I looked here and elsewhere, but I cannot find something that does exactly what I'm looking to accomplish using R.
I have data similar to below, where col1 is a unique ID, col2 is a group ID variable, col3 is a status code. I need to flag all rows with the same group ID, and where any of those rows have a specific status code, X in this case, as == 1, otherwise 0.
ID GroupID Status Flag
1 100 A 1
2 100 X 1
3 102 A 0
4 102 B 0
5 103 B 1
6 103 X 1
7 104 X 1
8 104 X 1
9 105 A 0
10 105 C 0
I have tried writing some ifelse where groupID == groupID and status == X then 1 else 0, but that doesn't work. The pattern of Status is random. In this example, the GroupID is exclusively pairs, but I don't want to assume that in the code, b/c I have other instance where there are 3 or more rows in a GroupID.
It would be helpful if this were open ended IE I could add other conditions if necessary, like, for each matching group ID, where Status == X, and other or other, etc.
Thank you !
Group-based operations like this are easy to do with the dplyr package.
The data:
library(dplyr)
txt <- 'ID GroupID Status
1 100 A
2 100 X
3 102 A
4 102 B
5 103 B
6 103 X
7 104 X
8 104 X
9 105 A
10 105 C '
df <- read.table(text = txt, header = T)
Once we have the data frame, we establish dplyr groups with the group_by function. The mutate command will then be applied per each group, creating a new column entry for each row.
df.new <- df %>%
group_by(GroupID) %>%
mutate(Flag = as.numeric(any(Status == 'X')))
# A tibble: 10 x 4
# Groups: GroupID [5]
ID GroupID Status Flag
<int> <int> <fct> <dbl>
1 1 100 A 1
2 2 100 X 1
3 3 102 A 0
4 4 102 B 0
5 5 103 B 1
6 6 103 X 1
7 7 104 X 1
8 8 104 X 1
9 9 105 A 0
10 10 105 C 0
From base R
ave(df$Status=='X',df$GroupID,FUN=any)
[1] TRUE TRUE FALSE FALSE TRUE TRUE TRUE TRUE FALSE FALSE
Data.table way:
library(data.table)
setDT(df)
df[ , flag := sum(Status == "X") > 0, by=GroupID]
An alternative using data.table
library(data.table)
dt <- read.table(stringsAsFactors = FALSE,text = "ID GroupID Status
1 100 A
2 100 X
3 102 A
4 102 B
5 103 B
6 103 X
7 104 X
8 104 X
9 105 A
10 105 C", header=T)
setDT(dt)[,.(ID,Status, Flag=ifelse("X"%in% Status,1,0)),by=GroupID]
#returns
GroupID ID Status Flag
1: 100 1 A 1
2: 100 2 X 1
3: 102 3 A 0
4: 102 4 B 0
5: 103 5 B 1
6: 103 6 X 1
7: 104 7 X 1
8: 104 8 X 1
9: 105 9 A 0
10: 105 10 C 0
A base R option with rowsum
i1 <- with(df1, rowsum(+(Status == "X"), group = GroupID) > 0)
transform(df1, Flag = +(GroupID %in% row.names(i1)[i1]))
Or using table
df1$Flag <- +(with(df1, GroupID %in% names(which(table(GroupID,
Status == "X")[,2]> 0))))
Related
I have some values in
df:
# A tibble: 7 × 1
var1
<dbl>
1 0
2 10
3 20
4 210
5 230
6 266
7 267
that I would like to compare to a second dataframe called
value_lookup
# A tibble: 4 × 2
var1 value
<dbl> <dbl>
1 0 0
2 200 10
3 230 20
4 260 30
In particual I would like to make a join based on >= meaning that a value that is greater or equal to the number in var1 gets a values of x. E.g. take the number 210 of the orginal dataframe. Since it is >= 200 and <230 it would get a value of 10.
Here is the expected output:
var1 value
1 0 0
2 10 0
3 20 0
4 210 10
5 230 20
6 266 30
7 267 30
I thought it should be doable using {fuzzyjoin} but I cannot get it done.
value_lookup <- tibble(var1 = c(0, 200,230,260),
value = c(0,10,20,30))
df <- tibble(var1 = c(0,10,20,210,230,266,267))
library(fuzzyjoin)
fuzzyjoin::fuzzy_left_join(
x = df,
y = value_lookup ,
by = "var1",
match_fun = list(`>=`)
)
An option is also findInterval:
df$value <- value_lookup$value[findInterval(df$var1, value_lookup$var1)]
Output:
var1 value
1 0 0
2 10 0
3 20 0
4 210 10
5 230 20
6 266 30
7 267 30
As you're mentioning joins, you could also do a rolling join via data.table with the argument roll = T which would look for same or closest value preceding var1 in your df:
library(data.table)
setDT(value_lookup)[setDT(df), on = 'var1', roll = T]
You can use cut:
df$value <- value_lookup$value[cut(df$var1,
c(value_lookup$var1, Inf),
right=F)]
# # A tibble: 7 x 2
# var1 value
# <dbl> <dbl>
# 1 0 0
# 2 10 0
# 3 20 0
# 4 210 10
# 5 230 20
# 6 266 30
# 7 267 30
I have a data frame (datadf) with 3 columns, 'x', 'y, and z. Several 'x' values are missing (NA). 'y' and 'z' are non measured variables.
