I have two data frames
df1:
01.2020 02.2020 03.2020
11190 4 1 2
12345 3 3 1
11323 1 2 2
df2
08.2020 04.2020 09.2020
11190 1 2 2
12345 1 2 3
11324 1 2 2
Dummy Data -
df1 <- structure(list(`01.2020` = c(4L, 3L, 1L), `02.2020` = c(1L, 3L, 2L), `03.2020` = c(2L, 1L, 2L)), class = "data.frame", row.names = c("11190","12345", "11323"))
df2 <- structure(list(`08.2020` = c(1L, 1L, 1L), `04.2020` = c(2L, 2L, 2L), `09.2020` = c(2L, 3L, 2L)), class = "data.frame", row.names = c("11190", "12345", "11324"))
I want to "outer merge" these two dataframes by key = index
How can we do that? what should be there in the place of by=
merge(x = sheet1_UN, y = sheet2_UN, by = "" , all = TRUE)
I want my final dataframe to look something like this
01.2020 02.2020 03.2020 08.2020 04.2020 09.2020
11190 4 1 2 1 1 2
12345 3 3 1 1 2 3
11323 1 2 2 - - -
11324 - - - 1 2 2
Thanks in advance.
another method
df3 <- merge(df1, df2, by = "row.names", all = TRUE)
output:
Row.names 01.2020 02.2020 03.2020 08.2020 04.2020 09.2020
1 11190 4 1 2 1 2 2
2 11323 1 2 2 NA NA NA
3 11324 NA NA NA 1 2 2
4 12345 3 3 1 1 2 3
This should do:
df1 %>% rownames_to_column('id') %>%
full_join(df2 %>% rownames_to_column('id'), by='id')
output:
id 01.2020 02.2020 03.2020 08.2020 04.2020 09.2020
1 11190 4 1 2 1 2 2
2 12345 3 3 1 1 2 3
3 11323 1 2 2 NA NA NA
4 11324 NA NA NA 1 2 2
You might use replace_na('-') if you want no NA values, like this:
df1 %>% rownames_to_column('id') %>%
full_join(df2 %>% rownames_to_column('id'), by='id') %>%
mutate(across(everything(), ~.x %>% as.character %>% replace_na('-')))
Related
I have data that look like these:
Subject Site Date
1 2 '2020-01-01'
1 2 '2020-01-01'
1 2 '2020-01-02'
2 1 '2020-01-02'
2 1 '2020-01-03'
2 1 '2020-01-03'
And I'd like to create an order variable for unique dates by Subject and Site. i.e.
Want
1
1
2
1
2
2
I define a little wrapper:
rle <- function(x) cumsum(!duplicated(x))
and I notice inconsistent behavior when I supply:
have1 <- unlist(tapply(val$Date, val[, c( 'Site', 'Subject')], rle))
versus
have2 <- unlist(tapply(val$Date, val[, c('Subject', 'Site')], rle))
> have1
[1] 1 1 2 1 2 2
> have2
[1] 1 2 2 1 1 2
Is there any way to ensure that the natural ordering of the dataset is followed regardless of the specific columns supplied to the INDEX argument?
library(dplyr)
val %>%
group_by(Subject, Site) %>%
mutate(Want = match(Date, unique(Date))) %>%
ungroup
-output
# A tibble: 6 × 4
Subject Site Date Want
<int> <int> <chr> <int>
1 1 2 2020-01-01 1
2 1 2 2020-01-01 1
3 1 2 2020-01-02 2
4 2 1 2020-01-02 1
5 2 1 2020-01-03 2
6 2 1 2020-01-03 2
val$Want <- with(val, ave(as.integer(as.Date(Date)), Subject, Site,
FUN = \(x) match(x, unique(x))))
val$Want
[1] 1 1 2 1 2 2
data
val <- structure(list(Subject = c(1L, 1L, 1L, 2L, 2L, 2L), Site = c(2L,
2L, 2L, 1L, 1L, 1L), Date = c("2020-01-01", "2020-01-01", "2020-01-02",
"2020-01-02", "2020-01-03", "2020-01-03")),
class = "data.frame", row.names = c(NA,
-6L))
I have a question for the community and hoping for some help.
I am trying to duplicate a data frame like the one below:
ID Time Solve
1 0 1
1 2 2
1 4 3
1 6 1
I am trying to duplicate the above data frame 100 times so, it would read as below:
ID Time Solve
1 0 1
1 2 2
1 4 3
1 6 1
2 0 1
2 2 2
2 4 3
2 6 1
3 0 1
3 2 2
3 4 3
3 6 1
4 0 1
4 2 2
4 4 3
4 6 1
.....
100 0 1
100 2 2
100 4 3
100 6 1
Does anyone have a good solution for this or a resource to read up on this?
