我有一个数据表，想要使用查找表对其进行修改。我想循环遍历数据中的代码列，并根据匹配的value列以及匹配正确行值的代码列的名称为每个列添加新的相应datayear列。查找表的field列。
我尝试过使用lapply和left_join，但我无法研究如何使用数据列名称来引用查找的field列中的正确值。我还考虑过查找表在宽格式中是否会更好，所以你至少会有匹配的列名，但我仍然无法生成可行的函数。
示例数据和所需的输出：
数据（编辑：实际数据将包含更多代码列）：
structure(list(id = 1:10, datayear = c(2007L, 2007L, 2007L, 2007L,
2007L, 2008L, 2008L, 2008L, 2008L, 2008L), nationalitycode = c(1L,
1L, 1L, 2L, 3L, 5L, 4L, 3L, 2L, 1L), subjectcode = c(2L, 5L,
5L, 5L, 2L, 5L, 4L, 2L, 1L, 4L)), .Names = c("id", "datayear",
"nationalitycode", "subjectcode"), class = "data.frame", row.names = c(NA,
-10L))
id datayear nationalitycode subjectcode
1 1 2007 1 2
2 2 2007 1 5
3 3 2007 1 5
4 4 2007 2 5
5 5 2007 3 2
6 6 2008 5 5
7 7 2008 4 4
8 8 2008 3 2
9 9 2008 2 1
10 10 2008 1 4
查找表：
structure(list(datayear = c(2007L, 2007L, 2007L, 2007L, 2007L,
2007L, 2007L, 2007L, 2007L, 2007L, 2008L, 2008L, 2008L, 2008L,
2008L, 2008L, 2008L, 2008L, 2008L, 2008L), field = structure(c(1L,
1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 2L,
2L, 2L, 2L), .Label = c("nationalitycode", "subjectcode"), class = "factor"),
code = c(1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L,
3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L), lookupvalue = structure(c(10L,
16L, 9L, 4L, 5L, 2L, 7L, 13L, 1L, 14L, 5L, 16L, 4L, 6L, 11L,
17L, 3L, 15L, 8L, 12L), .Label = c("Algebra", "Art", "Beekeeping",
"Chinese", "English", "French", "Geography", "H.E.", "Indian",
"Irish", "Italian", "Latin", "Maths", "P.E.", "Rivetting",
"Scottish", "Sewing"), class = "factor")), class = "data.frame", row.names = c(NA,
-20L), .Names = c("datayear", "field", "code", "lookupvalue"))
datayear field code lookupvalue
1 2007 nationalitycode 1 Irish
2 2007 nationalitycode 2 Scottish
3 2007 nationalitycode 3 Indian
4 2007 nationalitycode 4 Chinese
5 2007 nationalitycode 5 English
6 2007 subjectcode 1 Art
7 2007 subjectcode 2 Geography
8 2007 subjectcode 3 Maths
9 2007 subjectcode 4 Algebra
10 2007 subjectcode 5 P.E.
11 2008 nationalitycode 1 English
12 2008 nationalitycode 2 Scottish
13 2008 nationalitycode 3 Chinese
14 2008 nationalitycode 4 French
15 2008 nationalitycode 5 Italian
16 2008 subjectcode 1 Sewing
17 2008 subjectcode 2 Beekeeping
18 2008 subjectcode 3 Rivetting
19 2008 subjectcode 4 H.E.
20 2008 subjectcode 5 Latin
期望的输出：
id datayear nationalitycode subjectcode nationalityvalue subjectvalue
1 1 2007 1 2 Irish Geography
2 2 2007 1 5 Irish P.E.
3 3 2007 1 5 Irish P.E.
4 4 2007 2 5 Scottish P.E.
5 5 2007 3 2 Indian Geography
6 6 2008 5 5 Italian Latin
7 7 2008 4 4 French H.E.
8 8 2008 3 2 Chinese Beekeeping
9 9 2008 2 1 Scottish Sewing
10 10 2008 1 4 English H.E.
