is information in the row names that I can't discard because there isn't otherwise a way to distinguish between the column values.
This is a simple long-to-wide conversion; the twist is that we need to generate the key variable from the rownames by splitting the string appropriately.
In the data you present, the rowname consists of the concatination of a "position" (ie. 'first', 'second') and an id (ie. 'a', 'b'), which is stuck at the end. The structure of this makes splitting it complicated: ideally, you'd use a separator (ie. first_a, first_b) to make the separation unambiguous. Without a separator, our only option is to split on position, but that requires the splitting position to be a fixed distance from the start or end of the string.
In your example, the id is always the last single character, so we can pass -1 to the sep argument of separate to split off the last character as the ID column. If that wasn't always true, you would need to some up with a more complex solution to resolve the rownames.
Once you have converted the rownames into a "position" and "id" column, it's a simple matter to use spread to spread the position column into the wide format:
library(tidyverse)
df %>%
rownames_to_column('row') %>%
separate(row, into = c('num', 'id'), sep = -1) %>%
spread(num, data)
id first second
1 a 1 5
2 b 2 6
3 c 3 7
4 d 4 8
If row ids could be of variable length, the above solution wouldn't work. If you have a known and limited number of "position" values, you could use a regex solution to split the rowname:
Here, we extract the position value by matching to a regex containing all possible values (| is the OR operator).
We match the "id" value by putting that same regex in a positive lookahead operator. This regex will match 1 or more lowercase letters that come immediately after a match to the position value. The downside of this approach is that you need to specify all possible values of "position" in the regex -- if there are many options, this could quickly become too long and difficult to maintain:
df2
data
firsta 1
firstb 2
firstc 3
firstd 4
seconda 5
secondb 6
secondc 7
secondd 8
secondee 9
df2 %>%
rownames_to_column('row') %>%
mutate(num = str_extract(row, 'first|second'),
id = str_match(row, '(?<=first|second)[a-z]+')) %>%
select(-row) %>%
spread(num, data)
id first second
1 a 1 5
2 b 2 6
3 c 3 7
4 d 4 8
5 ee NA 9