Question

我很沮丧。我试图根据两列中的值来隔离某些df行。像往常一样，我首先在练习数据中尝试这个。我的代码运行正常。

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

当我在我的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”，但忽略了我的日期范围请求。

 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"), ]

我的练习数据和实际数据中的变量结构是精确相同。 Fruits / hai_atc是dfs中的两个因子，dfs中的日期是as.Dates。

为了解决这个问题，我尝试对我的数据进行子集化，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，我的代码看起来完全相同！

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")

**实际数据太大而无法输入，这里是str而不是

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的评论非常重要。您遇到的问题是由于避免使用%in%的所有OR语句缺少括号，而OR语句在提取函数[中不起作用。你的玩具示例实际上并没有按照你想要的方式工作 - 有一个orange水果的日期为2010-12-01。没有出现其他问题只是机会。

在此代码中读取布尔逻辑的方法

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

是：

我想要所有df2行，其中水果是猕猴桃，所有行都是水果橙色，以及水果是苹果和日期的所有行 2010年3月31日至10/1/2010之间。

这就是你得到的 - 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"), ]

此代码说：

我想要df2的所有行，其中水果是猕猴桃，橙子或苹果日期是2010年3月31日至10/1/2010之间。

尽管如此，%in%绝对是可行的方式。

Answer 2

这有用吗？

data1 <- subset(lti_first,
  (hai_atc %in% c("C10AA01", "C10AA03", "C10AA04", "C10AA05", "C10AA07",
                  "C10AB02", "C10AA04", "C10AB08", "C10AX09")) & 
  (date_of_claim >= as.Date("2010-04-01") & date_of_claim < as.Date("2010-10-01")))

请注意使用%in%和as.Date。

方括号多列R.

2 个答案: