I have a spreadsheet Y that has thousands of rows
I would like to extract a few hundred specific rows identified by key X
I used intersect function to generate object Z as below, but do not know how to proceed.
Many thanks
Z<-intersect(X$PatientID, Y$Patient.ID)
Try using %in%:
Sample Data
Y <- data.frame(PatientID = LETTERS,
Var1 = 1:26)
key <- data.frame(PatientID = c("A", "W", "V"),
Var1 = c(1, 22:23))
Using %in%:
want <- Y[Y$PatientID %in% key$PatientID, ]
Output:
# PatientID Var1
# 1 A 1
# 22 V 22
# 23 W 23
Note if you needed to use intersect, you would just do this:
Z <- intersect(Y$PatientID, key$PatientID)
want <- Y[Y$PatientID %in% Z,]
And it would give you the same output
You can merge:
base R
merge(Y,unique(X[,"PatientID",drop=F]))
dplyr:
dplyr::inner_join(Y, dplyr::distinct(X,PatientID))
Related
I have a vector containing "potential" column names:
col_vector <- c("A", "B", "C")
I also have a data frame, e.g.
library(tidyverse)
df <- tibble(A = 1:2,
B = 1:2)
My goal now is to create all columns mentioned in col_vector that don't yet exist in df.
For the above exmaple, my code below works:
df %>%
mutate(!!sym(setdiff(col_vector, colnames(.))) := NA)
# A tibble: 2 x 3
A B C
<int> <int> <lgl>
1 1 1 NA
2 2 2 NA
Problem is that this code fails as soon as a) more than one column from col_vector is missing or b) no column from col_vector is missing. I thought about some sort of if_else, but don't know how to make the column creation conditional in such a way - preferably in a tidyverse way. I know I can just create a loop going through all the missing columns, but I'm wondering if there is a more direc approach.
Example data where code above fails:
df2 <- tibble(A = 1:2)
df3 <- tibble(A = 1:2,
B = 1:2,
C = 1:2)
This should work.
df[,setdiff(col_vector, colnames(df))] <- NA
Solution
This base operation might be simpler than a full-fledged dplyr workflow:
library(tidyverse) # For the setdiff() function.
# ...
# Code to generate 'df'.
# ...
# Find the subset of missing names, and create them as columns filled with 'NA'.
df[, setdiff(col_vector, names(df))] <- NA
# View results
df
Results
Given your sample col_vector and df here
col_vector <- c("A", "B", "C")
df <- tibble(A = 1:2, B = 1:2)
this solution should yield the following results:
# A tibble: 2 x 3
A B C
<int> <int> <lgl>
1 1 1 NA
2 2 2 NA
Advantages
An advantage of my solution, over the alternative linked above by #geoff, is that you need not code by hand the set of column names, as symbols and strings within the dplyr workflow.
df %>% mutate(
#####################################
A = ifelse("A" %in% names(.), A, NA),
B = ifelse("B" %in% names(.), B, NA),
C = ifelse("C" %in% names(.), B, NA)
# ...
# etc.
#####################################
)
My solution is by contrast more dynamic
##############################
df[, setdiff(col_vector, names(df))] <- NA
##############################
if you ever decide to change (or even dynamically calculate!) your variable names midstream, since it determines the setdiff() at runtime.
Note
Incredibly, #AustinGraves posted their answer at precisely the same time (2021-10-25 21:03:05Z) as I posted mine, so both answers qualify as original solutions.
I want to extract a column from a dataframe in R based on a condition for another column in the same dataframe, the dataframe is given below.
b <- c(1,2,3,4)
g <- c("a", "b" ,"b", "c")
df <- data.frame(b,g)
row.names(df) <- c("aa", "bb", "cc" , "dd")
I want to extract all values for column b as a dataframe (with rownames) where column g has value 'b',
My required output is given below:
df
b
cc 3
dd 4
I have tried several methods like which or subset but it does not work. I have also tried to find the answer to this question on stackoverflow but I was not able to find it. Is there a way to do it?
Thanks,
You can use the subset function in base R -
subset(df, g == 'b', select = b)
# b
#bb 2
#cc 3
Using data.table
library(data.table)
setDT(df, key = 'g')['b', .(b)]
b
1: 2
2: 3
Or with collapse
library(collapse)
sbt(df, g == 'b', b)
b
1 2
2 3
This is the basic way of slicing data in r
df[df$g == 'b',]['b']
Or the tidyverse answer
df %>%
filter(g == 'b') %>%
select(b)
I'm trying to figure out how to replace rows in one dataframe with another by matching the values of one of the columns. Both dataframes have the same column names.
