Create a sequence of strings - r

Given a data set similar to the following
dat = structure(list(OpportunityId = c("006a000000zLXtZAAW", "006a000000zLXtZAAW",
"006a000000gst", "006a000000gstg", "006a000000gstg",
"006a000000zLXtZAAW"), IsWon = c(1, 1, 1, 1, 1, 1),
sequence = c("LLLML", "LHHHL", "LLLML", "HMLLL", "LLLLL", "LLLLL")),
.Names = c("OpportunityId","IsWon", "sequence"), row.names = c(NA, 6L), class = "data.frame")
dat
How would one go about adding each sequence that is associated with a particular opportunity id, such that the final looks like.
oppid sequence
006... LLL, LML, MMM
007... MMM, MML, MMH, LLL, HHH
007... LML, MMM
Any ideas?

We can paste the 'sequence' after grouping by 'OpportunityId'
library(data.table)
setDT(dat)[, .(sequence = toString(unique(sequence))) ,
by = .(oppid = OpportunityId)]

Maybe a combination of aggregate and unique could help.
aggregate(sequence ~ OpportunityId, dat, unique)
# OpportunityId sequence
#1 006a000000gst LLLML
#2 006a000000gstg HMLLL, LLLLL
#3 006a000000zLXtZAAW LLLML, LHHHL, LLLLL
As pointed out by #akrun in a comment, the sequence column is stored as a list in this case.
If necessary, the list in the sequence column can be converted into character format (a single string for each row) by means of:
dat$sequence <- sapply(dat$sequence, paste, collapse=", ")

With dplyr
library(dplyr)
dat_new <- dat %>%
group_by(OpportunityId) %>%
summarise(sequence = toString(sequence)) %>%
distinct(.keep_all = TRUE)
Output
# OpportunityId IsWon sequence
# 1 006a000000zLXtZAAW 1 LLLML, LHHHL, LLLLL
# 2 006a000000gst 1 LLLML
# 3 006a000000gstg 1 HMLLL, LLLLL

Related

find a row that has a string that contains a certain string, then take the row on top, the strong row and row under and move it to a new dataframe

So i have a table that looks like this:
I want to search though the first column for every time i see nl.audio take the row on top, take the nl.audio row and the row right under it and move them to a new column so it looks like this:
not sure how to go about doing this.
the table comes from trying to get nested json values into a dataframe. like this
library(jsonlite)
library(tidyverse)
files <- list.files(path=".", pattern=".json", all.files=FALSE,
full.names=FALSE)
data <- fromJSON(files[1])
dat2 <- unlist(data$translation_map)
dat2 <- as.data.frame(dat2)
dput:
structure(list(dat2 = c("Iraat.",
" _1645805605.mp3",
"Ie.", "wn", "", "Wdis.",
"ewdewf.mp3",
"wedew.", "[k]ws.[/k]",
" _1645805740.mp3",
"edwedwedw.", "Ik ewwewe[/k].",
"we45805760.mp3",
"I h89.", "ewd3n", "", "ad23dt", "",
"Ik d2. ", "I d2d3.",
"Ha3d3d/k] 20.", "H3d20.",
"id3n", "", "straat")), row.names = c("str-5e854867d9c6.nl.value",
"str_f15f7751-227dc6.nl.audio", "str_f15f7751.en.value",
"str.nl.value", "str_172a516ca.en.value",
"str_4567f686.nl.value", "str_4.nl.audio",
"stcb0ca14.en.value", "str_622f99395.nl.value",
"str_622f9395.nl.audio", "str_622f90de9395.en.value",
"str_f25afe16.nl.value", "str_f2fad09045afe16.nl.audio",
"str_f2fad89045afe16.en.value", "s9e844c432e80.nl.value",
"str_b0c1b42e80.en.value", "str_e6d847f3-60b7-.nl.value",
"str_.en.value", "str_b61f9404-.nl.value",
"str_ b.en.value", "str_76e28ea6.nl.value",
"str-61a1b83bf1ba.en.value", "str_6280d5a49c42a24.nl.value",
"str5-0d5a49c42a24.en.value", "str_5e6b2202e748.nl.value"
), class = "data.frame")
Something like this:
library(dplyr)
library(stringr)
df %>%
mutate(across(,str_squish)) %>%
mutate(A = ifelse(str_detect(V1, 'nl.audio'), lag(V2), NA_character_),
# B = str_extract(V2, '\\d+.mp3'),
B = str_extract(V2, '.*.mp3$'),
C = ifelse(str_detect(V1, 'nl.audio'), lead(V2), NA_character_),
.keep= "unused") %>%
na.omit()
A B C
2 nstraat. 1645805605.mp3 constraat.
7 tihdhis. 645805622.mp3 use.
df <- structure(list(V1 = c("str_f15d9c6.nl.value", "47c-5e854867d9c6.nl.audio",
"5e854867d9c6.en.value", "92bd-91b8f180bd3a.nl.value", "4-92bd-91b8f180bd3a.en.value",
"40a8-88ef-5890ecbOca14.nl.value", "890ecbOca14.nl.audio", "ca14.en.value"
), V2 = c("\tnstraat.", "\t1645805605.mp3", "\tconstraat.", "\tlemons",
" \t", "\ttihdhis.", "\t645805622.mp3", "\tuse.")), class = "data.frame", row.names = c(NA,
-8L))
We may need grep to find the index. Then add and subtract 1 to the index and extract the values from the second column based on that index (assuming data.frame columns)
i1 <- grep("nl.audio", df1[[1]], fixed = TRUE)
prev_ind <- i1-1
next_ind <- i1 + 1
data.frame(col1 = df1[[2]][prev_ind],
col2 = df1[[2]][next_ind],
col3 = df1[[2]][next_ind + 1])

