I'm trying to create a dummy variable based on if df1 is contained within df2. Note that df2 has columns more than just the columns in df1.
e.g.:
df1:
A
B
C
1
2
3
4
5
6
7
8
0
df2:
A
B
C
D
1
2
3
E
4
5
6
F
7
8
9
G
Resulting in:
df2:
A
B
C
D
Dummy
1
2
3
E
1
4
5
6
F
1
7
8
9
G
0
Any good approaches I should consider?
I've tried using an ifelse function applied to the dataframe, but I suspect I've coded it wrong. Any tips would be appreciated!
One approach would be to add a column called "dummy" to df1, then join with df2 on all variables of df1.
df1$dummy <- 1
library(dplyr)
dplyr::left_join(df2, df1) %>%
mutate(dummy = ifelse(is.na(dummy), 0, dummy))
# Joining, by = c("A", "B", "C")
# A B C D dummy
# 1 2 3 E 1
# 4 5 6 F 1
# 7 8 9 G 0
By default left_join joins using all commonly named variables, but this can be modified as required.
Related
I am writing a code for analysis a set of dplyr data.
here is how my table_1 looks:
1 A B C
2 5 2 3
3 9 4 1
4 6 3 8
5 3 7 3
And my table_2 looks like this:
1 D E F
2 2 9 3
I would love to based on table 1 column"A", if A>6, then create a column "G" in table1, equals to "C*D+C*E"
Basically, it's like make table 2 as a factor...
Is there any way I can do it?
I can apply a filter to Column "A" and multiply Column"C" with a set number instead of a factor from table_2
table_1_New <- mutate(Table_1,G=if_else(A<6,C*2+C*9))
You could try
#Initialize G column with 0
df1$G <- 0
#Get index where A value is greater than 6
inds <- df1$A > 6
#Multiply those values with D and E from df2
df1$G[inds] <- df1$C[inds] * df2$D + df1$C[inds] * df2$E
df1
# A B C G
#2 5 2 3 0
#3 9 4 1 11
#4 6 3 8 0
#5 3 7 3 0
Using dplyr, we can do
df1 %>% mutate(G = ifelse(A > 6, C*df2$D + C*df2$E, 0))
This question already has answers here:
Adding values in two data.tables
(2 answers)
Closed 5 years ago.
I've 2 different data.tables. I need to merge and sum based on a row values. The examples of two tables are given as Input below and expected output shown below.
Input
Table 1
X A B
A 3
B 4 6
C 5
D 9 12
Table 2
X A B
A 1 5
B 6 8
C 7 14
D 5
E 1 1
F 2 3
G 5 6
Expected Output:
X A B
A 4 5
B 10 14
C 12 14
D 14 12
E 1 1
F 2 3
G 5 6
We can do this by rbinding the two tables and then do a group by sum
library(data.table)
rbindlist(list(df1, df2))[, lapply(.SD, sum, na.rm = TRUE), by = X]
# X A B
#1: A 4 5
#2: B 10 14
#3: C 12 14
#4: D 14 12
#5: E 1 1
#6: F 2 3
#7: G 5 6
Or using a similar approach with dplyr
library(dplyr)
bind_rows(df1, df2) %>%
group_by(X) %>%
summarise_all(funs(sum(., na.rm = TRUE)))
Note: Here, we assume that the blanks are NA and the 'A' and 'B' columns are numeric/integer class
Merge your tables together first, then do the sum. If you later want to drop the individual values you can do so easily.