Ex:
df1 <- data.frame(x = c(1,2,3,4), y = c("a", "b", "c", "d"))
df2 <- data.frame(x = c(1,2), y = c("f", "g"))
Is there a way to replace the rows of df1 with the same row in df2 where they share the same x variable? It would look like this.
data.frame(x = c(1,2,3,4), y = c("f","g","c","d")
I've been working on this for a while and this is the closest I've gotten -
df1[which(df1$x %in% df2$x),]$y <- df2[which(df1$x %in% df2$x),]$y
But it just replaces the values with NA.
Does anyone know how to do this?
We can use match. :
inds <- match(df1$x, df2$x)
df1$y[!is.na(inds)] <- df2$y[na.omit(inds)]
df1
# x y
#1 1 f
#2 2 g
#3 3 c
#4 4 d
First off, well done in producing a nice reproducible example that's directly copy-pastable. That always helps, specially with an example of expected output. Nice one!
You have several options, but lets look at why your solution doesn't quite work:
First of all, I tried copy-pasting your last line into a new session and got the dreaded factor-error:
Warning message:
In `[<-.factor`(`*tmp*`, iseq, value = 1:2) :
invalid factor level, NA generated
If we look at your data frames df1 and df2 with the str function, you will see that they do not contain text but factors. These are not text - in short they represent categorical data (male vs. female, scores A, B, C, D, and F, etc.) and are really integers that have a text as label. So that could be your issue.
Running your code gives a warning because you are trying to import new factors (labels) into df1 that don't exist. And R doesn't know what to do with them, so it just inserts NA-values.
As r2evens answered, he used the stringsAsFactors to disable using strings as Factors - you can even go as far as disabling it on a session-wide basis using options(stringsAsFactors=FALSE) (and I've heard it will be disabled as default in forthcoming R4.0 - yay!).
After disabling stringsAsFactors, your code works - or does it? Try this on for size:
df2 <- df2[c(2,1),]
df1[which(df1$x %in% df2$x),]$y <- df2[which(df1$x %in% df2$x),]$y
What's in df1 now? Not quite right anymore.
In the first line, I swapped the two rows in df2 and lo and behold, the replaced values in df1 were swapped. Why is that?
Let's deconstruct your statement df2[which(df1$x %in% df2$x),]$y
Call df1$x %in% df2$x returns a logical vector (boolean) of which elements in df1$x are found ind df2 - i.e. the first two and not the second two. But it doesn't relate which positions in the first vector corresponds to which in the second.
Calling which(df1$x %in% df2$x) then reduces the logical vector to which indices were TRUE. Again, we do not now which elements correspond to which.
For solutions, I would recommend r2evans, as it doesn't rely on extra packages (although data.table or dplyr are two powerful packages to get to know).
In his solution, he uses merge to perform a "full join" which matches rows based on the value, rather than - well, what you did. With transform, he assigns new variables within the context of the data.frame returned from the merge function called in the first argument.
I think what you need here is a "merge" or "join" operation.
(I add stringsAsFactors=FALSE to the frames so that the merging and later work is without any issue, as factors can be disruptive sometimes.)
Base R:
df1 <- data.frame(x = c(1,2,3,4), y = c("a", "b", "c", "d"), stringsAsFactors = FALSE)
# df2 <- data.frame(x = c(1,2), y = c("f", "g"), stringsAsFactors = FALSE)
merge(df1, df2, by = "x", all = TRUE)
# x y.x y.y
# 1 1 a f
# 2 2 b g
# 3 3 c <NA>
# 4 4 d <NA>
transform(merge(df1, df2, by = "x", all = TRUE), y = ifelse(is.na(y.y), y.x, y.y))
# x y.x y.y y
# 1 1 a f f
# 2 2 b g g
# 3 3 c <NA> c
# 4 4 d <NA> d
transform(merge(df1, df2, by = "x", all = TRUE), y = ifelse(is.na(y.y), y.x, y.y), y.x = NULL, y.y = NULL)
# x y
# 1 1 f
# 2 2 g
# 3 3 c
# 4 4 d
Dplyr:
library(dplyr)
full_join(df1, df2, by = "x") %>%
mutate(y = coalesce(y.y, y.x)) %>%
select(-y.x, -y.y)
# x y
# 1 1 f
# 2 2 g
# 3 3 c
# 4 4 d
A join option with data.table where we join on the 'x' column, assign the values of 'y' in second dataset (i.y) to the first one with :=
library(data.table)
setDT(df1)[df2, y := i.y, on = .(x)]
NOTE: It is better to use stringsAsFactors = FALSE (in R 4.0.0 - it is by default though) or else we need to have all the levels common in both datasets
This question already has answers here:
How to delete multiple values from a vector?
(9 answers)
Closed 3 years ago.
I have a vector of values and a data frame.
I would like to filter out the rows of the data frame which contain (in specific column) any of the values in my vector.
I'm trying to figure out if a person in the survey has a child who was also questioned in the survey - if so I would like to remove them from my data frame.
