I'm trying to merge informations in two different data frames, but problem begins with uneven dimensions and trying to use not the column index but the information in the column. merge function in R or join's (dplyr) don't work with my data.
I have to dataframes (One is subset of the others with updated info in the last column):
df1=data.frame(Name = print(LETTERS[1:9]), val = seq(1:3), Case = c("NA","1","NA","NA","1","NA","1","NA","NA"))
Name val Case
1 A 1 NA
2 B 2 1
3 C 3 NA
4 D 1 NA
5 E 2 1
6 F 3 NA
7 G 1 1
8 H 2 NA
9 I 3 NA
Some rows in the Case column in df1 have to be changed with the info in the df2 below:
df2 = data.frame(Name = c("A","D","H"), val = seq(1:3), Case = "1")
Name val Case
1 A 1 1
2 D 2 1
3 H 3 1
So there's nothing important in the val column, however I added it into the examples since I want to indicate that I have more columns than two and also my real data is way bigger than the examples.
Basically, I want to change specific rows by checking the information in the first columns (in this case, they're unique letters) and in the end I still want to have df1 as a final data frame.
for a better explanation, I want to see something like this:
Name val Case
1 A 1 1
2 B 2 1
3 C 3 NA
4 D 1 1
5 E 2 1
6 F 3 NA
7 G 1 1
8 H 2 1
9 I 3 NA
Note changed information for A,D and H.
Thanks.
%in% from base-r is there to rescue.
df1=data.frame(Name = print(LETTERS[1:9]), val = seq(1:3), Case = c("NA","1","NA","NA","1","NA","1","NA","NA"), stringsAsFactors = F)
df2 = data.frame(Name = c("A","D","H"), val = seq(1:3), Case = "1", stringsAsFactors = F)
df1$Case <- ifelse(df1$Name %in% df2$Name, df2$Case[df2$Name %in% df1$Name], df1$Case)
df1
Output:
> df1
Name val Case
1 A 1 1
2 B 2 1
3 C 3 NA
4 D 1 1
5 E 2 1
6 F 3 NA
7 G 1 1
8 H 2 1
9 I 3 NA
Here is what I would do using dplyr:
df1 %>%
left_join(df2, by = c("Name")) %>%
mutate(val = if_else(is.na(val.y), val.x, val.y),
Case = if_else(is.na(Case.y), Case.x, Case.y)) %>%
select(Name, val, Case)
Related
I have two tables; "Reference" and "TableA".
I am looking through TableA which is an incomplete table and would like to turn it into a "complete" table by referencing the "Reference" table, filling in missing values, and/or adding rows where there are multiple matches are found.
Reproducible example of "Reference" and "TableA" are below:
A <- c(1,1,1,2,4,4,5,5,7,6,2,1)
B <- c(1,2,2,2,4,4,9,5,8,6,2,9)
C <- c(1,1,3,3,4,5,5,5,7,6,3,3)
D <- c(1,2,1,1,2,1,2,1,2,2,2,1)
Reference <- data.frame(A,B,C,D)
A <- c(NA,1,5,2,4,1)
B <- c(NA,2,NA,2,NA,1)
C <- c(3,NA,5,NA,NA,1)
D <- c(1,1,2,2,1,1)
TableA <- data.frame(A,B,C,D)
I have attempted to resolve this by doing the following:
for (i in 1:dim(TableA)[1])
{
tmp<-TableA[i,]
repet<-ifelse(is.na(TableA$D[i]), Reference, 1 )
for (j in 1:repet) {
tmp$D<-ifelse(repet>1, Reference$D[j,], tmp$D)
collector<-rbind(collector, tmp)
}
}
collector
However, this solution will return the entirety of Reference$D, but I would only like to return those records from Reference$D whose columns A,B,C match (or partially match) what is on TableA.
For example, in Row 1 of TableA, I would like to replace Row 1 with the Reference table's rows 3,4, and 12.
Expected output below.
Note that the Reference table combination 1,2,3,1 appears twice on the expected output as it is a match for both rows 1 & 2 of TableA.
A
B
C
D
1
2
3
1
2
2
3
1
1
9
3
1
1
2
3
1
5
9
5
2
2
2
3
2
4
4
5
1
1
1
1
1
I'll first create an extra column "string" in both TableA and Reference, with NA replaced with a dot . in TableA, which would be used in regex matching.
Then find out which string in TableA appeared in Reference, and store them in a matrix.
