subset a data.frame with multiple conditions

Question

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?

Related questions: stackoverflow.com/questions/1536590/… and stackoverflow.com/questions/2844669/… — Shane
– Shane, Commented Jun 16, 2010 at 11:09

Jyotirmoy Bhattacharya · Accepted Answer · 2010-06-16 09:29:52Z

3

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)),]

answered Jun 16, 2010 at 9:29

Jyotirmoy Bhattacharya

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Apprentice Queue · Accepted Answer · 2010-06-16 07:18:01Z

2

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']

answered Jun 16, 2010 at 7:18

Apprentice Queue

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3 Comments

Marek Over a year ago

With subset you could skip $ references, e.g.: subset(data, Analyte=="ATRAZINE" & format(Date, "%Y")=="2006"). And in your first solution as.Date is needed.

Apprentice Queue Over a year ago

as.Date isn't needed because R automatically converts it to Date.

Marek Over a year ago

My mistake. I miss in which version of R that was changed. Once I had an error in R-2.2.0 and from that moment I always used as.Date. Time to rewrite all of my code :)

Indi · Accepted Answer · 2016-03-17 07:51:55Z

0

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")

answered Mar 17, 2016 at 7:51

Indi

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