英文:
Filter non-NAs on numeric column name
问题
我正在尝试筛选包含数值/日期名称的列，使用as.Date变量。
例如，考虑这个小数据库：
dt <- data.table(names = c("A", "B", "C"),`2020-01-01` = c(1, NA, 2),`2020-01-02` = c(3, 4, 5),`2020-01-03` = c(6, 7, 8))
我当前正在按以下方式筛选所需的日期列：
dt1 <- dt %>% filter(!is.na(`2020-01-01`)) %>% select(names)
我的想法是创建一个meeting_date变量，这个变量应该作为我的所有R代码的日期参考。
meeting_date <- as.Date("2020-01-01")
但是当然，以下代码不起作用：
dt1 <- dt %>% filter(!is.na(meeting_date)) %>% select(names)
这是因为缺少反引号，所以我尝试了以下代码但未成功：
dt1 <- dt %>% filter(!is.na(paste("`", meeting_date, "`", sep=""))) %>% select(names)dt1 <- dt %>% filter(!is.na(noquote(paste("`", meeting_date, "`", sep="")))) %>% select(names)
有人知道如何继续吗？谢谢！
英文:
I am trying to filter a column which contains a numeric/date name using a as.Date variable.
As an example, consider this small database:
dt <- data.table(names = c("A", "B", "C"),`2020-01-01` = c(1, NA, 2),`2020-01-02` = c(3, 4, 5),`2020-01-03` = c(6, 7, 8))
I am currently filtering the desired date column as follows:
dt1 <- dt %>% filter(!is.na(`2020-01-01`)) %>% select(names)
My idea is to create a meeting_date variable, this variable should be used as a date reference for all my R code.
meeting_date <- as.Date("2020-01-01")
But of course the code:
dt1 <- dt %>% filter(!is.na(meeting_date)) %>% select(names)
Does not work. The reason for this is the missing backticks, so without success I tried the following codes:
dt1 <- dt %>% filter(!is.na(paste("`", meeting_date, "`", sep=""))) %>% select(names)dt1 <- dt %>% filter(!is.na(noquote(paste("`", meeting_date, "`", sep="")))) %>% select(names)
Does anyone knows how to proceed? Thanks!
答案1
得分: 2
可以这样做：
meeting_date <- as.Date("2020-01-01")dt %>%filter_at(vars(one_of(as.character(meeting_date))), ~ !is.na(.))names 2020-01-01 2020-01-02 2020-01-031 A 1 3 62 C 2 5 8
英文:
You can do:
meeting_date <- as.Date("2020-01-01")dt %>%filter_at(vars(one_of(as.character(meeting_date))), ~ !is.na(.))names 2020-01-01 2020-01-02 2020-01-031 A 1 3 62 C 2 5 8
答案2
得分: 0
你可以使用以下的代码来使用 subset + is.na：
meeting_date <- "2020-01-01"dtout <- subset(dt, as.vector(!is.na(dt[, ..meeting_date])))
这将得到如下的结果：
> dtoutnames 2020-01-01 2020-01-02 2020-01-031: A 1 3 62: C 2 5 8
英文:
You can use subset + is.naas below
meeting_date <- "2020-01-01"dtout <- subset(dt,as.vector(!is.na(dt[, ..meeting_date])))
such that
> dtoutnames 2020-01-01 2020-01-02 2020-01-031: A 1 3 62: C 2 5 8
答案3
得分: 0
长数据应该更容易处理：
library(data.table)dt <- data.table(names = c("A", "B", "C"),`2020-01-01` = c(1, NA, 2),`2020-01-02` = c(3, 4, 5),`2020-01-03` = c(6, 7, 8))#将数据转换为“长”格式并将新的“name”列更改为日期#在此过程中更改令人困惑的列“name”为日期。dt_long <- dt %>% pivot_longer(-names) %>%mutate(date = lubridate::ymd(name)) %>%select(-name)meeting_date <- as.Date("2020-01-01")dt_long %>% filter(date == meeting_date & (!is.na(value)))
英文:
Long data should be easier to work with:
library(data.table)dt <- data.table(names = c("A", "B", "C"),`2020-01-01` = c(1, NA, 2),`2020-01-02` = c(3, 4, 5),`2020-01-03` = c(6, 7, 8))#Make data 'long' & change the new 'name' column to dates# change confusing column 'name' to date while we're at it.dt_long <- dt %>% pivot_longer(-names) %>%mutate(date = lubridate::ymd(name)) %>%select(-name)meeting_date <- as.Date("2020-01-01")dt_long %>% filter(date == meeting_date & (!is.na(value)))
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