Finally, replicate the lgl_matrix row number by the number of matches, and use those row numbers as index in Reference.
library(tidyverse)
TableA <- TableA %>%
mutate(across(A:D, ~ replace_na(as.character(.x), "."))) %>%
rowwise() %>%
mutate(string = paste0(c_across(A:D), collapse = ""))
Reference <- Reference %>%
rowwise() %>%
mutate(string = paste0(c_across(A:D), collapse = ""))
lgl_matrix <- sapply(TableA$string, grepl, x = Reference$string)
Reference[rep(1:nrow(lgl_matrix), rowSums(lgl_matrix)), -5]
# A tibble: 8 x 4
# Rowwise:
A B C D
<dbl> <dbl> <dbl> <dbl>
1 1 1 1 1
2 1 2 3 1
3 1 2 3 1
4 2 2 3 1
5 4 4 5 1
6 5 9 5 2
7 2 2 3 2
8 1 9 3 1
I have a dataframe in R that I'd like to repeat several times, and I want to add in a new variable to index those repetitions. The best I've come up with is using mutate + rbind over and over, and I feel like there has to be an efficient dataframe method I could be using here.
Here's an example: df <- data.frame(x = 1:3, y = letters[1:3]) gives us the dataframe
x
y
1
a
2
b
3
c
I'd like to repeat that say 3 times, with an index that looks like this:
x
y
index
1
a
1
2
b
1
3
c
1
1
a
2
2
b
2
3
c
2
1
a
3
2
b
3
3
c
3
Using the rep function, I can get the first two columns, but not the index column. The best I've come up with so far (using dplyr) is:
df2 <-
df %>%
mutate(index = 1) %>%
rbind(df %>% mutate(index = 2)) %>%
rbind(df %>% mutate(index = 3))
This obviously doesn't work if I need to repeat my dataframe more than a handful of times. It feels like the kind of thing that should be easy to do using dataframe methods, but I haven't been able to find anything.
Grateful for any tips!
You can use this code for as many data frames as you would like. You just have to set the n argument:
replicate function takes 2 main arguments. We first specify the number of time we would like to reproduce our data set by n. Then we specify our data set as expr argument. The result would be a list whose elements are instances of our data set
After that we pass it along to imap function from purrr package to define the unique id for each of our data set. .x represents each element of our list (here a data frame) and .y is the position of that element which amounts to the number of instances we created. So for example we assign value 1 to the first id column of the first data set as .y is equal to 1 for that and so on.
library(dplyr)
library(purrr)
replicate(3, df, simplify = FALSE) %>%
imap_dfr(~ .x %>%
mutate(id = .y))
x y id
1 1 a 1
2 2 b 1
3 3 c 1
4 1 a 2
5 2 b 2
6 3 c 2
7 1 a 3
8 2 b 3
9 3 c 3
In base R you can use the following code:
do.call(rbind,
mapply(function(x, z) {
x$id <- z
x
}, replicate(3, df, simplify = FALSE), 1:3, SIMPLIFY = FALSE))
x y id
1 1 a 1
2 2 b 1
3 3 c 1
4 1 a 2
5 2 b 2
6 3 c 2
7 1 a 3
8 2 b 3
9 3 c 3
You can use rerun to repeat the dataframe n times and add an index column using bind_rows -
library(dplyr)
library(purrr)
n <- 3
df <- data.frame(x = 1:3, y = letters[1:3])
bind_rows(rerun(n, df), .id = 'index')
# index x y
#1 1 1 a
#2 1 2 b
#3 1 3 c
#4 2 1 a
#5 2 2 b
#6 2 3 c
#7 3 1 a
#8 3 2 b
#9 3 3 c
In base R, we can repeat the row index 3 times.
transform(df[rep(1:nrow(df), n), ], index = rep(1:n, each = nrow(df)))
One more way
n <- 3
map_dfr(seq_len(n), ~ df %>% mutate(index = .x))
x y index
1 1 a 1
2 2 b 1
3 3 c 1
4 1 a 2
5 2 b 2
6 3 c 2
7 1 a 3
8 2 b 3
9 3 c 3
Not sure if this is a duplicate but I couldn't find anything that either solves my original problem or the issue I'm running into with the partial I did find.
The goal is to sort a dataframe independently by column.