x y z
153 a 1
163 b 1
NA d 1
123 a 2
145 e 2
NA c 2
NA b 1
199 a 2
I have another data frame (imputeddf) with the same three columns:
x y z
123 a 1
145 a 2
124 b 1
168 b 2
123 c 1
176 c 2
184 d 1
101 d 2
I wish to replace NA in 'x' in 'datadf' with values from 'imputeddf' where 'y' and 'z' matches between the two data sets (each combo of 'y' and 'z' has its own value of 'x' to fill in).
The desired result:
x y z
153 a 1
163 b 1
184 d 1
123 a 2
145 e 2
176 c 2
124 b 1
199 a 2
I am trying things like:
finaldf <- datadf
finaldf$x <- if(datadf[!is.na(datadf$x)]){ddply(datadf, x=imputeddf$x[datadf$y == imputeddf$y & datadf$z == imputeddf$z])}else{datadf$x}
but it's not working.
What is the best way for me to fill in the NA in the using my imputed value df?
I would do this:
library(data.table)
setDT(DF1); setDT(DF2)
DF1[DF2, x := ifelse(is.na(x), i.x, x), on=c("y","z")]
which gives
x y z
1: 153 a 1
2: 163 b 1
3: 184 d 1
4: 123 a 2
5: 145 e 2
6: 176 c 2
7: 124 b 1
8: 199 a 2
Comments. This approach isn't so great, since it merges the whole of DF1, while we only need to merge the subset where is.na(x). Here, the improvement looks like (thanks, #Arun):
DF1[is.na(x), x := DF2[.SD, x, on=c("y", "z")]]
This way is analogous to #RHertel's answer.
From #Jakob's comment:
does this work for more than one x variable? If I want to fill up entire datasets with several columns?
You can enumerate the desired columns:
DF1[DF2, `:=`(
x = ifelse(is.na(x), i.x, x),
w = ifelse(is.na(w), i.w, w)
), on=c("y","z")]
The expression could be constructed using lapply and substitute, probably, but if the set of columns is fixed, it might be cleanest just to write it out as above.
Here's an alternative with base R:
df1[is.na(df1$x),"x"] <- merge(df2,df1[is.na(df1$x),][,c("y","z")])$x
> df1
# x y z
#1 153 a 1
#2 163 b 1
#3 124 b 1
#4 123 a 2
#5 145 e 2
#6 176 c 2
#7 184 d 1
#8 199 a 2
A dplyr solution, conceptually identical to the answers above. To pull out just the rows of imputeddf that correspond to NAs in datadf, use semi_join. Then, use another join to match back to datadf. (This step is not very clean, unfortunately.)
library(dplyr)
replacement_rows <- imputeddf %>%
semi_join(datadf %>% filter(is.na(x)), by = c("y", "z"))
datadf <- datadf %>%
left_join(replacement_rows, by = c("y", "z")) %>%
mutate(x = if_else(is.na(x.x), x.y, x.x)) %>%
select(x, y, z)
This gets what you want:
> datadf
# A tibble: 8 x 3
x y z
<dbl> <chr> <dbl>
1 153 a 1
2 163 b 1
3 184 d 1
4 123 a 2
5 145 e 2
6 176 c 2
7 124 b 1
8 199 a 2
In dplyr, you can use rows_patch to update NAs:
rows_patch(datadf, imputeddf, by = c("y", "z"), unmatched = "ignore")
# x y z
# 1 153 a 1
# 2 163 b 1
# 3 184 d 1
# 4 123 a 2
# 5 145 e 2
# 6 176 c 2
# 7 124 b 1
# 8 199 a 2
data:
datadf <- read.table(header = T, text = "x y z
153 a 1
163 b 1
NA d 1
123 a 2
145 e 2
NA c 2
NA b 1
199 a 2")
imputeddf <- read.table(header = T, text = " x y z
123 a 1
145 a 2
124 b 1
168 b 2
123 c 1
176 c 2
184 d 1
101 d 2")
I'd like to create an efficient ifelse statement such that if columns from df2 match columns from df1, then that row in df2 is coded a specific way. My code works but is very inefficient.
