我有以下数据集（示例）。显然，我使用的真实数据集要大得多：
gvkey tic stko year
001689 AEP1 1 2011
017096 BA3 1 2011
001440 AEP 0 2011
002285 BA 0 2011
001689 AEP1 1 2012
017096 BA3 1 2012
001440 AEP 0 2012
002285 BA 0 2012
以下是生成数据的代码：
dat <- data.frame(gvkey=c("001689", "017096", "001440", "002285"), tic=c("AEP1", "BA3", "AEP", "BA"), stko=c(1, 1, 0, 0), year=c(2011,2011,2011,2011,2012,2012,2012,2012))
以下是我想要做的事情：每行代表一年与公司配对，tic是公司的股票代码。 stko等于1的公司是子公司，与其母公司共享tic，但在股票代码中附有一个数字，例如AEP1属于AEP。基本上，我想创建一个新变量parent，它为每个子公司（stko=1行）指示母公司的gvkey。我想每年都这样做。最终数据集应如下所示：
gvkey tic stko year parent
001689 AEP1 1 2011 001440
017096 BA3 1 2011 002285
001440 AEP 0 2011
002285 BA 0 2011
001689 AEP1 1 2012 001440
017096 BA3 1 2012 002285
001440 AEP 0 2012
002285 BA 0 2012
现在，我最初的方法是编写几个for循环，这些循环在给定年份迭代行。无论何时，stko=1，然后在末尾提取没有数字的股票代码部分（例如，对于第一行AEP），并在给定年份中找到具有此确切股票代码的行（例如，第3行为第一年） 2011）并使用gvkey将该行的stko=1复制到初始观察结果。
但是，考虑到我的数据集的大小，这个过程会非常慢。如果有人能想到更快更容易的方法，我将不胜感激。
非常感谢!!
使用我的主数据集，dput(droplevels(head(dat)))的输出为：
structure(list(gvkey = c("176017", "128663", "61586", "278120",
"14062", "285313"), datadate = structure(c(4L, 4L, 1L, 3L, 2L,
1L), .Label = c("31dec2010", "31dec2011", "31dec2012", "31dec2013"
), class = "factor"), fyear = c(2013, 2013, 2010, 2012, 2011,
2010), indfmt = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = "INDL", class = "factor"),
consol = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = "C", class = "factor"),
popsrc = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = "D", class = "factor"),
datafmt = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = "STD", class = "factor"),
tic = c("ATHX", "SQNM", "IMH", "FNLIF", "CCDBF", "BSBR"),
cusip = structure(c(1L, 6L, 5L, 4L, 3L, 2L), .Label = c("04744L106",
"05967A107", "124900309", "33564P103", "45254P508", "817337405"
), class = "factor"), conm = structure(c(1L, 6L, 5L, 4L,
3L, 2L), .Label = c("ATHERSYS INC", "BANCO SANTANDER BRASIL -ADR",
"CCL INDUSTRIES -CL B", "FIRST NATIONAL FINL CORP", "IMPAC MORTGAGE HOLDINGS INC",
"SEQUENOM INC"), class = "factor"), curcd = structure(c(2L,
2L, 2L, 1L, 1L, 2L), .Label = c("CAD", "USD"), class = "factor"),
costat = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = "A", class = "factor"),
stko = c(0, 0, 0, 0, 0, 0)), .Names = c("gvkey", "datadate",
"fyear", "indfmt", "consol", "popsrc", "datafmt", "tic", "cusip",
"conm", "curcd", "costat", "stko"), row.names = c(NA, 6L), class = "data.frame")
答案 0 :(得分：2)
dplyr的另一个选项：
require(dplyr)
dat %>%
mutate(tic2 = gsub("[0-9]", "", tic)) %>%
group_by(tic2, year) %>%
mutate(parent = ifelse(stko == 1, as.character(gvkey[stko == 0][1]), "")) %>%
ungroup() %>%
select(-tic2)
#Source: local data frame [8 x 5]
#
# gvkey tic stko year parent
#1 001689 AEP1 1 2011 001440
#2 017096 BA3 1 2011 002285
#3 001440 AEP 0 2011
#4 002285 BA 0 2011
#5 001689 AEP1 1 2012 001440
#6 017096 BA3 1 2012 002285
#7 001440 AEP 0 2012
#8 002285 BA 0 2012
编辑：如果可能有没有匹配父级的公司，请尝试以下代码：
dat %>%
mutate(tic2 = gsub("[0-9]", "", tic)) %>%
group_by(tic2, year) %>%
mutate(parent = ifelse(stko == 1 & sum(stko == 0) > 0,
as.character(gvkey[stko == 0][1]), "")) %>%
ungroup() %>%
select(-tic2)
答案 1 :(得分：1)
看起来你有一个可行的解决方案，但你正在寻找速度。为此，您将要使用子集和向量化操作而不是循环（请参阅this post）。我还会找到使用dplyr或data.table的解决方案，因为它们都比基础R快得多。
这是我尝试基于这些想法的解决方案。
require(dplyr)
parents <- dat %>%
filter(stko == 0) %>%
select(tic, gvkey) %>%
unique(.)
rownames(parents) <- parents$tic
dat2 <- dat %>%
mutate(
parentTic = sub("[1-9]$", "", tic),
parentGvkey = parents[parentTic, "gvkey"])
答案 2 :(得分：0)
这是base::merge函数
rbind(transform(dat[dat$stko == 0, ], parent = ''),
merge(
transform(dat[dat$stko != 0, ], tic.parent = gsub('[0-9]', '', tic)),
unique(transform(dat[dat$stko == 0, ], parent = gvkey)[, c('tic', 'parent', 'year')]),
by.x = c('tic.parent', 'year'), by.y = c('tic', 'year')
)[, -1]
)
# gvkey tic stko year parent
# 3 001440 AEP 0 2011
# 4 002285 BA 0 2011
# 7 001440 AEP 0 2012
# 8 002285 BA 0 2012
# 5 001689 AEP1 1 2011 001440
# 6 001689 AEP1 1 2012 001440
# 71 017096 BA3 1 2011 002285
# 81 017096 BA3 1 2012 002285