I am new to R. I am trying to search the columns using grep multiple times within an apply loop. I use grep to specify which rows are summed based on the vector individuals
individuals <-c("ID1","ID2".....n)
bcdata_total <- sapply(individuals, function(x) {
apply(bcdata_clean[,grep(individuals, colnames(bcdata_clean))], 1, sum)
})
bcdata is of random size and contains random data but contains columns that have individuals in part of the string
>head(bcdata)
ID1-4 ID1-3 ID2-5
A 3 2 1
B 2 2 3
C 4 5 5
grep(individuals[1],colnames(bcdata_clean)) returns a vector that looks like
[1] 1 2, a list of the column names containing ID1. That vector is used to select columns to be summed in bcdata_clean. This should occur n number of times depending on the length of individuals
However this returns the error
In grep(individuals, colnames(bcdata)) :
argument 'pattern' has length > 1 and only the first element will be used
And results in all the columns of bcdata being identical
Ideally individuals would increment each time the function is run like this for each iteration
apply(bcdata_clean[,grep(individuals[1,2....n], colnames(bcdata_clean))], 1, sum)
and would result in something like this
>head(bcdata_total)
ID1 ID2
A 5 1
B 4 3
C 9 5
But I'm not sure how to increment individuals. What is the best way to do this within the function?
You can use split.default to split data on similarly named columns and sum them row-wise.
sapply(split.default(df, sub('-.*', '', names(df))), rowSums, na.rm. = TRUE)
# ID1 ID2
#A 5 1
#B 4 3
#C 9 5
data
df <- structure(list(`ID1-4` = c(3L, 2L, 4L), `ID1-3` = c(2L, 2L, 5L
), `ID2-5` = c(1L, 3L, 5L)), class = "data.frame", row.names = c("A", "B", "C"))
Passing individuals as my argument in function(x) fixed my issue
bcdata_total <- sapply(individuals, function(individuals) {
apply(bcdata_clean[,grep(individuals, colnames(bcdata_clean))], 1, sum)
})
An option with tidyverse
library(dplyr)
library(tidyr)
library(tibble)
df %>%
rownames_to_column('rn') %>%
pivot_longer(cols = -rn, names_to = c(".value", "grp"), names_sep="-") %>%
group_by(rn) %>%
summarise(across(starts_with('ID'), sum, na.rm = TRUE), .groups = 'drop') %>%
column_to_rownames('rn')
# ID1 ID2
#A 5 1
#B 4 3
#C 9 5
data
df <- df <- structure(list(`ID1-4` = c(3L, 2L, 4L), `ID1-3` = c(2L, 2L, 5L
), `ID2-5` = c(1L, 3L, 5L)), class = "data.frame", row.names = c("A", "B", "C"))
Related
I have a df like
ProjectID Dist
1 x
1 y
2 z
2 x
2 h
3 k
.... ....
and a vector of indices of lengthunique(df$ProjectID) like
2
3
1
....
I would like to get Dist by ProjectID whose index is the element vector corresponding to project ID. So the result I want looks like
ProjectID Dist
1 y
2 h
3 k
.... ....
I tried
aggregate(XRKL ~ ID, FUN=..?, data=df)
but I'm not sure where I can put the vector of indices. Is there a way to get the right result from dply ftns, tapply, or aggregate? Or do I need to make a function of my own? Thank you.
You can add the indices in the dataframe itself and then select that row from each group.
inds <- c(2, 3, 1)
df %>%
mutate(inds = inds[match(ProjectID, unique(ProjectID))]) %>%
#If ProjectID is sequential like 1, 2, 3
#mutate(inds = inds[ProjectID]) %>%
group_by(ProjectID) %>%
slice(first(inds)) %>%
ungroup() %>%
select(-inds)
# ProjectID Dist
# <int> <chr>
#1 1 y
#2 2 h
#3 3 k
data
df <- structure(list(ProjectID = c(1L, 1L, 2L, 2L, 2L, 3L), Dist = c("x",
"y", "z", "x", "h", "k")), class = "data.frame", row.names = c(NA, -6L))
Good evening,
I have a two columns tab separated .txt file, as the following:
number letter
1 a
1 b
2 a
2 b
3 b
I would like to collapse rows where the column "number" has identical value, by creating a comma separated value in the corresponding column "letter".
