Question

我有这些数据：

structure(list(type = c("journal", "all", "similar_age_1m", "similar_age_3m", 
"similar_age_journal_1m", "similar_age_journal_3m"), count = c("13972", 
"754555", "22408", "56213", "508", "1035"), rank = c("13759", 
"754043", "22339", "56074", "459", "947"), pct = c("98.48", "99.93", 
"99.69", "99.75", "90.35", "91.50")), .Names = c("type", "count", 
"rank", "pct"), row.names = c(NA, -6L), class = "data.frame")

我想把它变成一行，列2:4的名称以相应的类型为前缀。例如journal.count，journal.rank ...最快的方法是什么？由于某种原因dcast和reshape没有为我做这些，我的解决方案有点过于繁琐。

Answer 1

你提到reshape2，所以这是一种方法：

library("reshape2")
dcast(melt(dat, id.var="type"), 1~variable+type)

这给出了：

  1 count_all count_journal count_similar_age_1m count_similar_age_3m
1 1    754555         13972                22408                56213
  count_similar_age_journal_1m count_similar_age_journal_3m rank_all
1                          508                         1035   754043
  rank_journal rank_similar_age_1m rank_similar_age_3m
1        13759               22339               56074
  rank_similar_age_journal_1m rank_similar_age_journal_3m pct_all pct_journal
1                         459                         947   99.93       98.48
  pct_similar_age_1m pct_similar_age_3m pct_similar_age_journal_1m
1              99.69              99.75                      90.35
  pct_similar_age_journal_3m
1                      91.50

type和变量用_分隔，而不是.。

Answer 2

这是另一种方式：

y <- as.numeric(as.matrix(x[-1])) # flatten the data.frame
names(y) <- as.vector(outer(x[['type']], names(x)[-1], paste, sep='.'))

Answer 3

假设您可以为重塑添加虚拟“时间”变量，您也可以使用基数R轻松完成此操作。假设您的data.frame被调用：

mydf$id <- 1
(mydfw <- reshape(mydf, direction = "wide", idvar="id", timevar="type"))
#   id count.journal rank.journal pct.journal count.all rank.all pct.all
# 1  1         13972        13759       98.48    754555   754043   99.93
#   count.similar_age_1m rank.similar_age_1m pct.similar_age_1m
# 1                22408               22339              99.69
#   count.similar_age_3m rank.similar_age_3m pct.similar_age_3m
# 1                56213               56074              99.75
#   count.similar_age_journal_1m rank.similar_age_journal_1m
# 1                          508                         459
#   pct.similar_age_journal_1m count.similar_age_journal_3m
# 1                      90.35                         1035
#   rank.similar_age_journal_3m pct.similar_age_journal_3m
# 1                         947                      91.50

如果要重新排序列，清理也不算太糟糕。

mydfw <- mydfw[, unlist(sapply(names(mydf), grep, names(mydfw)))]

Answer 4

以下是使用expand.grid获取姓名的解决方案。

要获取数据，首先是子集以删除包含名称的第一列。然后，转置并转换为数字。

> eg <- expand.grid(colnames(x[, -1]), x[, 1])
> setNames(as.numeric(t(x[, -1])), paste(eg[[2]], eg[[1]], sep="."))
               journal.count                 journal.rank 
                    13972.00                     13759.00 
                 journal.pct                    all.count 
                       98.48                    754555.00 
                    all.rank                      all.pct 
                   754043.00                        99.93 
        similar_age_1m.count          similar_age_1m.rank 
                    22408.00                     22339.00 
          similar_age_1m.pct         similar_age_3m.count 
                       99.69                     56213.00 
         similar_age_3m.rank           similar_age_3m.pct 
                    56074.00                        99.75 
similar_age_journal_1m.count  similar_age_journal_1m.rank 
                      508.00                       459.00 
  similar_age_journal_1m.pct similar_age_journal_3m.count 
                       90.35                      1035.00 
 similar_age_journal_3m.rank   similar_age_journal_3m.pct 
                      947.00                        91.50

Answer 5

#assuming your data is called "test"
result <- as.data.frame(matrix(t(test[-1]),nrow=1),stringsAsFactors=FALSE)
names(result) <- as.vector(t(outer(unique(test$type),names(test[-1]),paste,sep=".")))

str(result)
'data.frame':   1 obs. of  18 variables:
 $ journal.count               : chr "13972"
 $ journal.rank                : chr "13759"
 $ journal.pct                 : chr "98.48"
 $ all.count                   : chr "754555"
 $ all.rank                    : chr "754043"
 $ all.pct                     : chr "99.93"
 $ similar_age_1m.count        : chr "22408"
 $ similar_age_1m.rank         : chr "22339"
 $ similar_age_1m.pct          : chr "99.69"
 $ similar_age_3m.count        : chr "56213"
 $ similar_age_3m.rank         : chr "56074"
 $ similar_age_3m.pct          : chr "99.75"
 $ similar_age_journal_1m.count: chr "508"
 $ similar_age_journal_1m.rank : chr "459"
 $ similar_age_journal_1m.pct  : chr "90.35"
 $ similar_age_journal_3m.count: chr "1035"
 $ similar_age_journal_3m.rank : chr "947"
 $ similar_age_journal_3m.pct  : chr "91.50"

Answer 6

假设您的数据框称为dat，这是一个解决方案。这有点粗糙，可能不是你想要的：

dat2 <- data.frame(matrix(unlist(lapply(1:nrow(dat), function(i) dat[i, -1])), nrow=1))
colnames(dat2) <- paste0(rep(dat[, 1], each=ncol(dat)-1), ".", 1:(ncol(dat)-1)) 
dat2

如果它不一定是数据框，这也可以起作用：

dat3 <- as.numeric(unlist(lapply(1:nrow(dat), function(i) dat[i, -1])))
names(dat3) <- paste0(rep(dat[, 1], each=ncol(dat)-1), ".", 1:(ncol(dat)-1)) 
dat3

将data.frame转换为单行

6 个答案: