r - 方括号多列 R

Question

我很困惑。我试图根据两列中的值隔离 df 的某些行。一如既往，我首先在实践数据中尝试这一点。我的代码工作正常。

data1<-df2[df2$fruit=="kiwi" |  df2$fruit=="orange" | df2$fruit=="apple"  & (df2$dates>= "2010-04-01" & df2$dates<  "2010-10-01"), ]

当我在真实数据上尝试相同的代码时，它不起作用。它收集我需要的“水果”，但忽略我的日期范围请求。

 data1<-lti_first[lti_first$hai_atc=="C10AA01" | lti_first$hai_atc=="C10AA03" | lti_first$hai_atc=="C10AA04" | lti_first$hai_atc=="C10AA05" | lti_first$hai_atc=="C10AA07" | lti_first$hai_atc=="C10AB02" |lti_first$hai_atc=="C10AA04" |lti_first$hai_atc=="C10AB08" | lti_first$hai_atc=="C10AX09" & (lti_first$date_of_claim >= "2010-04-01" & lti_first$date_of_claim<"2010-10-01"), ]

我的实践数据和真实数据中的变量结构完全相同。 Fruits/hai_atc 是两个 dfs 中的因子，日期是两个 dfs 中的 as.Dates。

为了解决这个问题，我尝试对数据进行子集化，但这对我来说也不起作用(但对练习数据有效)

x<-subset(lti_first, hai_atc=="V07AY03" | hai_atc=="A11JC94" & (date_of_claim>="2010-04-01" & date_of_claim<"2010-10-01"))

我做错了什么？对我来说，我的代码看起来是一样的!

示例 df

names<-c("tom", "mary", "tom", "john", "mary",
 "tom", "john", "mary", "john", "mary", "tom", "mary", "john", "john")
dates<-as.Date(c("2010-02-01", "2010-05-01", "2010-03-01", 
"2010-07-01", "2010-07-01", "2010-06-01", "2010-09-01",
 "2010-07-01", "2010-11-01", "2010-09-01", "2010-08-01", 
"2010-11-01", "2010-12-01", "2011-01-01"))
fruit<-as.character(c("apple", "orange", "banana", "kiwi",
 "apple", "apple", "apple", "orange", "banana", "apple",
 "kiwi", "apple", "orange", "apple"))
age<-as.numeric(c(60,55,60,57,55,60,57,55,57,55,60,55, 57,57))
sex<-as.character(c("m","f","m","m","f","m","m",
 "f","m","f","m","f","m", "m"))
df2<-data.frame(names,dates, age, sex, fruit)
df2


dput(df2)
structure(list(names = structure(c(3L, 2L, 3L, 1L, 2L, 3L, 1L, 
2L, 1L, 2L, 3L, 2L, 1L, 1L), .Label = c("john", "mary", "tom"
), class = "factor"), dates = structure(c(14641, 14730, 14669, 
14791, 14791, 14761, 14853, 14791, 14914, 14853, 14822, 14914, 
14944, 14975), class = "Date"), age = c(60, 55, 60, 57, 55, 60, 
57, 55, 57, 55, 60, 55, 57, 57), sex = structure(c(2L, 1L, 2L, 
2L, 1L, 2L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 2L), .Label = c("f", 
"m"), class = "factor"), fruit = structure(c(1L, 4L, 2L, 3L, 
1L, 1L, 1L, 4L, 2L, 1L, 3L, 1L, 4L, 1L), .Label = c("apple", 
"banana", "kiwi", "orange"), class = "factor")), .Names = c("names", 
"dates", "age", "sex", "fruit"), row.names = c(NA, -14L), class = "data.frame")

**真实数据太大而无法放入 dput，这里是一个 str

str(sample_lti_first)
'data.frame':   20 obs. of  5 variables:
 $ hai_dispense_number: Factor w/ 53485 levels "Patient HAI0000017",..: 22260 22260 2527 24311 24311 24311 24311 13674 13674 13674 ...
 $ sex                : Factor w/ 4 levels "F","M","U","X": 2 2 2 1 1 1 1 1 1 1 ...
 $ hai_age            : int  18 18 27 40 40 40 40 28 28 28 ...
 $ date_of_claim      : Date, format: "2009-10-09" "2009-10-09" "2009-10-18" ...
 $ hai_atc            : Factor w/ 1038 levels "","A01AA01","A01AB03",..: 144 76 859 80 1009 1009 859 81 1008 859 ...

Answer 1

我认为扩展@Aaron 的评论很重要。您遇到的问题是由于使用 %in% 的所有 OR 语句周围缺少括号引起的。避免，并不是 OR 语句在提取函数中不起作用 [ 。您的玩具示例实际上并没有完全按照您想要的方式工作 - 有一个 orange带有日期的水果2010-12-01 。其他问题没有出现只是偶然。

这段代码中 bool 逻辑的读取方式

df2[df2$fruit=="kiwi" |  df2$fruit=="orange" | df2$fruit=="apple"  & (df2$dates>= "2010-04-01" & df2$dates<  "2010-10-01"), ]

是:

I want all rows of df2 where fruit is kiwi, all rows where fruit is orange, and all rows where fruit is apple and the dates are between 3/31/2010 and 10/1/2010.

这就是您得到的结果 - 只有苹果被截断到适当的日期范围。实际上，玩具数据集中的日期范围之外没有猕猴桃。

现在添加一对括号:

df2[(df2$fruit=="kiwi" |  df2$fruit=="orange" | df2$fruit=="apple")  & (df2$dates >= "2010-04-01" & df2$dates <  "2010-10-01"), ]

这段代码说:

I want all the rows of df2 where fruit is either kiwi, orange, or apple and the dates are between 3/31/2010 and 10/1/2010.

话虽如此，%in%绝对是一条出路。