In other words, this should be the output:
number letter
1 a,b
2 a,b
3 b
I have looked up the web but I did not find an actual solution.
Thank you in advance,
Giuseppe
We can use aggregate in base R
aggregate(letter ~ number, df1, FUN = paste, collapse=",")
-output
# number letter
#1 1 a,b
#2 2 a,b
#3 3 b
Or with tidyverse
library(dplyr)
library(stringr)
df1 %>%
group_by(number) %>%
summarise(letter = str_c(letter, collapse=","))
data
df1 <- structure(list(number = c(1L, 1L, 2L, 2L, 3L), letter = c("a",
"b", "a", "b", "b")), class = "data.frame", row.names = c(NA,
-5L))
We can also combine aggregate() with toString:
#Code
newdf <- aggregate(letter~.,df,toString)
Output:
number letter
1 1 a, b
2 2 a, b
3 3 b
Some data:
#Data
df <- structure(list(number = c(1L, 1L, 2L, 2L, 3L), letter = c("a",
"b", "a", "b", "b")), class = "data.frame", row.names = c(NA,
-5L))
I have a dataset that has two columns. One is userid, the other is company type, like below:
userid company.type
1 A
2 A
3 C
1 B
2 B
3 B
4 A
I want to know how many unique userid's there are that have company.type of A and B or A and C, (but not B and C).
I'm assuming it's some sort of aggregate function, but I'm not sure how to place the qualifier that company.type has to be A and B or A and C only.
We can do this with base R using table
tbl <- table(df1) > 0
sum(((tbl[, 1] & tbl[,2]) | (tbl[,1] & tbl[,3])) & (!(tbl[,2] & tbl[,3])))
#[1] 2
Here's an idea with dplyr. setequal checks if two vectors are composed of the same elements, regardless of ordering:
library(dplyr)
df %>%
group_by(userid) %>%
summarize(temp = setequal(company.type, c("A", "B")) |
setequal(company.type, c("A", "C"))) %>%
pull(temp) %>%
sum()
# [1] 2
Data:
df <- structure(list(userid = c(1L, 2L, 3L, 1L, 2L, 3L, 4L), company.type = c("A",
"A", "C", "B", "B", "B", "A")), .Names = c("userid", "company.type"
), class = "data.frame", row.names = c(NA, -7L))
See: Check whether two vectors contain the same (unordered) elements in R
Sort DF and reduce it to one row per userid with a types column consisting of a comma-separated string of company types. Then filter it using the indicated condition. Finally use tally to get the number of rows left after filtering. To get the details omit the tally line.
library(dplyr)
DF %>%
arrange(userid, company.type) %>%
group_by(userid) %>%
summarize(types = toString(company.type)) %>%
ungroup %>%
filter(grepl("A.*B|A.*C", types) & ! grepl("B.*C", types)) %>%
tally
giving:
# A tibble: 1 x 1
n
<int>
1 2
Note
The input used, in reproducible form, is:
Lines <- "userid company.type
1 A
2 A
3 C
1 B
2 B
3 B
4 A"
DF <- read.table(text = Lines, header = TRUE)
I am trying to iterate through columns, and if the column is a whole year, it should be duplicated four times, and renamed to quarters
So this
2000 Q1-01 Q2-01 Q3-01
1 2 3 3
Should become this:
Q1-00 Q2-00 Q3-00 Q4-00 Q1-01 Q2-01 Q3-01
1 1 1 1 2 3 3
Any ideas?
We can use stringr::str_detect to look for colnames with 4 digits then take the last two digits from those columns
library(dplyr)
library(tidyr)
library(stringr)
df %>% gather(key,value) %>% group_by(key) %>%
mutate(key_new = ifelse(str_detect(key,'\\d{4}'),paste0('Q',1:4,'-',str_extract(key,'\\d{2}$'),collapse = ','),key)) %>%
ungroup() %>% select(-key) %>%
separate_rows(key_new,sep = ',') %>% spread(key_new,value)
PS: I hope you don't have a large dataset
Since you want repeated columns, you can just re-index your data frame and then update the column names
df <- structure(list(`2000` = 1L, Q1.01 = 2L, Q2.01 = 3L, Q3.01 = 3L,
`2002` = 1L, Q1.03 = 2L, Q2.03 = 3L, Q3.03 = 3L), row.names = c(NA,
-1L), class = "data.frame")
#> df
#2000 Q1.01 Q2.01 Q3.01 2002 Q1.03 Q2.03 Q3.03
#1 1 2 3 3 1 2 3 3
# Get indices of columns that consist of 4 numbers
col.ids <- grep('^[0-9]{4}$', names(df))
# For each of those, create new names, and for the rest preserve the old names
new.names <- lapply(seq_along(df), function(i) {
if (i %in% col.ids)
return(paste(substr(names(df)[i], 3, 4), c('Q1', 'Q2', 'Q3', 'Q4'), sep = '.'))