R: adding matching vector values from two dataframes in one column

I have a data frame which is configured roughly like this:
df <- cbind(c('hello', 'yes', 'example'),c(7,8,5),c(0,0,0))
words
frequency
count
hello
7
0
yes
8
0
example
5
0
What I'm trying to do is add values to the third column from a different data frame, which is similiar but looks like this:
df2 <- cbind(c('example','hello') ,c(5,6))
words
frequency
example
5
hello
6
My goal is to find matching values for the first column in both data frames (they have the same column name) and add matching values from the second data frame to the third column of the first data frame.
The result should look like this:
df <- cbind(c('hello', 'yes', 'example'),c(7,8,5),c(6,0,5))
words
frequency
count
hello
7
6
yes
8
0
example
5
5
What I've tried so far is:
df <- merge(df,df2, by = "words", all.x=TRUE)
However, it doesn't work.
I could use some help understanding how could it be done. Any help will be welcome.
This is an "update join". My favorite way to do it is in dplyr:
library(dplyr)
df %>% rows_update(rename(df2, count = frequency), by = "words")
In base R you could do the same thing like this:
names(df2)[2] = "count2"
df = merge(df, df2, by = "words", all.x=TRUE)
df$count = ifelse(is.na(df$coutn2), df$count, df$count2)
df$count2 = NULL
Here is an option with data.table:
library(data.table)
setDT(df)[setDT(df2), on = "words", count := i.frequency]
Output
words frequency count
<char> <num> <num>
1: hello 7 6
2: yes 8 0
3: example 5 5
Or using match in base R:
df$count[match(df2$words, df$words)] <- df2$frequency
Or another option with tidyverse using left_join and coalesce:
library(tidyverse)
left_join(df, df2 %>% rename(count.y = frequency), by = "words") %>%
mutate(count = pmax(count.y, count, na.rm = T)) %>%
select(-count.y)
Data
df <- structure(list(words = c("hello", "yes", "example"), frequency = c(7,
8, 5), count = c(0, 0, 0)), class = "data.frame", row.names = c(NA,
-3L))
df2 <- structure(list(words = c("example", "hello"), frequency = c(5, 6)), class = "data.frame", row.names = c(NA,
-2L))

