[英]Using nested function with lapply
此代碼有效(耗時數小時和數秒,僅轉換為秒):
library(lubridate)
library(tidyverse)
original_date_time<-"2018-01-3111:59:59"
period_to_seconds(hms(paste(hour(original_date_time), minute(original_date_time),second(original_date_time), sep = ":")))
我有這個小標題:
df<-data.frame("id"=c(1,2,3,4,5), "Time"=c("1999-12-31 10:10:10","1999-12-31 09:05:13","1999-12-31 00:05:25","1999-12-31 07:04","1999-12-31 03:05:07"))
tib<-as_tibble(df)
tib
結果:
# A tibble: 5 x 2
id Time
<dbl> <fct>
1 1 1999-12-31 10:10:10
2 2 1999-12-31 09:05:13
3 3 1999-12-31 00:05:25
4 4 1999-12-31 07:04
5 5 1999-12-31 03:05:07
現在,我想將更改時間的代碼應用於tib$Time
每個單元格。 我嘗試過:
time_converted_data_<-lapply(tib$Time, period_to_seconds(hms(paste(hour(tib$Time), minute(tib$Time),second(tib$Time), sep = ":"))))
但這給了我錯誤:
Error in match.fun(FUN) :
c("'period_to_seconds(hms(paste(hour(tib$Time), minute(tib$Time), ' is not a function, character or symbol", "' second(tib$Time), sep = \":\")))' is not a function, character or symbol")
如何解決? 我想要R基本版本和tidyverse版本。
基數R
您的功能已向量化。 所以,你可以做
period_to_seconds(hms(paste(hour(tib$Time), minute(tib$Time),second(tib$Time), sep = ":")))
#[1] 36600 32700 300 25440 11100
對於非矢量化函數,您可以嘗試類似
foo = function(x){
period_to_seconds(hms(paste(hour(x), minute(x),second(x), sep = ":")))
}
lapply(tib$Time, foo)
#[[1]]
#[1] 36610
#[[2]]
#[1] 32713
#[[3]]
#[1] 325
#[[4]]
#[1] 25440
#[[5]]
#[1] 11107
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