I have a list of respondent IDs, and vectors of mother/father personal IDs. If the ID appears in the mother/father column I would like to remove it.
df <- data.frame(ID= c(101,102,103,104,105), Name = (Martin, Sammie, Reg, Seamus, Aine)
vec <- c(103,105,108,120,150)
Output should be a dataframe with three rows - Martin, Sammie, Seamus.
ID Name
1 101 Martin
2 102 Sammie
3 104 Seamus
df[!(df$ID %in% vec), ] # Or subset(df, !(ID %in% vec))
# ID Name
# 1 101 Martin
# 2 102 Sammie
# 4 104 Seamus
Data
df <- data.frame(ID= c(101,102,103,104,105), Name = c("Martin", "Sammie", "Reg", "Seamus", "Aine"))
vec <- c(103,105,108,120,150)
You can do this with filter from dplyr
library(tidyverse)
df2 <- df%>%
filter(!ID %in% vec)
If you create this as a data.table (and load data.table package, and fix the errors in the example data):
library(data.table)
df <- data.table(ID= c(101,102,103,104,105), Name = c("Martin", "Sammie", "Reg", "Seamus", "Aine"))
vec <- c(103,105,108,120,150)
# solution, slightly different from base R
df[!(ID %in% vec)]
Data.table is likely going to run a bit quicker than base R so very useful with large datasets. Microbenchmarking with a large dataset using base R, tidyverse and data.table shows data.table to be a bit quicker than tidyverse and a lot faster than base.
library(tidyverse)
library(data.table)
library(microbenchmark)
n <- 10000000
df <- data.frame("ID" = c(1:n), "Name" = sample(LETTERS, size = n, replace = TRUE))
dt <- data.table(df)
vec <- sample(1:n, size = n/10, replace = FALSE)
microbenchmark(dt[!(ID %in% vec)], df[!(df$ID %in% vec),], df%>% filter(!ID %in% vec))
This seems like it should be really simple. Ive 2 data frames of unequal length in R. one is simply a random subset of the larger data set. Therefore, they have the same exact data and a UniqueID that is exactly the same. What I would like to do is put an indicator say a 0 or 1 in the larger data set that says this row is in the smaller data set.
I can use which(long$UniqID %in% short$UniqID) but I can't seem to figure out how to match this indicator back to the long data set
Made same sample data.
long<-data.frame(UniqID=sample(letters[1:20],20))
short<-data.frame(UniqID=sample(letters[1:20],10))
You can use %in% without which() to get values TRUE and FALSE and then with as.numeric() convert them to 0 and 1.
long$sh<-as.numeric(long$UniqID %in% short$UniqID)
I'll use #AnandaMahto's data to illustrate another way using duplicated which also works if you've a unique ID or not.
Case 1: Has unique id column
set.seed(1)
df1 <- data.frame(ID = 1:10, A = rnorm(10), B = rnorm(10))
df2 <- df1[sample(10, 4), ]
transform(df1, indicator = 1 * duplicated(rbind(df2, df1)[, "ID",
drop=FALSE])[-seq_len(nrow(df2))])
Case 2: Has no unique id column
set.seed(1)
df1 <- data.frame(A = rnorm(10), B = rnorm(10))
df2 <- df1[sample(10, 4), ]
transform(df1, indicator = 1 * duplicated(rbind(df2, df1))[-seq_len(nrow(df2))])
The answers so far are good. However, a question was raised, "what if there wasn't a "UniqID" column?
At that point, perhaps merge can be of assistance:
Here's an example using merge and %in% where an ID is available:
set.seed(1)
df1 <- data.frame(ID = 1:10, A = rnorm(10), B = rnorm(10))
df2 <- df1[sample(10, 4), ]
temp <- merge(df1, df2, by = "ID")$ID
df1$matches <- as.integer(df1$ID %in% temp)
And, a similar example where an ID isn't available.
set.seed(1)
df1_NoID <- data.frame(A = rnorm(10), B = rnorm(10))
df2_NoID <- df1_NoID[sample(10, 4), ]
temp <- merge(df1_NoID, df2_NoID, by = "row.names")$Row.names
df1_NoID$matches <- as.integer(rownames(df1_NoID) %in% temp)
You can directly use the logical vector as a new column:
long$Indicator <- 1*(long$UniqID %in% short$UniqID)
See if this can get you started:
long <- data.frame(UniqID=sample(1:100)) #creating a long data frame
short <- data.frame(UniqID=long[sample(1:100, 30), ]) #creating a short one with the same ids.
long$indicator <- long$UniqID %in% short$UniqID #creating an indicator column in long.
> head(long)
UniqID indicator
1 87 TRUE
2 15 TRUE
3 100 TRUE
4 40 FALSE
5 89 FALSE
6 21 FALSE