[英]column full of NA using lapply in a data.table R
I have a problem using lapply
in a data.table. 我在lapply
中使用lapply
有问题。 Here are two examples: 这是两个示例:
library(data.table)
library(lubridate)
test <- function(x)
{
if(is.na(x)) return(NA)
if(x=="") return(NA)
if(substr(x,3,3)=="/") return(as_date(x,"%d/%m/%Y"))
return(2)
}
x1<-data.table(v1=c("","07/06/2016","",NA), v2=c("2004-06-18","","2004-06-18","2004-06-18"))
x1[,lapply(.SD,test)]
x2<-data.table(v1=c("2004-06-19","2004-06-18","",NA),v2=c("2004-06-18","","2004-06-18","2004-06-18"))
x2[,lapply(.SD,test)]
In the first example, the first column after the lapply
is full of NA
, but I wanted to obtain is NA, 2016-06-07, NA, NA
. 在第一个示例中, lapply
之后的第一列充满了NA
,但我想获取的是NA, 2016-06-07, NA, NA
。
In the second example, the last two rows of the first column are wrong, because each row contains 2 but in my opinion should contains NA
. 在第二个示例中,第一列的最后两行是错误的,因为每行包含2,但我认为应该包含NA
。
I don't understand how R considers the NA
here. 我不明白R在这里如何考虑NA
。 What do I miss to get what I want? 我想得到我想要的东西吗?
After a lot of tries, the answer is that data.table
considers columns as variables, and .SD
is a list whose elements are the columns as variables, and so when applying a function, as test here, this function must take as argument a list. 经过大量尝试,答案是data.table
将列视为变量,而.SD
是一个列表,其元素是列作为变量,因此在应用函数时(如此处测试),该函数必须将a作为参数清单。
Here is what you should change: 这是您应该更改的内容:
testList <- function(x)
{
lapply(x,test)
}
x1[,lapply(.SD,testList)]
If someone knows about another solution, please don't hesitate to share. 如果有人知道其他解决方案,请随时分享。
First, I can't run your example without throwing an error. 首先,我不能在没有抛出错误的情况下运行您的示例。 The second columns of your data.tables are of class "Date", but the ""
entry isn't a date. data.tables的第二列属于“日期”类,但""
项不是日期。 When it prints it's formatted to look like NA
. 打印时,其格式看起来像NA
。 Try running is.na(x1$v2[2])
and x1$v2[2] == ""
. 尝试运行is.na(x1$v2[2])
和x1$v2[2] == ""
。
Also, it looks like you have a problem with vectorization. 同样,您似乎在向量化方面遇到了问题。
Try running test(x1$v1)
. 尝试运行test(x1$v1)
。 Pay attention to the warning messages. 请注意警告消息。 is.na(x)
returns a logical vector, but if
only uses the first element in the vector. is.na(x)
返回逻辑向量,但是if
仅使用向量中的第一个元素。
In addition: Warning message:
In if (is.na(x)) return(NA) :
the condition has length > 1 and only the first element will be used
You might be able to fix it by applying to each row: 您可能可以通过应用到每一行来修复它:
x1[, lapply(.SD, test), by = 1:nrow(x1)]
Otherwise you'll need to modify your test
function to accept a vector of strings and return a vector of results. 否则,您将需要修改test
函数以接受字符串向量并返回结果向量。 But you should really consider returning a vector of a single type. 但是,您实际上应该考虑返回单一类型的向量。
Finally, I don't understand the purpose of lubridate
in this example. 最后,在此示例中,我不了解lubridate
的目的。 Why not use as.Date(x,"%d/%m/%Y")
. 为什么不使用as.Date(x,"%d/%m/%Y")
。 What do you gain from as_date
? 您从as_date
获得什么?
You can rewrite your function to work on vectors: 您可以重写函数以处理矢量:
test <- function(x)
{
ans <- rep.int(2, length(x))
ans[is.na(x) | x == ""] <- NA
dates <- grepl('../', x)
ans[dates] <- as_date(x[dates], "%d/%m/%Y")
return(ans)
}
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