out <- merge(df1, df2, by.x="X", by.y="X", all.x=T, all.y=T)
out$sum <- rowSums(out[2:3])
out$A <- out$B <- NULL # drop original values
Below code will help you to do required job for all numeric columns at once
library(dplyr)
Table = Table1 %>% full_join(Table2) %>%
group_by(X) %>% summarise_all(funs(sum(.,na.rm = T)))
I have a dataset with several variables, but I want to keep the rows that are the same based on two columns. Here is an example of what I want to do:
a <- c(rep('A',3), rep('B', 3), rep('C',3))
b <- c(1,1,2,4,4,4,5,5,5)
df <- data.frame(a,b)
a b
1 A 1
2 A 1
3 A 2
4 B 4
5 B 4
6 B 4
7 C 5
8 C 5
9 C 5
I know that if I use the duplicated function I can get:
df[!duplicated(df),]
a b
1 A 1
3 A 2
4 B 4
7 C 5
But since the level 'A' on column a does not have a unique value in b, I want to drop both observations to get a new data.frame as this:
a b
4 B 4
7 C 5
I don't mind to have repeated values across b, as long as for every same level on a there is the same value in b.
Is there a way to do this? Thanks!
This one maybe?
ag <- aggregate(b~a, df, unique)
ag[lengths(ag$b)==1,]
# a b
#2 B 4
#3 C 5
Maybe something like this:
> ind <- apply(sapply(with(df, split(b,a)), diff), 2, function(x) all(x==0) )
> out <- df[!duplicated(df),]
> out[out$a %in% names(ind)[ind], ]
a b
4 B 4
7 C 5
Here is another option with data.table
library(data.table)
setDT(df)[, if(uniqueN(b)==1) .SD[1L], by = a]
# a b
#1: B 4
#2: C 5
For example, now I get the table
A B C
A 0 4 1
B 2 1 3
C 5 9 6
I like to order the columns and rows by my own defined order, to achieve
B A C
B 1 2 3
A 4 0 1
C 9 5 6
This can be accomplished in base R. First we make the example data:
# make example data
df.text <- 'A B C
0 4 1
2 1 3
5 9 6'
df <- read.table(text = df.text, header = T)
rownames(df) <- LETTERS[1:3]
A B C
A 0 4 1
B 2 1 3
C 5 9 6
Then we simply re-order the columns and rows using a vector of named indices:
# re-order data
defined.order <- c('B', 'A', 'C')
df <- df[, defined.order]
df <- df[defined.order, ]
B A C
B 1 2 3
A 4 0 1
C 9 5 6
If the defined order is given as
defined_order <- c("B", "A", "C")
and the initial table is created by
library(data.table)
# create data first
dt <- fread("
id A B C
A 0 4 1
B 2 1 3
C 5 9 6")
# note that row names are added as own id column
then you could achieve the desired result using data.table as follows:
# change column order
setcolorder(dt, c("id", defined_order))
# change row order
dt[order(defined_order)]
# id B A C
# 1: B 1 2 3
# 2: A 4 0 1
# 3: C 9 5 6
I need to find a way to sum columns by their index,I'm working on a bigread.csv file, I'll show here a sample of the problem; I'd like for example to sum from the 2nd to the 5th and from the 6th to the 7h the following matrix:
a 1 3 3 4 5 6
b 2 1 4 3 4 1
c 1 3 2 1 1 5
d 2 2 4 3 1 3
The result has to be like this:
a 11 11
b 10 5
c 7 6
d 8 4
The columns have all different names
We can use rowSums on the subset of columns i.e 2:5 and 6:7 separately and then create a new data.frame with the output.
data.frame(df1[1], Sum1=rowSums(df1[2:5]), Sum2=rowSums(df1[6:7]))
# id Sum1 Sum2
#1 a 11 11
#2 b 10 5
#3 c 7 6
#4 d 11 4
The package dplyr has a function exactly made for that purpose:
require(dplyr)
df1 = data.frame(a=c(1,2,3,4,3,3),b=c(1,2,3,2,1,2),c=c(1,2,3,21,2,3))
df2 = df1 %>% transmute(sum1 = a+b , sum2 = b+c)
df2 = df1 %>% transmute(sum1 = .[[1]]+.[[2]], sum2 = .[[2]]+.[[3]])