How to use seq() function inside ifelse statement in R

I am facing an issue while using the seq() function inside ifelse statement. I have a dataframe which contains the following columns.
Dataframe(df): newmodel id
NewModel_1 30
NewModel_2 30
i need to increase the id value for these 2 rows since id should not be same for a model. There is constant value(99) from which we have to increment the id values based on the condition.
When i am trying to implement the below code
df %>% mutate(id=ifelse(any(grepl("NewModel_", df$newmodel)), seq(from =99+1, by =1, length.out=2) , id))
I am getting the output as
newmodel id
NewModel_1 100
NewModel_1 100
Where as the expected one is
newmode1 id
NewModel_1 100
NewModel_1 101
Can someone explain me why it is happening??
Thanks in Advance
Are you looking for something like this?
inds <- grepl('NewModel_', df$newmodel)
df$id[inds] <- seq(100, by = 1, length.out = sum(inds))
df
# newmodel id
#1 NewModel_1 100
#2 NewModel_2 101
data
df <- structure(list(newmodel = c("NewModel_1", "NewModel_2"), id = c(30L,
30L)), class = "data.frame", row.names = c(NA, -2L))
I guess is because somehow the function is getting only the first item of the seq.
You can try this way, it works here.
if(any(grepl("NewModel_", df$newmodel))) {
df$id <- seq(from = 99 + 1, length.out = (length(df$id)))
}
UPDATE: The return of ifelse statement is only one value, so you are trying to input a vector in a single element. An alternative is to use an apply function.
The reason your ifelse(.) is failing is that ifelse keys its output length based on the input length of the conditional vector; if it is shorter than either of the yes= or no= vectors, the extra length is silently ignored. In your case, any(grepl("NewModel_", df$newmodel)) will never be other than length 1, so the output will be length 1.
For example:
ifelse(TRUE, 1:2, 3:4)
# [1] 1
ifelse(c(TRUE, FALSE), 1:2, 3:4)
# [1] 1 4
### and for an example of how R's overly-permissive recycling can go "wrong"
ifelse(c(TRUE, FALSE, TRUE), 1:2, 3:4)
# [1] 1 4 1
Here's a quick method using match to assign a unique integer to each of the models.
base R
dat$newid <- 99 + match(dat$newmodel, unique(dat$newmodel))
dat
# newmodel id newid
# 1 NewModel_1 30 100
# 2 NewModel_2 30 101
dplyr
library(dplyr)
dat %>%
mutate(newid = 99 + match(newmodel, unique(newmodel)))
# newmodel id newid
# 1 NewModel_1 30 100
# 2 NewModel_2 30 101
Data
dat <- structure(list(newmodel = c("NewModel_1", "NewModel_2"), id = c(30L, 30L), newid = c(100, 101)), row.names = c(NA, -2L), class = "data.frame")

If string has a certain character, fill an empty cell in the same row with a certain value

Say I have the following data frame:
# S/N a b
# 1 L1-S2 <blank>
# 2 T1-T3 <blank>
# 3 T1-L2 <blank>
How do I turn the above data frame into this:
# S/N a b
# 1 L1-S2 LS
# 2 T1-T3 T
# 3 T1-L2 TL
I am thinking of writing a loop, where
For x in column a,
If first character in x == L AND 4th character in x == S,
fill the corresponding cell in b with LS
and so on...
However, I am not sure how to implement it, or if there is a more elegant way of doing this.
We can extract the upper case letters and remove the repeated ones
library(stringr)
library(dplyr)
df1 %>%
mutate(b = str_replace(str_replace(a, "^([A-Z])\\d+-([A-Z])\\d+",
"\\1\\2"), "(.)\\1+", "\\1"))
-output
# S_N a b
#1 1 L1-S2 LS
#2 2 T1-T3 T
#3 3 T1-L2 TL
Or another option is str_extract_all to extract the upper case letters, loop over the list with map, paste the unique elements
library(purrr)
df1 %>%
mutate(b = str_extract_all(a, "[A-Z]") %>%
map_chr(~ str_c(unique(.x), collapse="")))
Or using a corresponding base R option for the first tidyverse option
df1$b <- sub("(.)\\1+", "\\1", gsub("[0-9-]+", "", df1$a))
Or with strsplit
df1$b <- sapply(strsplit(df1$a, "[0-9-]+"),
function(x) paste(unique(x), collapse=""))
data
df1 <- structure(list(S_N = 1:3, a = c("L1-S2", "T1-T3", "T1-L2"),
b = c(NA,
NA, NA)), class = "data.frame", row.names = c(NA, -3L))