return(names(df)[i])
})
# Now repeat each of those columns 4 times
df <- df[rep(seq_along(df), ifelse(seq_along(df) %in% col.ids, 4, 1))]
# ...and finally set the column names to the desired new names
names(df) <- unlist(new.names)
#> df
#00.Q1 00.Q2 00.Q3 00.Q4 Q1.01 Q2.01 Q3.01 02.Q1 02.Q2 02.Q3 02.Q4 Q1.03 Q2.03 Q3.03
#1 1 1 1 1 2 3 3 1 1 1 1 2 3 3
I have two data frames. dfOne is made like this:
X Y Z T J
3 4 5 6 1
1 2 3 4 1
5 1 2 5 1
and dfTwo is made like this
C.1 C.2
X Z
Y T
I want to obtain a new dataframe where there are simultaneously X, Y, Z, T Values which are major than a specific threshold.
Example. I need simultaneously (in the same row):
X, Y > 2
Z, T > 4
I need to use the second data frame to reach my objective, I expect something like:
dfTwo$C.1>2
so the result would be a new dataframe with this structure:
X Y Z T J
3 4 5 6 1
How could I do it?
Here is a base R method with Map and Reduce.
# build lookup table of thresholds relative to variable name
vals <- setNames(c(2, 2, 4, 4), unlist(dat2))
# subset data.frame
dat[Reduce("&", Map(">", dat[names(vals)], vals)), ]
X Y Z T J
1 3 4 5 6 1
Here, Map returns a list of length 4 with logical variables corresponding to each comparison. This list is passed to Reduce which returns a single logical vector with length corresponding to the number of rows in the data.frame, dat. This logical vector is used to subset dat.
data
dat <-
structure(list(X = c(3L, 1L, 5L), Y = c(4L, 2L, 1L), Z = c(5L,
3L, 2L), T = c(6L, 4L, 5L), J = c(1L, 1L, 1L)), .Names = c("X",
"Y", "Z", "T", "J"), class = "data.frame", row.names = c(NA,
-3L))
dat2 <-
structure(list(C.1 = structure(1:2, .Label = c("X", "Y"), class = "factor"),
C.2 = structure(c(2L, 1L), .Label = c("T", "Z"), class = "factor")), .Names = c("C.1",
"C.2"), class = "data.frame", row.names = c(NA, -2L))
We can use the purrr package
Here is the input data.
# Data frame from lmo's solution
dat <-
structure(list(X = c(3L, 1L, 5L), Y = c(4L, 2L, 1L), Z = c(5L,
3L, 2L), T = c(6L, 4L, 5L), J = c(1L, 1L, 1L)), .Names = c("X",
"Y", "Z", "T", "J"), class = "data.frame", row.names = c(NA,
-3L))
# A numeric vector to show the threshold values
# Notice that columns without any requirements need NA
vals <- c(X = 2, Y = 2, Z = 4, T = 4, J = NA)
Here is the implementation
library(purrr)
map2_dfc(dat, vals, ~ifelse(.x > .y | is.na(.y), .x, NA)) %>% na.omit()
# A tibble: 1 x 5
X Y Z T J
<int> <int> <int> <int> <int>
1 3 4 5 6 1
map2_dfc loop through each column in dat and each value in vals one by one with a defined function. ~ifelse(.x > .y | is.na(.y), .x, NA) means if the number in each column is larger than the corresponding value in vals, or vals is NA, the output should be the original value from the column. Otherwise, the value is replaced to be NA. The output of map2_dfc(dat, vals, ~ifelse(.x > .y | is.na(.y), .x, NA)) is a data frame with NA values in some rows indicating that the condition is not met. Finally, na.omit removes those rows.
Update
Here I demonstrate how to covert the dfTwo dataframe to the vals vector in my example.
