CODEWITHSUNDEEP
rdatedataframe
Majid
Majid

Reputation: 1844

Retrieve values of the data.frame by matching ID and column name

I have a dataframe named df1 which has four columns (i.e. id, s, date and value). The value column is empty and I want to fill it using a second dataframe that is named df2. df2 is filled with id column and many other columns that are named using dates which they belong. All I need is to find corresponding values of df1$value in df2, where both dates and id numbers are matching.

Example data:

set.seed(123)

#df1
df1 <- data.frame(id = 1:100, 
                      s = runif(100,100,1000), 
                      date = sample(seq(as.Date('1999/01/01'), as.Date('2001/01/01'), by="day"), 100), 
                      value = NA)

#df2
df2 <- data.frame(matrix(runif(80000,1,100), ncol=800, nrow=100))[-1]
    names(df2) <- seq(as.Date("1999-01-01"),as.Date("2002-12-31"),1)[c(1:799)]  
    df2 <- cbind(id =  1:100, df2)

Upvotes: 3

Views: 1326

Answers (3)

akrun
akrun

Reputation: 886948

We can use efficient way with data.table join. It should be fast for big datasets

library(data.table)
setDT(df1)[melt(setDT(df2), id.var = 'id')[, 
       date := as.IDate(variable, '%Y-%m-%d')], on = .(id, date)]

Upvotes: 3

Harshal Gajare
Harshal Gajare

Reputation: 615

You can also use melt and then left join using both the keys:

library(dplyr)
library(reshape2)
set.seed(123)

#df1
df1 <- data.frame(id = 1:100, 
                  s = runif(100,100,1000), 
                  date = sample(seq(as.Date('1999/01/01'), as.Date('2001/01/01'), by="day"), 100), 
                  value = NA)
#df2
df2 <- data.frame(matrix(runif(80000,1,100), ncol=800, nrow=100))[-1]
names(df2) <- seq(as.Date("1999-01-01"),as.Date("2002-12-31"),1)[c(1:799)]  
df2 <- cbind(id =  1:100, df2)

df2<-melt(df2, id.vars = "id", value.name = "Value", variable.name = "date")

df2$date<-as.Date(df2$date, format = "%Y-%m-%d")
df1<-left_join(df1, df2, by = c("id", "date"))

head(df1)
  id        s       date value    Value
1  1 358.8198 2000-03-15    NA 48.31799
2  2 809.4746 1999-09-01    NA 62.15760
3  3 468.0792 1999-12-23    NA 16.41291
4  4 894.7157 2000-11-26    NA 32.70024
5  5 946.4206 1999-12-18    NA  5.83607
6  6 141.0008 2000-10-09    NA 74.64832

Upvotes: 4

Ronak Shah
Ronak Shah

Reputation: 388817

One way is to convert df2 into long format using gather and then do left_join

library(dplyr)
library(tidyr)

df1 %>%
  left_join(df2 %>% 
             gather(date, value, -id) %>% 
              mutate(date = as.Date(date)), by = c("id", "date"))

#     id   s       date value
#1     1 359 2000-03-15 48.32
#2     2 809 1999-09-01 62.16
#3     3 468 1999-12-23 16.41
#4     4 895 2000-11-26 32.70
#5     5 946 1999-12-18  5.84
#6     6 141 2000-10-09 74.65
#7     7 575 2000-10-25  9.22
#8     8 903 2000-03-17  6.46
#9     9 596 1999-10-25 73.48
#10   10 511 1999-04-17 62.43
#...

data

set.seed(123)
df1 <- data.frame(id = 1:100, 
              s = runif(100,100,1000), 
 date = sample(seq(as.Date('1999/01/01'), as.Date('2001/01/01'), by="day"), 100))


df2 <- data.frame(matrix(runif(80000,1,100), ncol=800, nrow=100))[-1]
names(df2) <- seq(as.Date("1999-01-01"),as.Date("2002-12-31"),1)[c(1:799)]  
df2 <- cbind(id =  1:100, df2)

Upvotes: 6

Related Questions