非常感谢任何帮助！
答案 0 :(得分：1)
诀窍是根据查找表的相应子集加入。这是通过使用正确的字段值进行分段。
library(dplyr)
dt1 = structure(list(id = 1:10, datayear = c(2007L, 2007L, 2007L, 2007L,
2007L, 2008L, 2008L, 2008L, 2008L, 2008L), nationalitycode = c(1L,
1L, 1L, 2L, 3L, 5L, 4L, 3L, 2L, 1L), subjectcode = c(2L, 5L,
5L, 5L, 2L, 5L, 4L, 2L, 1L, 4L)), .Names = c("id", "datayear",
"nationalitycode", "subjectcode"), class = "data.frame", row.names = c(NA, -10L))
dt2 = structure(list(datayear = c(2007L, 2007L, 2007L, 2007L, 2007L,
2007L, 2007L, 2007L, 2007L, 2007L, 2008L, 2008L, 2008L, 2008L,
2008L, 2008L, 2008L, 2008L, 2008L, 2008L), field = structure(c(1L,
1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 2L,
2L, 2L, 2L), .Label = c("nationalitycode", "subjectcode"), class = "factor"),
code = c(1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L,
3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L), lookupvalue = structure(c(10L,
16L, 9L, 4L, 5L, 2L, 7L, 13L, 1L, 14L, 5L, 16L, 4L, 6L, 11L,
17L, 3L, 15L, 8L, 12L), .Label = c("Algebra", "Art", "Beekeeping",
"Chinese", "English", "French", "Geography", "H.E.", "Indian",
"Irish", "Italian", "Latin", "Maths", "P.E.", "Rivetting",
"Scottish", "Sewing"), class = "factor")), class = "data.frame", row.names = c(NA,
-20L), .Names = c("datayear", "field", "code", "lookupvalue"))
dt1 %>%
left_join(dt2 %>% filter(field == "nationalitycode"), by=c("datayear"="datayear","nationalitycode"="code")) %>%
left_join(dt2 %>% filter(field == "subjectcode"), by=c("datayear"="datayear","subjectcode"="code")) %>%
rename(nationalityvalue = lookupvalue.x,
subjectvalue = lookupvalue.y) %>%
select(-field.x, -field.y)
# id datayear nationalitycode subjectcode nationalityvalue subjectvalue
# 1 1 2007 1 2 Irish Geography
# 2 2 2007 1 5 Irish P.E.
# 3 3 2007 1 5 Irish P.E.
# 4 4 2007 2 5 Scottish P.E.
# 5 5 2007 3 2 Indian Geography
# 6 6 2008 5 5 Italian Latin
# 7 7 2008 4 4 French H.E.
# 8 8 2008 3 2 Chinese Beekeeping
# 9 9 2008 2 1 Scottish Sewing
# 10 10 2008 1 4 English H.E.
对于你要求使用循环的更一般的情况，我需要重塑你的查找表，以便我可以使用列名。该过程将自动检测您的查找表中有多少个唯一字段，并使用for循环执行连接（顺序）。
library(dplyr)
library(tidyr)
dt1 = structure(list(id = 1:10, datayear = c(2007L, 2007L, 2007L, 2007L,
2007L, 2008L, 2008L, 2008L, 2008L, 2008L), nationalitycode = c(1L,
1L, 1L, 2L, 3L, 5L, 4L, 3L, 2L, 1L), subjectcode = c(2L, 5L,
5L, 5L, 2L, 5L, 4L, 2L, 1L, 4L)), .Names = c("id", "datayear",
"nationalitycode", "subjectcode"), class = "data.frame", row.names = c(NA, -10L))
dt2 = structure(list(datayear = c(2007L, 2007L, 2007L, 2007L, 2007L,
2007L, 2007L, 2007L, 2007L, 2007L, 2008L, 2008L, 2008L, 2008L,
2008L, 2008L, 2008L, 2008L, 2008L, 2008L), field = structure(c(1L,
1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 2L,
2L, 2L, 2L), .Label = c("nationalitycode", "subjectcode"), class = "factor"),
code = c(1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L,
3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L), lookupvalue = structure(c(10L,
16L, 9L, 4L, 5L, 2L, 7L, 13L, 1L, 14L, 5L, 16L, 4L, 6L, 11L,
17L, 3L, 15L, 8L, 12L), .Label = c("Algebra", "Art", "Beekeeping",
"Chinese", "English", "French", "Geography", "H.E.", "Indian",
"Irish", "Italian", "Latin", "Maths", "P.E.", "Rivetting",
"Scottish", "Sewing"), class = "factor")), class = "data.frame", row.names = c(NA,
-20L), .Names = c("datayear", "field", "code", "lookupvalue"))
# reshape your lookup data
dt2 %>%
spread(field, code) -> dt2_reshaped
# start dataset (to join every field you have)
dt_temp = dt1
# for every field you have do the join
for (fld in as.character(unique(dt2$field))) {
dt_temp %>% left_join(dt2_reshaped %>% select_("datayear", "lookupvalue", fld), by=c("datayear",fld)) -> dt_temp
names(dt_temp)[names(dt_temp) == "lookupvalue" ] = gsub("code","value",fld)
