CODEWITHSUNDEEP
rsubsetdataframe
pslice
pslice

Reputation: 503

subset a data.frame with multiple conditions

Suppose my data looks like this:

2372  Kansas KS2000111 HUMBOLDT, CITY OF    ATRAZINE    1.3 05/07/2006
9104  Kansas KS2000111 HUMBOLDT, CITY OF    ATRAZINE   0.34 07/23/2006
9212  Kansas KS2000111 HUMBOLDT, CITY OF    ATRAZINE   0.33 02/11/2007
2094  Kansas KS2000111 HUMBOLDT, CITY OF    ATRAZINE    1.4 05/06/2007
16763 Kansas KS2000111 HUMBOLDT, CITY OF    ATRAZINE   0.61 05/11/2009
1076  Kansas KS2000111 HUMBOLDT, CITY OF METOLACHLOR   0.48 05/12/2002
1077  Kansas KS2000111 HUMBOLDT, CITY OF METOLACHLOR    0.3 05/07/2006

I want to be able to subset by the Analyte and a partial match on the date(namely I just want the year). I have been trying this, but I know it isn't quite right.

 data[data$Analyte=="ATRAZINE" & grep("2006",as.character(data$Date)),]

Any suggestions?

Upvotes: 3

Views: 17866

Answers (3)

Indi
Indi

Reputation: 1449

Realize this question has been asked quite some years back, hopefully should help some one in the future.

Used dplyr for sub-setting using multiple conditions, and checking the year after converting into Date type

library(dplyr)

data %>% filter( Analyte=="ATRAZINE" & format(as.Date(Date,format = "%m/%d/%Y"),"%Y") == "2006")

Upvotes: 0

Jyotirmoy Bhattacharya
Jyotirmoy Bhattacharya

Reputation: 9587

For this problem I would go with the approach in Apprentice Queue's answer of extracting the year from the date rather than doing generic string matching. I would suggest:

data[data$Analyte =="ATRAZINE"
     & as.POSIXlt(data$Date, format="%m/%d/%Y")$year == 106]

But if you really had to do regexp matching, you could use grepl which returns a logical vector rather than grep which returns a vector of indices.

data[data$Analyte=="ATRAZINE" & grepl("2006",as.character(data$Date)),]

Upvotes: 3

Apprentice Queue
Apprentice Queue

Reputation: 2036

One way using date literals:

data[data$Analyte =="ATRAZINE"
     & (data$Date >= '2006-01-01' & data$Date < '2007-01-01')]

Another way using format

data[data$Analyte =="ATRAZINE"
     & format(data$Date, "%Y") == '2006']

Upvotes: 2

Related Questions