Reproducible example
a <- data.frame(name = c("a","a","a","b","b","b"),date1 = c(2,3,1,3,1,2),date2 = c(0,2,3,1,2,0),date3 = c(0,2,0,3,2,1))
a
name date1 date2 date3
1 a 2 0 0
2 a 3 2 2
3 a 1 3 0
4 b 3 1 3
5 b 1 2 2
6 b 2 0 1
b <- ddply(a, "name", function(x) { as.data.frame(lapply(x, sort))
b
name date1 date2 date3
1 a 1 0 0
2 a 2 2 0
3 a 3 3 2
4 b 1 0 1
5 b 2 1 2
6 b 3 2 3
Now this works as expected, but is the opposite of what I'm looking to do.
Desired output
b
name date1 date2 date3
1 a 3 3 2
2 a 2 2 0
3 a 1 0 0
4 b 3 2 3
5 b 2 1 2
6 b 1 0 1
I've tried to add in the decreasing=T parameter but haven't had any luck with the variations I've tried and usually end up with an error about missing arguments or undefined columns being selected. How does one correctly implement a decreasing sort with this syntax and/or otherwise achieve the end result without relying on explicitly naming the columns (they names are dates so change often)
Bonus
How could this code be adapted to account for NA's with na.last
Thank you!
I think you nuked the data.frame rows with your code, not a very good practice standard dplyr use the arrange() function like this
library(tidyverse)
a <- data.frame(name = c("a","a","a","b","b","b"),date1 = c(2,3,1,3,1,2),date2 = c(0,2,3,1,2,0),date3 = c(0,2,0,3,2,1))
a %>%
arrange(name,-date1)
If you want to live a dangerous life here is the code for it
a %>%
group_by(name) %>%
mutate_all(sort,decreasing = TRUE)
name date1 date2 date3
<fct> <dbl> <dbl> <dbl>
1 a 3 3 2
2 a 2 2 0
3 a 1 0 0
4 b 3 2 3
5 b 2 1 2
6 b 1 0 1
A solution with the data.table package is the following
library(data.table)
a <- data.table(name = c("a","a","a","b","b","b"),date1 = c(2,3,1,3,1,2),date2 = c(0,2,3,1,2,0),date3 = c(0,2,0,3,2,1))
# alternatively:
# a <- data.frame(name = c("a","a","a","b","b","b"),date1 = c(2,3,1,3,1,2),date2 = c(0,2,3,1,2,0),date3 = c(0,2,0,3,2,1))
# setDT(a)
b <- a[, lapply(.SD, sort, decreasing = TRUE), by = name]
.SD returns the subset of data, in this case created with the by = name. It splits the original data.table by the values in the given column.
This also fulfills your bonus requirement, the na.last can be supplied.
aa <- data.table(name = c("a","a","a","b","b","b"),date1 = c(NA,3,1,3,1,NA),date2 = c(0,2,NA,1,2,0),date3 = c(0,2,0,3,2,NA))
bb <- aa[, lapply(.SD, sort, decreasing = TRUE, na.last = TRUE), by = name]
I am working in R with a dataset that is created from mongodb with the use of mongolite.
I am getting a list that looks like so:
_id A B A B A B NA NA
1 a 1 b 2 e 5 NA NA
2 k 4 l 3 c 3 d 4
I would like to merge the datasetto look like this:
_id A B
1 a 1
2 k 4
1 b 2
2 l 3
1 e 5
2 c 3
1 NA NA
2 d 4
The NAs in the last columns are there because the columns are named from the first entry and if a later entry has more columns than that they don't get names assigned to them, (if I get help for this as well it would be awesome but it's not the reason I am here).
Also the number of columns might differ for different subsets of the dataset.
I have tried melt() but since it is a list and not a dataframe it doesn't work as expected, I have tried stack() but it dodn't work because the columns have the same name and some of them don't even have a name.
I know this is a very weird situation and appreciate any help.
Thank you.
using library(magrittr)
data:
df <- fread("
_id A B A B A B NA NA
1 a 1 b 2 e 5 NA NA
2 k 4 l 3 c 3 d 4 ",header=T)
setDF(df)
Code:
df2 <- df[,-1]
odds<- df2 %>% ncol %>% {(1:.)%%2} %>% as.logical
even<- df2 %>% ncol %>% {!(1:.)%%2}
cbind(df[,1,drop=F],
A=unlist(df2[,odds]),
B=unlist(df2[,even]),
row.names=NULL)
result:
# _id A B
# 1 1 a 1
# 2 2 k 4
# 3 1 b 2
# 4 2 l 3
# 5 1 e 5
# 6 2 c 3
# 7 1 <NA> NA
# 8 2 d 4
We can use data.table. Assuming A and B are always following each other. I created an example with 2 sets of NA's in the header. With grep we can find the ones fread has named V8 etc. Using R's recycling of vectors, you can rename multiple headers in one go. If in your case these are named differently change the pattern in the grep command. Then we melt the data in via melt
library(data.table)
df <- fread("
_id A B A B A B NA NA NA NA
1 a 1 b 2 e 5 NA NA NA NA
2 k 4 l 3 c 3 d 4 e 5",
header = TRUE)
