对具有多个条件的数据框进行子集 [英] subset a data.frame with multiple conditions

问题描述

假设我的数据如下所示：

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

我想要能够通过Analyte进行子集，并且部分匹配日期（即我只想要一年）。我一直在尝试这个，但我知道这是不对的。

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)),]

任何建议？

推荐答案

对于这个问题，我将在Apprentice Queue的解答中提取年份的答案，而不是使用通用字符串匹配。我建议：

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]

但是，如果你真的要做正则表达式匹配，你可以使用 grepl 返回一个逻辑向量，而不是 grep 它返回一个索引向量。

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)),]

这篇关于对具有多个条件的数据框进行子集的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！