Example data:
Df1
A B C
111 2 1
111 5 2
111 7 3
112 2 4
112 8 5
113 2 6
Df2
A B
112 2
111 2
113 2
111 5
111 7
112 8
Desired Outcome:
Df2
A B C
112 2 4
111 2 1
113 2 6
111 5 2
111 7 3
112 8 5
What I've done is this:
Df2$C<- ifelse(Df2$A == 111 & Df2$B == 2, 1, 0)
Df2$C<- ifelse(Df2$A == 111 & Df2$B == 5, 2, 0)
Df2$C<- ifelse(Df2$A == 111 & Df2$B == 7, 3, 0)...
This works, but is there a way such that df2 could reference the columns in df1 and create column df2$C, so that each combination doesn't have to be manually typed out?
This would typically be done with a join. left_join from dplyr will connect each of the rows in your first table with the each of the matching rows in the second table.
https://dplyr.tidyverse.org/reference/join.html
library(dplyr)
Df2 %>% left_join(Df1)
Joining, by = c("A", "B")
A B C
1 112 2 4
2 111 2 1
3 113 2 6
4 111 5 2
5 111 7 3
6 112 8 5
merge from base R will give a similar result, but doesn't keep the original row order without some extra wrangling.
Merge two data frames while keeping the original row order
merge(Df2, Df1)
A B C
1 111 2 1
2 111 5 2
3 111 7 3
4 112 2 4
5 112 8 5
6 113 2 6
I have a data frame (datadf) with 3 columns, 'x', 'y, and z. Several 'x' values are missing (NA). 'y' and 'z' are non measured variables.
x y z
153 a 1
163 b 1
NA d 1
123 a 2
145 e 2
NA c 2
NA b 1
199 a 2
I have another data frame (imputeddf) with the same three columns:
x y z
123 a 1
145 a 2
124 b 1
168 b 2
123 c 1
176 c 2
184 d 1
101 d 2
I wish to replace NA in 'x' in 'datadf' with values from 'imputeddf' where 'y' and 'z' matches between the two data sets (each combo of 'y' and 'z' has its own value of 'x' to fill in).
The desired result:
x y z
153 a 1
163 b 1
184 d 1
123 a 2
145 e 2
176 c 2
124 b 1
199 a 2
I am trying things like:
finaldf <- datadf
finaldf$x <- if(datadf[!is.na(datadf$x)]){ddply(datadf, x=imputeddf$x[datadf$y == imputeddf$y & datadf$z == imputeddf$z])}else{datadf$x}
but it's not working.
What is the best way for me to fill in the NA in the using my imputed value df?
I would do this:
library(data.table)
setDT(DF1); setDT(DF2)
DF1[DF2, x := ifelse(is.na(x), i.x, x), on=c("y","z")]
which gives
x y z
1: 153 a 1
2: 163 b 1
3: 184 d 1
4: 123 a 2
5: 145 e 2
6: 176 c 2
7: 124 b 1
8: 199 a 2
Comments. This approach isn't so great, since it merges the whole of DF1, while we only need to merge the subset where is.na(x). Here, the improvement looks like (thanks, #Arun):
DF1[is.na(x), x := DF2[.SD, x, on=c("y", "z")]]
This way is analogous to #RHertel's answer.
From #Jakob's comment:
does this work for more than one x variable? If I want to fill up entire datasets with several columns?
You can enumerate the desired columns:
DF1[DF2, `:=`(
x = ifelse(is.na(x), i.x, x),
w = ifelse(is.na(w), i.w, w)
), on=c("y","z")]
The expression could be constructed using lapply and substitute, probably, but if the set of columns is fixed, it might be cleanest just to write it out as above.
Here's an alternative with base R:
df1[is.na(df1$x),"x"] <- merge(df2,df1[is.na(df1$x),][,c("y","z")])$x
> df1
# x y z
#1 153 a 1
#2 163 b 1
#3 124 b 1
#4 123 a 2
#5 145 e 2
#6 176 c 2
#7 184 d 1
#8 199 a 2
A dplyr solution, conceptually identical to the answers above. To pull out just the rows of imputeddf that correspond to NAs in datadf, use semi_join. Then, use another join to match back to datadf. (This step is not very clean, unfortunately.)
