Question

我有3个数据帧。

Data1 -
Name_description   Numbers 
ABC                23
DEF                34
GHI                45
XYZ                43
JVK                23
LMN                21

数据2只有名称列表

数据3再次具有名称和数字

Data 3
Name_desc           Numbers 
    ABC                56
    DEF                67
    GHI                89
    XYZ                60
    JVK                88
    LMN                65
    PQR                100
    KJL                85

我想做以下事情 -

Look for all names from data 2 are present in data 1
If any names are missing then 
{
get those names
get the numbers for those missing names from data 3
append above two things (missing names & numbers) to data 1
}
else
{data1<-data1
}

我只是想合并文件，但我还需要确保如果数据2中的数据1中没有名称丢失，那么数据1应保持不变。（上面代码中提到的相同内容）

在上述情况下，我的最终输出应为

Data 1- 

Name_description   Numbers 
    ABC                23
    DEF                34
    GHI                45
    XYZ                43
    JVK                23
    LMN                21
    PQR                100
    KJL                85

谢谢

Answer 1

首先，合并Data2和NA，然后在这个新的data.frame中找到Data3并将它们与Data3匹配，最后用{{替换它们1}}值。

> tmp <- merge(Data1, Data2, by.x="Name_description", by.y="Names", all=TRUE)
> ind <- match(tmp$Name_description[is.na(tmp$Numbers)], Data3$Name_desc)
> tmp$Numbers[ind] <- Data3$Numbers[ind]
> tmp
  Name_description Numbers
1              ABC      23
2              DEF      34
3              GHI      45
4              JVK      23
5              LMN      21
6              XYZ      43
7              KJL     100
8              PQR      85

Answer 2

我发现dplyr::coalesce在OP提到的情况下非常方便。加入3个数据框后，可以使用Numbers列NA列（包含coalesce），可以使用library(dplyr) Data1 %>% full_join(Data2, by=c("Name_description" = "Names")) %>% inner_join(Data3, by=c("Name_description" = "Name_desc")) %>% mutate(Numbers = coalesce( Numbers.x, Numbers.y)) %>% select(Name_description, Numbers) # Name_description Numbers # 1 ABC 23 # 2 DEF 34 # 3 GHI 45 # 4 XYZ 43 # 5 JVK 23 # 6 LMN 21 # 7 PQR 100 # 8 KJL 85合并为：

Data1 <- read.table(text = 
"Name_description   Numbers 
ABC                23
DEF                34
GHI                45
XYZ                43
JVK                23
LMN                21",
header = TRUE, stringsAsFactors = FALSE)

Data2 <- read.table(text = 
"Names            
ABC                
DEF                
GHI                
XYZ                
JVK                
LMN    
PQR
KJL",
header = TRUE, stringsAsFactors = FALSE)


Data3 <- read.table(text = 
"Name_desc           Numbers 
ABC                56
DEF                67
GHI                89
XYZ                60
JVK                88
LMN                65
PQR                100
KJL                85",
header = TRUE, stringsAsFactors = FALSE)

数据：

public static void main(String[] args) { LinkedList<String> list = new LinkedList<>();//declare your list Scanner input = new Scanner(System.in);//create a scanner System.out.println("How many participants? "); int nbr = input.nextInt();//read the number of element input.nextLine(); do { System.out.println("What is the name of the people?"); list.add(input.nextLine());//read and insert into your list in one shot nbr--;//decrement the index } while (nbr > 0);//repeat until the index will be 0 System.out.println(list);//print your list

Answer 3

使用dplyr，它应该类似于：

data1 %>% 
    bind_rows(
        data2 %>% 
        anti_join(data1) %>% 
        left_join(data3)
    )

Answer 4

我们可以使用dplyr和left_join在ifelse中实现这一目标。

library(dplyr)

Data4 <- Data2 %>%
  left_join(Data1, by = c("Names" = "Name_description")) %>%
  left_join(Data3, by = c("Names" = "Name_desc")) %>%
  mutate(Numbers = ifelse(is.na(Numbers.x), Numbers.y, Numbers.x)) %>%
  select(Names, Numbers)
Data4
#    Names Numbers
# 1   ABC      23
# 2   DEF      34
# 3   GHI      45
# 4   XYZ      43
# 5   JVK      23
# 6   LMN      21
# 7   PQR     100
# 8   KJL      85

数据

Data1 <- read.table(text = "Name_description Numbers ABC 23 DEF 34 GHI 45 XYZ 43 JVK 23 LMN 21", header = TRUE, stringsAsFactors = FALSE) Data2 <- read.table(text = "Names ABC DEF GHI XYZ JVK LMN PQR KJL", header = TRUE, stringsAsFactors = FALSE) Data3 <- read.table(text = "Name_desc Numbers ABC 56 DEF 67 GHI 89 XYZ 60 JVK 88 LMN 65 PQR 100 KJL 85", header = TRUE, stringsAsFactors = FALSE)

Answer 5

我们实际上根本不需要merge，你想要的是Number的第一个可用选择，从Data1然后Data3开始，当我Name在Data2而不在其他人中时，我想返回NA。

执行此操作的最快方法是使用data.table，但我也会提供其他选项。

<强> data.table

data.table::rbindlist默认情况下不使用名称（use.names=FALSE），因此在这种情况下非常方便。

library(data.table)
rbindlist(list(Data1,Data3,Data2))[,.SD[1,],by="Name_description"]

# 1:              ABC      23
# 2:              DEF      34
# 3:              GHI      45
# 4:              XYZ      43
# 5:              JVK      23
# 6:              LMN      21
# 7:              PQR     100
# 8:              KJL      85

tidyverse解决方案

.keep_all dplyr::distinct参数对于避免使用%>% filter(!duplicated(Names))或%>% group_by(Names) %>% Slice(1)的可读性较低非常有用。

library(tidyverse)
lst(Data1,Data3,cbind(Data2,NA)) %>%
  map(setNames,c("Names","Numbers")) %>%
  bind_rows %>%
  distinct(Names,.keep_all = TRUE) 

# Names Numbers
# 1   ABC      23
# 2   DEF      34
# 3   GHI      45
# 4   XYZ      43
# 5   JVK      23
# 6   LMN      21
# 7   PQR     100
# 8   KJL      85

基础解决方案

x <- do.call(rbind,lapply(list(Data1,Data3,cbind(Data2,NA)),setNames,c("Names","Numbers")))
x[!duplicated(x[[1]]),]  
#    Names Numbers
# 1    ABC      23
# 2    DEF      34
# 3    GHI      45
# 4    XYZ      43
# 5    JVK      23
# 6    LMN      21
# 13   PQR     100
# 14   KJL      85

比较3个数据帧中的值并附加缺失值

5 个答案: