Question

对于两个示例数据帧：

df1 <- structure(list(name = c("Katie", "Eve", "James", "Alexander", 
"Mary", "Barrie", "Harry", "Sam"), postcode = c("CB12FR", "CB12FR", 
"NE34TR", "DH34RL", "PE46YH", "IL57DS", "IP43WR", "IL45TR")), .Names = c("name", 
"postcode"), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, 
-8L), spec = structure(list(cols = structure(list(name = structure(list(), class = c("collector_character", 
"collector")), postcode = structure(list(), class = c("collector_character", 
"collector"))), .Names = c("name", "postcode")), default = structure(list(), class = c("collector_guess", 
"collector"))), .Names = c("cols", "default"), class = "col_spec"))

df2 <-structure(list(name = c("Katie", "James", "Alexander", "Lucie", 
"Mary", "Barrie", "Claire", "Harry", "Clare", "Hannah", "Rob", 
"Eve", "Sarah"), postcode = c("CB12FR", "NE34TR", "DH34RL", "DL56TH", 
"PE46YH", "IL57DS", "RE35TP", "IP43WQ", "BH35OP", "CB12FR", "DL56TH", 
"CB12FR", "IL45TR"), rating = c(1L, 1L, 1L, 2L, 3L, 1L, 4L, 2L, 
2L, 3L, 1L, 4L, 2L)), .Names = c("name", "postcode", "rating"
), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, 
-13L), spec = structure(list(cols = structure(list(name = structure(list(), class = c("collector_character", 
"collector")), postcode = structure(list(), class = c("collector_character", 
"collector")), rating = structure(list(), class = c("collector_integer", 
"collector"))), .Names = c("name", "postcode", "rating")), default = structure(list(), class = c("collector_guess", 
"collector"))), .Names = c("cols", "default"), class = "col_spec"))

我想在df1中增加一列，以提供df2中的评分。每种邮递区号可能有多个等级（这就是为什么直接合并不起作用的原因。

我只想在邮政编码和名称的前3个字符相同的情况下合并两个数据帧（假设它们在df1中是唯一的）。例如，如果有一个Katherine和Katie-（都具有相同的邮政编码），则这些不会合并

我很高兴在没有合并的地方留空白。

有什么想法吗？

Answer 1

用多列进行简单连接不会解决您的问题吗？像

SP.SOD.executeFunc('clientpeoplepicker.js', 'SPClientPeoplePicker', function() {
    var pickerDiv = $("[id^='Employee_x0020_Name'][title='Employee Name']");
    var picker = SPClientPeoplePicker.SPClientPeoplePickerDict[pickerDiv[0].id];
    picker.OnUserResolvedClientScript = function(peoplePickerId, selectedUsersInfo) {
 //It will get the desired display name of the people from picker div, similarly any other property can be accessed via selectedUsersInfo
        var empname = selectedUsersInfo[0].DisplayText;
        console.log(empname);
    }
});

如果列名不匹配的替代解决方案，

df<-merge(x=df1,y=df2,by=c('name','postcode'),all.x=T)

根据R中的匹配字符进行合并

1 个答案: