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R错误: Format=NA
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Stack Overflow用户
提问于 2015-11-11 04:25:00
回答 1查看 55关注 0票数 0

我试图将我的时间数据分配为15分钟的离散间隔。但是,它在最小值上显示了错误。有人知道这个吗?

THis是我的数据

代码语言:javascript
复制
"10-28-15","00:04:13","13-01","1"
"10-28-15","00:04:16","13-10","1"
"10-28-15","00:04:30","13-11","1"
"10-28-15","11:59:44","13-12","1"
"10-28-15","00:04:48","13-13","1"
"10-28-15","00:04:50","13-14","1"
"10-28-15","00:04:57","13-15","0"
"10-28-15","00:37:57","13-17","0"
"10-28-15","00:45:04","13-17","0"
"10-28-15","00:13:04","13-17","0"

这是我的R剧本:

代码语言:javascript
复制
sample<-read.csv("C:\\Users\\Toshiba\\Desktop\\Parkeasy\\Sunway\\parkeasy.csv",header=FALSE,sep=",")
sample$int<- strptime(paste(sample$V1,sample$V2),format="%m-%d-%y %H:%M:%S")
min_V2<-trunc(min(strptime("2015-10-28 00:00:01", "%d-%m-%y %H:%M:%S"),"min")
max_V2<-trunc(min(strptime("2015-10-28 23:59:59", "%d-%m-%y %H:%M:%S")),"min") + 900
out <- cut(sample$int, breaks = seq(min_V2, max_V2, by = "15 min"))

执行min_V2时显示错误

代码语言:javascript
复制
min_V2
POSIXlt[1:1] Format:NA

seq.int(0,to0 - from,by)中的错误:“to”不能是NA、NaN或无穷大

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回答 1

Stack Overflow用户

发布于 2015-11-11 04:39:03

您的日期格式不匹配。

代码语言:javascript
复制
min_V2<-trunc(min(strptime("2015-10-28 00:00:01", "%d-%m-%y %H:%M:%S"),"min")
max_V2<-trunc(min(strptime("2015-10-28 23:59:59", "%d-%m-%y %H:%M:%S")),"min") + 900

%d-%m-%y %H:%M:%S不同,您希望%Y-%m-%d %H:%M:%S匹配您的日期字符串。

票数 2
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/33644260

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