First, let's create the dfTwo data frame.
dfTwo <- read.table(text = "C.1 C.2
X Z
Y T",
header = TRUE, stringsAsFactors = FALSE)
dfTwo
C.1 C.2
1 X Z
2 Y T
To complete the task, I load the dplyr and tidyr package.
library(dplyr)
library(tidyr)
Now I begin the transformation of dfTwo. The first step is to use stack function to convert the format.
dfTwo2 <- dfTwo %>%
stack() %>%
setNames(c("Col", "Group")) %>%
mutate(Group = as.character(Group))
dfTwo2
Col Group
1 X C.1
2 Y C.1
3 Z C.2
4 T C.2
The second step is to add the threshold information. One way to do this is to create a look-up table showing the association between Group and Value
threshold_df <- data.frame(Group = c("C.1", "C.2"),
Value = c(2, 4),
stringsAsFactors = FALSE)
threshold_df
Group Value
1 C.1 2
2 C.2 4
And then we can use the left_join function to combine the data frame.
dfTwo3 <- dfTwo2 %>% left_join(threshold_dt, by = "Group")
dfTwo3
Col Group Value
1 X C.1 2
2 Y C.1 2
3 Z C.2 4
4 T C.2 4
Now it is the third step. Notice that there is a column called J which does not need any threshold. So we need to add this information to dfTwo3. We can use the complete function from tidyr. The following code completes the data frame by adding Col in dat but not in dfTwo3 and NA to the Value.
dfTwo4 <- dfTwo3 %>% complete(Col = colnames(dat))
dfTwo4
# A tibble: 5 x 3
Col Group Value
<chr> <chr> <dbl>
1 J <NA> NA
2 T C.2 4
3 X C.1 2
4 Y C.1 2
5 Z C.2 4
The fourth step is arrange the right order of dfTwo4. We can achieve this by turning Col to factor and assign the level based on the order of the column name in dat.
dfTwo5 <- dfTwo4 %>%
mutate(Col = factor(Col, levels = colnames(dat))) %>%
arrange(Col) %>%
mutate(Col = as.character(Col))
dfTwo5
# A tibble: 5 x 3
Col Group Value
<chr> <chr> <dbl>
1 X C.1 2
2 Y C.1 2
3 Z C.2 4
4 T C.2 4
5 J <NA> NA
We are almost there. Now we can create vals from dfTwo5.
vals <- dfTwo5$Value
names(vals) <- dfTwo5$Col
vals
X Y Z T J
2 2 4 4 NA
Now we are ready to use the purrr package to filter the data.
The aboved are the breakdown of steps. We can combine all these steps into the following code for simlicity.
library(dplyr)
library(tidyr)
threshold_df <- data.frame(Group = c("C.1", "C.2"),
Value = c(2, 4),
stringsAsFactors = FALSE)
dfTwo2 <- dfTwo %>%
stack() %>%
setNames(c("Col", "Group")) %>%
mutate(Group = as.character(Group)) %>%
left_join(threshold_df, by = "Group") %>%
complete(Col = colnames(dat)) %>%
mutate(Col = factor(Col, levels = colnames(dat))) %>%
arrange(Col) %>%
mutate(Col = as.character(Col))
vals <- dfTwo2$Value
names(vals) <- dfTwo2$Col
dfOne[Reduce(intersect, list(which(dfOne["X"] > 2),
which(dfOne["Y"] > 2),
which(dfOne["Z"] > 4),
which(dfOne["T"] > 4))),]
# X Y Z T J
#1 3 4 5 6 1
Or iteratively (so fewer inequalities are tested):
vals = c(X = 2, Y = 2, Z = 4, T = 4) # from #lmo's answer
dfOne[Reduce(intersect, lapply(names(vals), function(x) which(dfOne[x] > vals[x]))),]
# X Y Z T J
#1 3 4 5 6 1
I'm writing this assuming that the second DF is meant to categorize the fields in the first DF. It's way simpler if you don't need to use the second one to define the conditions:
dfNew = dfOne[dfOne$X > 2 & dfOne$Y > 2 & dfOne$Z > 4 & dfOne$T > 4, ]
Or, using dplyr:
library(dplyr)
dfNew = dfOne %>% filter(X > 2 & Y > 2 & Z > 4 & T > 4)
In case that's all you need, I'll save this comment while I poke at the more complicated version of the question.