Thanks!
We can use replicate
out <- do.call(rbind, replicate(100, df1, simplify = FALSE))
out$ID <- as.integer(gl(nrow(out), nrow(df1), nrow(out)))
Or another option is rep
out <- df1[rep(seq_len(nrow(df1)), 100),]
out$ID <- as.integer(gl(nrow(out), nrow(df1), nrow(out)))
Or make use of uncount
library(tidyr)
library(dplyr)
uncount(df1, 100) %>%
mutate(ID = as.integer(gl(n(), nrow(df1), n()))
Or another option is
df1 %>%
nest_by(ID) %>%
uncount(100) %>%
mutate(ID = row_number()) %>%
unnest(c(data))
data
df1 <- structure(list(ID = c(1L, 1L, 1L, 1L), Time = c(0L, 2L, 4L, 6L
), Solve = c(1L, 2L, 3L, 1L)), class = "data.frame", row.names = c(NA,
-4L))
I have the following table:
id_question id_event num_events
2015012713 49508 1
2015012711 49708 1
2015011523 41808 3
2015011523 44008 3
2015011523 44108 3
2015011522 41508 3
2015011522 43608 3
2015011522 43708 3
2015011521 39708 1
2015011519 44208 1
The third column gives the count of events by question. I want to create a variable that would index the events by question only where there are multiple events per question. It would look something like that:
id_question id_event num_events index_event
2015012713 49508 1
2015012711 49708 1
2015011523 41808 3 1
2015011523 44008 3 2
2015011523 44108 3 3
2015011522 41508 3 1
2015011522 43608 3 2
2015011522 43708 3 3
2015011521 39708 1
2015011519 44208 1
How can I do that?
We can use tidyverse to create an 'index_event' after grouping by 'id_question'. If the number of rows are greater than 1 (n() >1), then get the sequence of rows (row_number()) and the default option in case_when is NA
library(dplyr)
df1 %>%
group_by(id_question) %>%
mutate(index_event = case_when(n() >1 ~ row_number()))
# A tibble: 10 x 4
# Groups: id_question [6]
# id_question id_event num_events index_event
# <int> <int> <int> <int>
# 1 2015012713 49508 1 NA
# 2 2015012711 49708 1 NA
# 3 2015011523 41808 3 1
# 4 2015011523 44008 3 2
# 5 2015011523 44108 3 3
# 6 2015011522 41508 3 1
# 7 2015011522 43608 3 2
# 8 2015011522 43708 3 3
# 9 2015011521 39708 1 NA
#10 2015011519 44208 1 NA
Or with data.table, we use rowid on 'id_question' and change the elements that are 1 in 'num_events' to NA with NA^ (making use of NA^0, NA^1)
library(data.table)
setDT(df1)[, index_event := rowid(id_question) * NA^(num_events == 1)]
Or using base R, another option with the sequence of frequency from 'id_question' and change elements to NA as in the previous case
df1$index_event <- with(df1, sequence(table(id_question)) * NA^(num_events == 1))
df1$index_event
#[1] NA NA 1 2 3 1 2 3 NA NA
data
df1 <- structure(list(id_question = c(2015012713L, 2015012711L, 2015011523L,
2015011523L, 2015011523L, 2015011522L, 2015011522L, 2015011522L,
2015011521L, 2015011519L), id_event = c(49508L, 49708L, 41808L,
44008L, 44108L, 41508L, 43608L, 43708L, 39708L, 44208L), num_events = c(1L,
1L, 3L, 3L, 3L, 3L, 3L, 3L, 1L, 1L)), class = "data.frame", row.names = c(NA,
-10L))
If num_events = 1 you can return NA or create a row-index for each id_question.
