subset a data.frame with multiple conditions

偶尔善良 提交于 2019-12-17 19:29:24

问题


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?


回答1:


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)),]



回答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']



回答3:


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") 


来源:https://stackoverflow.com/questions/3051279/subset-a-data-frame-with-multiple-conditions

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