library(dplyr)
replacement_rows <- imputeddf %>%
semi_join(datadf %>% filter(is.na(x)), by = c("y", "z"))
datadf <- datadf %>%
left_join(replacement_rows, by = c("y", "z")) %>%
mutate(x = if_else(is.na(x.x), x.y, x.x)) %>%
select(x, y, z)
This gets what you want:
> datadf
# A tibble: 8 x 3
x y z
<dbl> <chr> <dbl>
1 153 a 1
2 163 b 1
3 184 d 1
4 123 a 2
5 145 e 2
6 176 c 2
7 124 b 1
8 199 a 2
In dplyr, you can use rows_patch to update NAs:
rows_patch(datadf, imputeddf, by = c("y", "z"), unmatched = "ignore")
# x y z
# 1 153 a 1
# 2 163 b 1
# 3 184 d 1
# 4 123 a 2
# 5 145 e 2
# 6 176 c 2
# 7 124 b 1
# 8 199 a 2
data:
datadf <- read.table(header = T, text = "x y z
153 a 1
163 b 1
NA d 1
123 a 2
145 e 2
NA c 2
NA b 1
199 a 2")
imputeddf <- read.table(header = T, text = " x y z
123 a 1
145 a 2
124 b 1
168 b 2
123 c 1
176 c 2
184 d 1
101 d 2")
I am trying to rank multiple numeric variables ( around 700+ variables) in the data and am not sure exactly how to do this as I am still pretty new to using R.
I do not want to overwrite the ranked values in the same variable and hence need to create a new rank variable for each of these numeric variables.
From reading the posts, I believe assign and transform function along with rank maybe able to solve this. I tried implementing as below ( sample data and code) and am struggling to get it to work.
The output dataset in addition to variables xcount, xvisit, ysales need to be populated
With variables xcount_rank, xvisit_rank, ysales_rank containing the ranked values.
input <- read.table(header=F, text="101 2 5 6
102 3 4 7
103 9 12 15")
colnames(input) <- c("id","xcount","xvisit","ysales")
input1 <- input[,2:4] #need to rank the numeric variables besides id
for (i in 1:3)
{
transform(input1,
assign(paste(input1[,i],"rank",sep="_")) =
FUN = rank(-input1[,i], ties.method = "first"))
}
input[paste(names(input)[2:4], "rank", sep = "_")] <-
lapply(input[2:4], cut, breaks = 10)
The problem with this approach is that it's creating the rank values as (101, 230] , (230, 450] etc whereas I would like to see the values in the rank variable to be populated as 1, 2 etc up to 10 categories as per the splits I did. Is there any way to achieve this? input[5:7] <- lapply(input[5:7], rank, ties.method = "first")
The approach I tried from the solutions provided below is:
input <- read.table(header=F, text="101 20 5 6
102 2 4 7
103 9 12 15
104 100 8 7
105 450 12 65
109 25 28 145
112 854 56 93")
colnames(input) <- c("id","xcount","xvisit","ysales")
input[paste(names(input)[2:4], "rank", sep = "_")] <-
lapply(input[2:4], cut, breaks = 3)
Current output I get is:
id xcount xvisit ysales xcount_rank xvisit_rank ysales_rank
1 101 20 5 6 (1.15,286] (3.95,21.3] (5.86,52.3]
2 102 2 4 7 (1.15,286] (3.95,21.3] (5.86,52.3]
3 103 9 12 15 (1.15,286] (3.95,21.3] (5.86,52.3]
4 104 100 8 7 (1.15,286] (3.95,21.3] (5.86,52.3]
5 105 450 12 65 (286,570] (3.95,21.3] (52.3,98.7]
6 109 25 28 145 (1.15,286] (21.3,38.7] (98.7,145]
7 112 854 56 93 (570,855] (38.7,56.1] (52.3,98.7]
Desired output:
id xcount xvisit ysales xcount_rank xvisit_rank ysales_rank
1 101 20 5 6 1 1 1
2 102 2 4 7 1 1 1
3 103 9 12 15 1 1 1
4 104 100 8 7 1 1 1
5 105 450 12 65 2 1 2
6 109 25 28 145 1 2 3
Would like to see the records in the group they would fall under if I try to rank the interval values.
Using dplyr
library(dplyr)
nm1 <- paste("rank", names(input)[2:4], sep="_")
input[nm1] <- mutate_each(input[2:4],funs(rank(., ties.method="first")))
input
# id xcount xvisit ysales rank_xcount rank_xvisit rank_ysales
#1 101 2 5 6 1 2 1
#2 102 3 4 7 2 1 2
#3 103 9 12 15 3 3 3
Update
Based on the new input and using cut
input[nm1] <- mutate_each(input[2:4], funs(cut(., breaks=3, labels=FALSE)))
input
# id xcount xvisit ysales rank_xcount rank_xvisit rank_ysales
#1 101 20 5 6 1 1 1
#2 102 2 4 7 1 1 1
#3 103 9 12 15 1 1 1
#4 104 100 8 7 1 1 1
#5 105 450 12 65 2 1 2
#6 109 25 28 145 1 2 3
#7 112 854 56 93 3 3 2