如何根据列匹配，高效地将这些imdb电影标题ID替换成实际标题？

Question

我一直在R中处理他们发布的一些IMDB数据，今天尴尬地卡在这上面很久了。

primaryName     tconst                 primaryTitle                          knownForTitles
1          Aaron Lim  tt2317744            My Friend Bernard tt0268228,tt0891369,tt2317744,tt3709694
2      Aaron Woodley  tt3228088          Spark: A Space Tail tt0326065,tt1650535,tt4426464,tt3228088
3 Abdelkader Belhedi tt11069302       The Carthage Castaways         tt11698758,tt11069302,tt0485746

我正在努力想出一个方法来匹配这些数据。knownForTitles 中的ID。tconst 列。匹配后，我想把在 knownForTitles 与实际标题名称来自 primaryTitle，像下面这样。

primaryName     tconst                 primaryTitle                          knownForTitles
1          Aaron Lim  tt2317744            My Friend Bernard Movie Title,Movie Title,Movie Title,Movie Title
2      Aaron Woodley  tt3228088          Spark: A Space Tail Movie Title,Movie Title,Movie Title,Movie Title
3 Abdelkader Belhedi tt11069302       The Carthage Castaways         Movie Title,Movie Title,Movie Title

我只能想到使用一堆for循环，这对于上千行来说可能效率很低。如果有人能给我指出一个更好的方向，那就太好了。

Answer 1

代码是这样的。解释如下。

代码是这样的.

df = data.frame(primaryName = c("Aaron Lim", "Aaron Woodley"), tconst = c("tt2317744", "tt3228088"), primaryTitle = c("My friend Ron", "Spark: Some Title"), knownForTitles = c("tt0268228,tt0891369,tt2317744,tt3709694", "tt0326065,tt1650535,tt4426464,tt3228088"))
df$tconst = as.character(df$tconst)
Names = df %>%  
  mutate(V2 = strsplit(as.character(knownForTitles), ",")) %>% 
  tidyr::unnest(V2) %>% 
  select(-knownForTitles) %>% 
  as.data.frame(.) 
Movies = df[,2:3]
Modi = left_join(Names, Movies, by = c("V2" = "tconst")) 
Modi$primaryTitle.y = as.character(Modi$primaryTitle.y)
Modi[is.na(Modi$primaryTitle.y), "primaryTitle.y"] = "Test"

Modi %>% 
  group_by(tconst) %>%  
  summarise(primNew = stringr::str_c(primaryTitle.y, collapse = ", ")) %>% 
  inner_join(df, .)

产量.

    primaryName    tconst      primaryTitle                          knownForTitles
1     Aaron Lim tt2317744     My friend Ron tt0268228,tt0891369,tt2317744,tt3709694
2 Aaron Woodley tt3228088 Spark: Some Title tt0326065,tt1650535,tt4426464,tt3228088
                              primNew
1     Test, Test, My friend Ron, Test
2 Test, Test, Test, Spark: Some Title

---

解释.

让我们定义一些玩具数据。

    df = data.frame(primaryName = c("Aaron Lim", "Aaron Woodley"), 
                    tconst = c("tt2317744", "tt3228088"), 
                    primaryTitle = c("My friend", "Spark"), 
                    knownForTitles = c("tt0268228,tt0891369,tt2317744,tt3709694", "tt0326065,tt1650535,tt4426464,tt3228088"))
    df$tconst = as.character(df$tconst)

然后你就可以用tidyr的 unnest 函数将所有的列串分割成行，就像这样。

Names = df %>%  
  mutate(V2 = strsplit(as.character(knownForTitles), ",")) %>% 
  tidyr::unnest(V2) %>% 
  select(-knownForTitles) %>% 
  as.data.frame(.) 

结果

> Names
    primaryName    tconst      primaryTitle        V2
1     Aaron Lim tt2317744     My friend Ron tt0268228
2     Aaron Lim tt2317744     My friend Ron tt0891369
3     Aaron Lim tt2317744     My friend Ron tt2317744
4     Aaron Lim tt2317744     My friend Ron tt3709694
5 Aaron Woodley tt3228088 Spark: Some Title tt0326065
6 Aaron Woodley tt3228088 Spark: Some Title tt1650535
7 Aaron Woodley tt3228088 Spark: Some Title tt4426464
8 Aaron Woodley tt3228088 Spark: Some Title tt3228088

然后你会得到所有的电影名称 tconstants 与

Movies = df[,2:3]
Modi = left_join(Names, Movies, by = c("V2" = "tconst")) 

而结果

    primaryName    tconst    primaryTitle.x        V2    primaryTitle.y
1     Aaron Lim tt2317744     My friend Ron tt0268228              <NA>
2     Aaron Lim tt2317744     My friend Ron tt0891369              <NA>
3     Aaron Lim tt2317744     My friend Ron tt2317744     My friend Ron
4     Aaron Lim tt2317744     My friend Ron tt3709694              <NA>
5 Aaron Woodley tt3228088 Spark: Some Title tt0326065              <NA>
6 Aaron Woodley tt3228088 Spark: Some Title tt1650535              <NA>
7 Aaron Woodley tt3228088 Spark: Some Title tt4426464              <NA>
8 Aaron Woodley tt3228088 Spark: Some Title tt3228088 Spark: Some Title

由于这是玩具数据，所以有 NA 值，会造成一些麻烦，所以我们做

Modi$primaryTitle.y = as.character(Modi$primaryTitle.y)
Modi[is.na(Modi$primaryTitle.y), "primaryTitle.y"] = "Test"

来应对。

最后，你修改匹配的电影，并将它们折叠成一行，用

Modi %>% 
  group_by(tconst) %>%  
  summarise(primNew = stringr::str_c(primaryTitle.y, collapse = ", ")) %>% 
  inner_join(df, .)

而结果

    primaryName    tconst      primaryTitle                          knownForTitles
1     Aaron Lim tt2317744     My friend Ron tt0268228,tt0891369,tt2317744,tt3709694
2 Aaron Woodley tt3228088 Spark: Some Title tt0326065,tt1650535,tt4426464,tt3228088
                              primNew
1     Test, Test, My friend Ron, Test
2 Test, Test, Test, Spark: Some Title

Answer 2

我们可以通过以下方式获取数据 separate_rows, match knownForTitles 与 tconst 得到相应 primaryTitle 值，并将每一个 Name.

library(dplyr)

df %>%
  tidyr::separate_rows(knownForTitles, sep = ',') %>%
  mutate(knownForTitles = primaryTitle[match(knownForTitles, tconst)]) %>%
  group_by(primaryName) %>%
  summarise(knownForTitles = toString(na.omit(knownForTitles)))

---

在基础R中，我们可以将字符串和 match

df$knownForTitles <- sapply(strsplit(df$knownForTitles, ','), function(x) 
                 with(df, toString(na.omit(primaryTitle[match(x, tconst)]))))