This can be done in base R :
df$index_event <- with(df, ave(num_events == 1, id_question,
FUN = function(x) replace(seq_along(x), x, NA)))
df
# id_question id_event num_events index_event
#1 2015012713 49508 1 NA
#2 2015012711 49708 1 NA
#3 2015011523 41808 3 1
#4 2015011523 44008 3 2
#5 2015011523 44108 3 3
#6 2015011522 41508 3 1
#7 2015011522 43608 3 2
#8 2015011522 43708 3 3
#9 2015011521 39708 1 NA
#10 2015011519 44208 1 NA
dplyr :
library(dplyr)
df %>%
group_by(id_question) %>%
mutate(index_event = if_else(num_events == 1, NA_integer_, row_number()))
Or data.table :
library(data.table)
setDT(df)
df[,index_event := ifelse(num_events == 1, NA_integer_, seq_len(.N)), id_question]
data
df <- structure(list(id_question = c(2015012713L, 2015012711L, 2015011523L,
2015011523L, 2015011523L, 2015011522L, 2015011522L, 2015011522L,
2015011521L, 2015011519L), id_event = c(49508L, 49708L, 41808L,
44008L, 44108L, 41508L, 43608L, 43708L, 39708L, 44208L), num_events = c(1L,
1L, 3L, 3L, 3L, 3L, 3L, 3L, 1L, 1L)),class = "data.frame",row.names = c(NA, -10L))
I have 200 columns and want to calculate mean and rank and then generate columns. Here is an example of data
df<-read.table(text="Q1a Q2a Q3b Q4c Q5a Q6c Q7b
1 2 4 2 2 0 1
3 2 1 2 2 1 1
4 3 2 1 1 1 1",h=T)
I want to sum a, b and c for each row, and then sum them together. Next I want to calculate the rank for each row. I want to generate the following table:
Q1a Q2a Q3b Q4c Q5a Q6c Q7b a b c Total Rank
1 2 4 2 2 0 1 5 5 2 12 2
3 2 1 2 2 1 1 7 2 3 12 2
4 3 2 1 1 1 1 8 3 2 13 1
library(dplyr)
df %>%
cbind(sapply(c('a', 'b', 'c'), function(x) rowSums(.[, grep(x, names(.)), drop=FALSE]))) %>%
mutate(Total = a + b + c,
Rank = match(Total, sort(Total, decreasing = T)))
Output is:
Q1a Q2a Q3b Q4c Q5a Q6c Q7b a b c Total Rank
1 1 2 4 2 2 0 1 5 5 2 12 2
2 3 2 1 2 2 1 1 7 2 3 12 2
3 4 3 2 1 1 1 1 8 3 2 13 1
Sample data:
df <- structure(list(Q1a = c(1L, 3L, 4L), Q2a = c(2L, 2L, 3L), Q3b = c(4L,
1L, 2L), Q4c = c(2L, 2L, 1L), Q5a = c(2L, 2L, 1L), Q6c = c(0L,
1L, 1L), Q7b = c(1L, 1L, 1L)), class = "data.frame", row.names = c(NA,
-3L))
You can also go with the tidyverse approach. However, it is longer.
library(tidyverse)
df %>%
rownames_to_column(var = "ID") %>%
gather(question, value, -ID) %>%
mutate(type = substr(question, 3,3)) %>%
group_by(ID, type) %>%
summarise(sumType = sum(value, na.rm = TRUE)) %>%
as.data.frame() %>%
spread(type, sumType) %>%
mutate(Total = a+b+c,
Rank = match(Total, sort(Total, decreasing = T)))
Results:
ID a b c Total Rank
1 1 5 5 2 12 2
2 2 7 2 3 12 2
3 3 8 3 2 13 1
I have a data frame that looks like this:
cat df1 df2 df3
1 1 NA 1 NA
2 1 NA 2 NA
3 1 NA 3 NA
4 2 1 NA NA
5 2 2 NA NA
6 2 3 NA NA
I want to populate df3 so that when cat = 1, df3 = df2 and when cat = 2, df3 = df1. However I am getting a few different error messages.
My current code looks like this:
df$df3[df$cat == 1] <- df$df2
df$df3[df$cat == 2] <- df$df1
Try this code:
df[df$cat==1,"df3"]<-df[df$cat==1,"df2"]
df[df$cat==2,"df3"]<-df[df$cat==1,"df1"]
The output:
df
cat df1 df2 df3
1 1 1 1 1
2 2 1 2 1
3 3 1 3 NA
4 4 2 NA NA
5 5 2 NA NA
6 5 2 NA NA
You can try
ifelse(df$cat == 1, df$df2, df$df1)
[1] 1 2 3 1 2 3
# saving
df$df3 <- ifelse(df$cat == 1, df$df2, df$df1)
# if there are other values than 1 and 2 you can try a nested ifelse
# that is setting other values to NA
ifelse(df$cat == 1, df$df2, ifelse(df$cat == 2, df$df1, NA))
# or you can try a tidyverse solution.
library(tidyverse)
df %>%
mutate(df3=case_when(cat == 1 ~ df2,
cat == 2 ~ df1))
cat df1 df2 df3
1 1 NA 1 1
2 1 NA 2 2
3 1 NA 3 3
4 2 1 NA 1
5 2 2 NA 2
6 2 3 NA 3
# data
df <- structure(list(cat = c(1L, 1L, 1L, 2L, 2L, 2L), df1 = c(NA, NA,
NA, 1L, 2L, 3L), df2 = c(1L, 2L, 3L, NA, NA, NA), df3 = c(NA,
NA, NA, NA, NA, NA)), .Names = c("cat", "df1", "df2", "df3"), class = "data.frame", row.names = c("1",
"2", "3", "4", "5", "6"))