}
dt_temp
# id datayear nationalitycode subjectcode nationalityvalue subjectvalue
# 1 1 2007 1 2 Irish Geography
# 2 2 2007 1 5 Irish P.E.
# 3 3 2007 1 5 Irish P.E.
# 4 4 2007 2 5 Scottish P.E.
# 5 5 2007 3 2 Indian Geography
# 6 6 2008 5 5 Italian Latin
# 7 7 2008 4 4 French H.E.
# 8 8 2008 3 2 Chinese Beekeeping
# 9 9 2008 2 1 Scottish Sewing
# 10 10 2008 1 4 English H.E.
答案 1 :(得分：1)
如果Sub delay()
Dim cell1 As Range
Dim nextrow1 As Long
Dim a1 As Double
Application.ScreenUpdating = False
a1 = Application.WorksheetFunction.CountA(Sheets("Data").Range("J:J"))
For Each cell1 In Sheets("Data").Range("J5:J" & a1)
If cell1.Value = "delayed " Then
nextrow1 = Application.WorksheetFunction.CountA(Sheets("Delayed").Range("J:J"))
Rows(cell1.Row).Copy Destination:=Sheets("Delayed").Range("A" & nextrow + 1)
End If
Next
Application.ScreenUpdating = False
End Sub
是您的第一个X而data.frame是您的第二个，LU和data.table会使这一点变得简单，重要的是，明确。
merge你可以采取一些措施来缩短这一点，但我认为这可以很清楚地发生了什么。
编辑：这是一个可以帮助您入门的功能：
library(data.table)
# Convert the data.frames into data.tables
setDT(X)
setDT(LU)
# Join the tables on datayear and the appropriate code, for the
# nationality data only.
X1 <- merge(X, LU[field == "nationalitycode"],
by.x=c("datayear", "nationalitycode"),
by.y=c("datayear", "code"))
# Now join the resulting table by subjectcode.
X2 <- merge(X1, LU[field == "subjectcode"],
by.x=c("datayear", "subjectcode"),
by.y=c("datayear", "code"))
# Now subset the data.table to the columns you want, set the key
# (order) by id, and rename some columns.
M <- X2[, c("id", "datayear", "nationalitycode", "subjectcode",
"lookupvalue.x", "lookupvalue.y"), with=FALSE]
setkey(M, "id")
setnames(M, c("lookupvalue.x", "lookupvalue.y"),
c("nationalityvalue", "subjectvalue"))
M
# id datayear nationalitycode subjectcode nationalityvalue subjectvalue
# 1: 1 2007 1 2 Irish Geography
# 2: 2 2007 1 5 Irish P.E.
# 3: 3 2007 1 5 Irish P.E.
# 4: 4 2007 2 5 Scottish P.E.
# 5: 5 2007 3 2 Indian Geography
# 6: 6 2008 5 5 Italian Latin
# 7: 7 2008 4 4 French H.E.
# 8: 8 2008 3 2 Chinese Beekeeping
# 9: 9 2008 2 1 Scottish Sewing
# 10: 10 2008 1 4 English H.E.