问如何像字典一样添加键值对？
EN

Stack Overflow用户

提问于 2017-09-03 01:47:02

回答 3查看 893关注 0票数 1

我的数据(总共8532个)如下所示：

Prd_Id  Weight
DRA24   19.35
DRA24   NA
DRA24   NA
DRA24   19.35
DRA24   19.35
DRA59   8.27
DRA59   8.27
DRA59   8.27
DRA59   8.27
DRA59   NA
DRA59   NA

基本上，问题是有很多对Prd_id和weight，其中有些Prd_id没有提到weight --例如，我在数据中显示，第二和第三没有，所以我知道weight的值，我只需要用它替换NA，所有相同的Prd_id都会有相同的weight，但是在R中没有像字典这样的东西，所以我发现很难解决这个问题。我试过使用for loop，但是需要很长时间，我的代码如下所示：

for(i in 1:nrow(bms)){
  for(j in 1:1555){
    if(spl$Prd_Id[j]==bms$Prd_Id[i]){
      bms$weight[i]=spl$weight[j]
    }
  }
}

bms是整个data (8532个)，spl (1555个)是bms的一个子集，其值为Prd_Id。

list

for-loop

missing-data

回答 3

Stack Overflow用户

回答已采纳

发布于 2017-09-03 03:04:10

正如@r2evans所建议的那样，您可以使用类似SQL的连接策略，再加上dplyr的coalesce，如下所示：

library(dplyr)

# create 'bms'.
bms <- data_frame(
  Prd_Id = c("DRA24", "DRA24", "DRA24", "DRA24", "DRA24", "DRA59", "DRA59", "DRA59", "DRA59", "DRA59", "DRA59"),
  Weight = c(19.35, NA, NA, 19.35, 19.35, 8.27, 8.27, 8.27, 8.27, NA, NA)
)

# create 'spl'
spl <- bms %>% filter(!is.na(Weight)) %>% filter(!duplicated(Prd_Id))

# SQL-like join and coalesce strategy
res <- bms %>% 
  left_join(spl, by = "Prd_Id", suffix = c("_bms", "_spl")) %>% 
  mutate(Weight = coalesce(Weight_bms, Weight_spl)) %>%
  select(-Weight_bms, -Weight_spl)

票数 1

Stack Overflow用户

发布于 2017-09-04 03:11:10

不需要left_join

bms %>% 
  group_by(Prd_Id) %>% 
  mutate(Weight = Weight[!is.na(Weight)][1])

另一种使用first的方法

bms %>% 
  group_by(Prd_Id) %>% 
  mutate(Weight = first(Weight[!is.na(Weight)]))

结果：

# A tibble: 11 x 2
# Groups:   Prd_Id [2]
   Prd_Id Weight
    <chr>  <dbl>
 1  DRA24  19.35
 2  DRA24  19.35
 3  DRA24  19.35
 4  DRA24  19.35
 5  DRA24  19.35
 6  DRA59   8.27
 7  DRA59   8.27
 8  DRA59   8.27
 9  DRA59   8.27
10  DRA59   8.27
11  DRA59   8.27

当然，你也可以在香草R中这样做：

transform(bms, Weight = ave(Weight, Prd_Id, FUN = function(x) x[!is.na(x)][1]))

结果是相同的：

   Prd_Id Weight
1   DRA24  19.35
2   DRA24  19.35
3   DRA24  19.35
4   DRA24  19.35
5   DRA24  19.35
6   DRA59   8.27
7   DRA59   8.27
8   DRA59   8.27
9   DRA59   8.27
10  DRA59   8.27
11  DRA59   8.27

票数 1

Stack Overflow用户

发布于 2017-09-04 03:07:14

这里有一个基本的R解

# example data
bms <- data.frame(
  Prd_Id = c("DRA24", "DRA24", "DRA24", "DRA24", "DRA24", "DRA59", "DRA59", "DRA59", "DRA59", "DRA59", "DRA59"),
  Weight = c(19.35, NA, NA, 19.35, 19.35, 8.27, 8.27, 8.27, 8.27, NA, NA)
)

# create key-value pairs
spl <- unique(bms[!is.na(bms[,"Weight"]),])
spl <- setNames(spl[,"Weight"], spl[,"Prd_Id"])

# fill NAs
idx <- which(is.na(bms[,"Weight"]))
bms[idx,"Weight"] <- spl[bms[idx, "Prd_Id"]]