df
_id A B A B A B A B A B
1: 1 a 1 b 2 e 5 <NA> NA <NA> NA
2: 2 k 4 l 3 c 3 d 4 e 5
# assuming A B are always following each other. Can be done in 1 statement.
cols <- names(df)
cols[grep(pattern = "^V", x = cols)] <- c("A", "B")
names(df) <- cols
# melt data (if df is a data.frame replace df with setDT(df)
df_melted <- melt(df, id.vars = 1,
measure.vars = patterns(c('A', 'B')),
value.name=c('A', 'B'))
df_melted
_id variable A B
1: 1 1 a 1
2: 2 1 k 4
3: 1 2 b 2
4: 2 2 l 3
5: 1 3 e 5
6: 2 3 c 3
7: 1 4 <NA> NA
8: 2 4 d 4
9: 1 5 <NA> NA
10: 2 5 e 5
Thank you for your help, they were great inspirations.
Even though #Andre Elrico gave a solution that worked in the reproducible example better #phiver gave a solution that worked better on my overall problem.
By using both those I came up with the following.
library(data.table)
#The data were in a list of lists called list for this example
temp <- as.data.table(matrix(t(sapply(list, '[', seq(max(sapply(list, lenth))))),
nrow = m))
# m here is the number of lists in list
cols <- names(temp)
cols[grep(pattern = "^V", x = cols)] <- c("B", "A")
#They need to be the opposite way because the first column is going to be substituted with id, and this way they fall on the correct column after that
cols[1] <- "id"
names(temp) <- cols
l <- melt.data.table(temp, id.vars = 1,
measure.vars = patterns(c("A", "B")),
value.name = c("A", "B"))
That way I can use this also if I have more than 2 columns that I need to manipulate like that.
I have a dataframe:
a <- c(rep("A", 3), rep("B", 3), rep("C",2))
b <- c(1,1,2,4,1,1,2,2)
c <- c(1,NA,2,4,NA,1,2,2)
df <-data.frame(a,b,c)
I have a dataframe with some duplicate variables in column 1 but when I use the duplicated function, it randomly chooses the row after de-duping using duplicate(function)
dedup_df = df[!duplicated(df$a), ]
How can I ensure that the output returns me the row that does not contain an NA on column c ?
I tried to use the dplyr package but the output prints only a result
library(dplyr)
options(dplyr.print_max = Inf )
df %>% ## source dataframe
group_by(a) %>% ## grouped by variable
filter(!is.na(c) ) %>% ## filter by Gross value
as.data.frame(dedup_df)
Your use of duplicated function to remove duplicate observations (lines) using a column as key from a data frame is correct.
But it seems that you are worried that it may keep a line that contains NA in another column and drop another line that contains a non NA value.
I'll use you example, but with a slight modification
a <- c(rep("A", 3), rep("B", 3), rep("C",2))
b <- c(1,1,2,4,1,1,2,2)
c <- c(NA,1,2,4,NA,1,2,2)
df <-data.frame(a,b,c)
> df
a b c
1 A 1 NA
2 A 1 1
3 A 2 2
4 B 4 4
5 B 1 NA
6 B 1 1
7 C 2 2
8 C 2 2
In this case, your dedup_df contains an NA for the first value.
> dedup_df = df[!duplicated(df$a), ]
> dedup_df
a b c
1 A 1 NA
4 B 4 4
7 C 2 2
Solution:
Reorder df by column c first and then use the same command. This reordering by column c will send all NAs to the end of the data frame. When the duplicated passes it will see these lines having NA last and will tag them as TRUE if there was a previous one without NA.
df = df[order(df$c),]
dedup_df = df[!duplicated(df$a), ]
> dedup_df
a b c
2 A 1 1
6 B 1 1
7 C 2 2
You can also reorder in descending order
df = df[order(df$c,decreasing = T),]
dedup_df = df[!duplicated(df$a), ]
> dedup_df
a b c
4 B 4 4
3 A 2 2
7 C 2 2