根據另一個數據幀中的列創建新的數據幀行

Question

我有2個數據幀，第一列是一個列表（df A），另一列的第一列包含列表中的項目，但在某些情況下每行有多個項目（df B）。 我想要做的就是去通過，並從一個DF每個項目創建新行什麼，發生在DF B的第一列

DF一

dfA 
    Index X 
1 1 alpha 
2 2 beta 
3 3 gamma 
4 4 delta 

DF乙

dfB 
    list X 
1 1 4 alpha 
2 3 2 1 beta 
3 4 1 2 gamma 
4 3  delta 

期望

dfC 
    Index x 
1 1  alpha 
2 4  alpha 
3 3  beta 
4 2  beta 
5 1  beta 
6 4  gamma 
7 1  gamma 
8 2  gamma 
9 3  delta 

我使用的實際數據： DF一

dput(head(allwines)) 
structure(list(Wine = c("Albariño", "Aligoté", "Amarone", "Arneis", 
"Asti Spumante", "Auslese"), Description = c("Spanish white wine grape that makes crisp, refreshing, and light-bodied wines.", 
"White wine grape grown in Burgundy making medium-bodied, crisp, dry wines with spicy character.", 
"From Italy’s Veneto Region a strong, dry, long- lived red, made from a blend of partially dried red grapes.", 
"A light-bodied dry wine the Piedmont Region of Italy", "From the Piedmont Region of Italy, A semidry sparkling wine produced from the Moscato di Canelli grape in the village of Asti", 
"German white wine from grapes that are very ripe and thus high in sugar" 
)), .Names = c("Wine", "Description"), row.names = c(NA, 6L), class = "data.frame") 

DF乙

> dput(head(cheesePairing)) 
structure(list(Wine = c("Cabernet Sauvignon\r\n        \r\n       \r\n      \r\n       \r\n        \r\n         Pinot Noir\r\n        \r\n       \r\n      \r\n       \r\n        \r\n         Sauvignon Blanc\r\n        \r\n       \r\n      \r\n       \r\n        \r\n         Zinfandel", 
"Chianti\r\n        \r\n       \r\n      \r\n       \r\n        \r\n         Pinot Noir\r\n        \r\n       \r\n      \r\n       \r\n        \r\n         Sangiovese", 
"Chardonnay", "Bardolino\r\n        \r\n       \r\n      \r\n       \r\n        \r\n         Malbec\r\n        \r\n       \r\n      \r\n       \r\n        \r\n         Riesling\r\n        \r\n       \r\n      \r\n       \r\n        \r\n         Rioja\r\n        \r\n       \r\n      \r\n       \r\n        \r\n         Sauvignon Blanc", 
"Tempranillo", "Asti Spumante"), Cheese = c("Abbaye De Belloc Cheese", 
"Ardrahan cheese", "Asadero cheese", "Asiago cheese", "Azeitao", 
"Baby Swiss Cheese"), Suggestions = c("Pair with apples, sliced pears OR a sampling of olives and thin sliced salami. Pass around slices of baguette.", 
"Serve with a substantial wheat cracker and apples or grapes.", 
"Rajas (blistered chile strips) fresh corn tortillas", "Table water crackers, raw nuts (almond, walnuts)", 
"Nutty brown bread, grapes", "Server with dried fruits, whole grain, nutty breads, nuts" 
)), .Names = c("Wine", "Cheese", "Suggestions"), row.names = c(NA, 
6L), class = "data.frame") 

Answer 1

爲了解決柯特的答案，我想我找到了一個更有效的解決方案......假設我正確地解釋了你的目標。

我的評論代碼是在下面。您應該能夠按原樣運行並獲得所需的dfC。有一點需要注意的是，我假設（根據您的實際數據）分隔符分裂dfB $索引是「\ r \ n」。

# set up fake data 
dfA<-data.frame(Index=c('1','2','3','4'), X=c('alpha','beta','gamma','delta')) 
dfB<-data.frame(Index=c('1 \r\n 4','3 \r\n 2 \r\n 1','4 \r\n 1 \r\n 2','3'), X=c('alpha','beta','gamma','delta')) 

dfA$Index<-as.character(dfA$Index) 
dfA$X<-as.character(dfA$X) 
dfB$Index<-as.character(dfB$Index) 
dfB$X<-as.character(dfB$X) 


dfB_index_parsed<-strsplit(x=dfB$Index,"\r\n") # split Index of dfB by delimiter "\r\n" and store in a list 
names(dfB_index_parsed)<-dfB$X 
dfB_split_num<-lapply(dfB_index_parsed, length) # find the number of splits per row of dfB and store in a list 
dfB_split_num_vec<-do.call('c', dfB_split_num) # convert number of splits above from list to vector 

g<-do.call('c',dfB_index_parsed) # store all split values in a single vector 
g<-gsub(' ','',g) # remove trailing/leading spaces that occur after split 
names(g)<-rep(names(dfB_split_num_vec), dfB_split_num_vec) # associate each split Index from dfB with X from dfB 
g<-g[g %in% dfA$Index] # check which dfB$Index are in dfA$Index 

dfC<-data.frame(Index=g, X=names(g)) # construct data.frame 

Answer 2

首先，讓我提供一個功能回答你的問題。我懷疑我的答案是非常有效的，但它有效。

# construct toy data 
dfA <- data.frame(index = 1:4, X = letters[1:4]) 

dfB <- data.frame(X = letters[1:4]) 
dfB$list_elements <- list(c(1, 4), c(3, 2, 1), c(4, 1, 2), c(3)) 

# define function that provides solution 

unlist_merge_df <- function(listed_df, reference_df){ 
    # reference_df assumed to have columns "X" and "index" 
    # listed_df assumed to have column "list_elements" 
    df_out <- data.frame(index = c(), X = c()) 
    my_list <- listed_df$list_elements 
    for(idx in 1:length(my_list)){ 
     df_out <- rbind(df_out, 
         data.frame(index = my_list[[idx]], 
            X = listed_df[idx, 'X']) 
         ) 
    } 
    return(df_out) 
} 

# call the function 
dfC <- unlist_merge_df(dfB, dfA) 

# show output in human and R-parseable formats 
dfC 

dput(dfC) 

---

輸出是：

index X 
1 1 a 
2 4 a 
3 3 b 
4 2 b 
5 1 b 
6 4 c 
7 1 c 
8 2 c 
9 3 d 

structure(list(index = c(1, 4, 3, 2, 1, 4, 1, 2, 3), X = structure(c(1L, 
1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L), .Label = c("a", "b", "c", "d" 
), class = "factor")), .Names = c("index", "X"), row.names = c(NA, 
9L), class = "data.frame") 

---

其次，讓我說，你所處的情況不是很desireable。如果你能避免它，你可能應該。要麼完全不使用數據框，只使用列表，或者完全避免構建列出的數據框（如果可以的話），並直接構造所需的輸出。