我有两个具有几乎相同数据的数据:
Test.Takers,有29260条观测和以下列名:
Paternal.Name,Maternal.Name,First.Name,Application.Number
和Every.Student.In.The.Country,有12000000个观察到的列名:
Paternal.Name,Maternal.Name,First.Name,Application.Number
Test.Takers$Application.Number中填充了NA值,我想用Every.Student.In.The.Country.中的Application.Numbers填充该字段。
我试图通过从Every.Student.In.The.Country.中删除Paternal.Names和Maternal.Names来做到这一点。然后,我将使用以下代码填充Test.Takers$Application.Number:
Test.Takers$Application.Number[i] <- subset$Application.Number[pmatch(as.character(Test.Taker$First.Name[i]), subset$First.Names)]这可以填补约2/3的Test.Takers$Application.Number。在试图找出为什么如此多的Every.Student.In.The.Country$First.Name仍然是NA之后,我发现来自该Test.Takers$Application.Number的一些名称中包含一个'#‘。我认为“#”抛出了pmatch函数,这样来自Test.Takers$First.Name的名称(如'TERESA DEL CA‘)与来自Every.Student.In.The.Country$First.Name的名称(如'TERESA DEL#CARMEN’)不匹配。
任何关于如何解决这个问题的建议都是很棒的。我觉得某种类似regex函数的东西可能会有所帮助,但我不太确定。
编辑:下面是一些复制问题的示例代码。请记住,我所处理的真实数据相当大--大约有30000和12000000个观测结果。如果你仔细看了这段代码,发现没有效率,请告诉我。
Test.Takers <- data.frame(
Paternal.Name = c('Last', 'Last', 'Paternal'),
Maternal.Name = c('Maternal', 'Last', 'Last'),
First.Name = c('First', 'Name', 'TERESA DEL CA'),
Application.Number = NA)
Every.Student.In.The.Country <- data.frame(
Paternal.Name = c('Last', 'Last', 'Last', 'Paternal', 'Paternal', 'Paternal'),
Maternal.Name = c('Maternal', 'Last', 'Maternal', 'Last', 'Maternal', 'Last'),
First.Name = c('First', 'Name', 'Whatever', 'TERESA DEL#CARMEN', 'Another', 'Something Else'),
Application.Number = c(123, 456, 789, 234, 567, 890)
)
#a place holder that will hold a subset of all a selected paternal last names
indexp <- data.frame(Paternal.Name='name')
for(i in 1:nrow(Test.Takers)) {
namep <- as.character(Test.Takers$Paternal.Name[i])
#below if statement prevents us from having to subset the paternal lastname unnecessarily
if(is.na(indexp$Paternal.Name[1]) == T | as.character(indexp$Paternal.Name[1]) != namep) {
indexp <- subset(Every.Student.In.The.Country, Paternal.Name == as.character(Test.Takers$Paternal.Name[i]))
}
#below if-statement prevents an error that arrises
#when a paternal last name does not exist
#in the Every.Student.In.The.Country file
if(is.na(indexp$Paternal.Name[1]) == F) {
#group paternal last names by maternal last names
indexm <- subset(indexp, Maternal.Name == as.character(Test.Takers$Maternal.Name[i]))
#find a partial string match to find an exact or similiar first name within the selected
#last name subset. Attaches a application.number if a match is found
Test.Takers$Application.Number[i] <- indexm$Application.Number[pmatch(as.character(Test.Takers$First.Name[i]), indexm$First.Name)]
}}发布于 2016-06-21 18:58:28
如果#符号是您可以在函数duplicates.ok=TRUE中添加Test.Takers$Application.Number[i] <- subset$Application.Number[pmatch(as.character(Test.Taker$First.Name[i]), subset$First.Names),dup=T]的唯一问题
或者您可以删除#符号Test.Takers$Application.Number[i] <- subset$Application.Number[pmatch(as.character(Test.Taker$First.Name[i]),gsub("#"," ",subset$First.Names))]
https://stackoverflow.com/questions/37951193
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