grades <- read.table("studentgrades.csv",header = TRUE,row.names="StudentID", sep = ",")
gradess <- read.csv("studentgrades.csv",header = TRUE,row.names="StudentID", sep = ",")read.table的结果是:
grades
[1] First Last Math Science Social.Studies
<0 rows> (or 0-length row.names)read.csv的结果是:
gradess
First Last Math Science Social.Studies
11 Bob Smith 90 80 67
12 Jane Weary 75 NA 80
10 Dan "Thornton" 65 75 70
40 Mary O'Leary 90 95 92我只是不知道为什么read.tables不能给我正确的结果。
发布于 2016-10-19 16:09:30
该问题是由于O'Leary of last name列中的quote (')引起的。您需要更改read.table中的默认quote选项,该选项在默认情况下设置为('),以获得所需的结果。
如果您在read.table中使用quote=NULL,如下所示
grades <- read.table("studentgrades.csv",header = TRUE,sep=",",quote=NULL,row.names="StudentID")然后你就会得到想要的结果。
> grades
First Last Math Science Social.Studies
11 Bob Smith 90 80 67
12 Jane Weary 75 NA 80
10 Dan "Thornton" 65 75 70
40 Mary O'Leary 90 95 92https://stackoverflow.com/questions/40122168
复制相似问题