Delete columns in an R loop

I have a dataframe where I want to replace the variables
age_1 with values of variable age1_corr_1 if age1_corr_1 is not NA
age_2 with values of variable age1_corr_2 if age1_corr_2 is not NA, ...,
age_n with values of variable age1_corr_n if age1_corr_n is not NA.
Then I'd like to delete the variables age1_corr_1, age1_corr_2, ..., age1_corr_n. I have figured out how to do the first part (change the values) in a loop but couldn't figure out how to delete the variables after. Any suggestion?
Sample data
y <- data.frame("age_1" = c(5,1,1,10), "age1_corr_1" = c(1,NA,NA,0), "age_2" = c(1,2,3,4), "age1_corr_2" = c(NA, NA, 10, 9),
"age_3" = c(4,3,2,5), "age1_corr_3" = c(NA,NA,NA,6), "age_4" = c(1,4,2,7), "age1_corr_4" = c(NA, NA, NA,NA))
The code that will change values of age_n based on age1_corr_n
for(i in 1:4){
cname1 <- paste0("age_",i)
cname2 <- paste0("age1_corr_",i)
y[,cname1] <- ifelse(!is.na(y[,cname2]), y[,cname2], y[,cname1])
}
The output I'd like to have is
age_1 age_2 age_3 age_4
1 1 1 4 1
2 1 2 3 4
3 1 10 2 2
4 0 9 6 7
You have several options if there is a pattern to the columns you want to remove (or conversely, the ones you want to keep).
Here's the data you provided:
y <- data.frame("age_1" = c(5,1,1,10), "age1_corr_1" = c(1,NA,NA,0), "age_2" = c(1,2,3,4), "age1_corr_2" = c(NA, NA, 10, 9),
"age_3" = c(4,3,2,5), "age1_corr_3" = c(NA,NA,NA,6), "age_4" = c(1,4,2,7), "age1_corr_4" = c(NA, NA, NA,NA))
Here's a dplyr example of how to get only those columns that follow the pattern age_N, where N is 1, 2, 3, or 4:
library(dplyr)
x <- select(y, paste("age", 1:4, sep = "_"))
Alternatively, you could choose the pattern for the columns you DON'T want:
x <- select(y, -grep("_corr_", current_vars()))
This uses the following strategy:
* you can select for everything BUT a column or set of columns by adding a minus sign first.
* current_vars() is a helper function in dplyr that evaluates to all the variable names for the data (here, y)
Do the real work with dplyr::coalesce() (description: "Given a set of vectors, coalesce() finds the first non-missing value at each position."). Then drop the columns with dplyr::select(), using a negative sign in front of the columns you don't need anymore.
library(magrittr)
y %>%
dplyr::mutate(
age1_corr_4 = as.numeric(age1_corr_4), # Delete this line if it's already a numeric/floating data type.
age_1 = dplyr::coalesce(age1_corr_1, age_1),
age_2 = dplyr::coalesce(age1_corr_2, age_2),
age_3 = dplyr::coalesce(age1_corr_3, age_3),
age_4 = dplyr::coalesce(age1_corr_4, age_4)
) %>%
dplyr::select(
-age1_corr_1, -age1_corr_2, -age1_corr_3, -age1_corr_4
)
Produces
age_1 age_2 age_3 age_4
1 1 1 4 1
2 1 2 3 4
3 1 10 2 2
4 0 9 6 7
Edit: I apologize, I focused on the coalesce part of the task and ignored the n part of the task.
Here are two other approaches that can handle an arbitrary number of columns. For this specific example dataset, make sure that the 4th column is correctly represented as a float with y$age1_corr_4 <- as.numeric(y$age1_corr_4)).
Like Dan Hall's response, one approach keeps the columns you want...
library(magrittr)
coalesce_corr1 <- function( index ) {
name_age <- paste0("age_" , index)
name_corr <- paste0("age1_corr_", index)
y %>%
dplyr::mutate(
!!name_age := dplyr::coalesce(.data[[name_corr]], .data[[name_age]])
) %>%
dplyr::select(!!name_age)
}
1:4 %>%
purrr::map(coalesce_corr) %>%
dplyr::bind_cols()
...and the other drops the columns you don't want.
z <- y
coalesce_corr2 <- function( index ) {
name_age <- paste0( "age_" , index)
name_corr <- paste0( "age1_corr_", index)
z <<- z %>%
dplyr::mutate(
!!name_age := dplyr::coalesce(.data[[!!name_corr]], .data[[!!name_age]])
)
z[[name_corr]] <<- NULL
}
1:4 %>%
purrr::walk(coalesce_corr2)
z
I wish this last one didn't require a global variable (that uses <<-), and for this reason, I actually recommend Dan's approaches, but I wanted to try out quosures for output variables.

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