-1

私はリストを持っています、mm：

 head(mm)
[[1]]
[1] "8  1901 - 1908  >>Primus<< sbk"

[[2]]
[1] "12  1901 - 1912  A & B:s skofabriks arbetares sbk."

[[3]]
[1] "5  1907 - 1911  A. B. Elevators arberates sbk"

[[4]]
[1] "5  1901 - 1905  Abk. N.K.B. (Nya Klöfverbladet)"

[[5]]
[1] "2  1904 - 1905  absolutisternas sbk"

[[6]]
[1] "12  1901 - 1912  Aftonbladets personals sbk"


length(mm)
[1] 429

dput(head(mm))
list("8  1901 - 1908  >>Primus<< sbk", "12  1901 - 1912  A & B:s skofabriks arbetares sbk.", 
    "5  1907 - 1911  A. B. Elevators arberates sbk", "5  1901 - 1905  Abk. N.K.B. (Nya Klöfverbladet)", 
    "2  1904 - 1905  absolutisternas sbk", "12  1901 - 1912  Aftonbladets personals sbk")

私は会社名も持っています：

head(unique(data$Name))
[1] ">>Primus<< sbk"                    "A & B:s skofabriks arbetares sbk." "A. B. Elevators arberates sbk"    
[4] "Abk. N.K.B. (Nya Klöfverbladet)"   "absolutisternas sbk"               "Aftonbladets personals sbk"

length(unique(data$Name))
[1] 429

私は新しいリストを作成しようとしています。ここでは、リストの各要素mmが、各会社で発生する回数だけ繰り返されますdata frame。

data[1:20,1:2]
     Year                              Name
1    1901                    >>Primus<< sbk
185  1902                    >>Primus<< sbk
382  1903                    >>Primus<< sbk
597  1904                    >>Primus<< sbk
822  1905                    >>Primus<< sbk
1059 1906                    >>Primus<< sbk
1310 1907                    >>Primus<< sbk
1567 1908                    >>Primus<< sbk
2    1901 A & B:s skofabriks arbetares sbk.
186  1902 A & B:s skofabriks arbetares sbk.
383  1903 A & B:s skofabriks arbetares sbk.
598  1904 A & B:s skofabriks arbetares sbk.
823  1905 A & B:s skofabriks arbetares sbk.
1060 1906 A & B:s skofabriks arbetares sbk.
1311 1907 A & B:s skofabriks arbetares sbk.
1568 1908 A & B:s skofabriks arbetares sbk.
1827 1909 A & B:s skofabriks arbetares sbk.
2090 1910 A & B:s skofabriks arbetares sbk.
2355 1911 A & B:s skofabriks arbetares sbk.
2602 1912 A & B:s skofabriks arbetares sbk.

dput(data[1:20,1:2])
structure(list(Year = c(1901L, 1902L, 1903L, 1904L, 1905L, 1906L, 
1907L, 1908L, 1901L, 1902L, 1903L, 1904L, 1905L, 1906L, 1907L, 
1908L, 1909L, 1910L, 1911L, 1912L), Name = c(">>Primus<< sbk", 
">>Primus<< sbk", ">>Primus<< sbk", ">>Primus<< sbk", ">>Primus<< sbk", 
">>Primus<< sbk", ">>Primus<< sbk", ">>Primus<< sbk", "A & B:s skofabriks arbetares sbk.", 
"A & B:s skofabriks arbetares sbk.", "A & B:s skofabriks arbetares sbk.", 
"A & B:s skofabriks arbetares sbk.", "A & B:s skofabriks arbetares sbk.", 
"A & B:s skofabriks arbetares sbk.", "A & B:s skofabriks arbetares sbk.", 
"A & B:s skofabriks arbetares sbk.", "A & B:s skofabriks arbetares sbk.", 
"A & B:s skofabriks arbetares sbk.", "A & B:s skofabriks arbetares sbk.", 
"A & B:s skofabriks arbetares sbk.")), .Names = c("Year", "Name"
), row.names = c(1L, 185L, 382L, 597L, 822L, 1059L, 1310L, 1567L, 
2L, 186L, 383L, 598L, 823L, 1060L, 1311L, 1568L, 1827L, 2090L, 
2355L, 2602L), class = "data.frame")

たとえば、会社>>Primus<< sbkが8回発生するため、「mm[[1]]」は8回繰り返されます。

length(data[data$Name==">>Primus<< sbk",2])
[1] 8

私のアプローチは次のとおりです。

mm=lapply(1:length(maxz),function(x) paste(diffz[[x]]+1,"",minz[[x]],"-",maxz[[x]],"",names(maxz)[[x]]))

hb=lapply(seq_along(mm),function(x,m) rep(mm[[x]],length(data[data$Name==m,2])),m=unique(data$Name))

しかし、私は上記warningを実行した後にこれを取得しhbます：

There were 50 or more warnings (use warnings() to see the first 50)

head(warnings())
$`longer object length is not a multiple of shorter object length`
data$Name == m

$`longer object length is not a multiple of shorter object length`
data$Name == m

私は何が間違っているのですか？:(

EDIT：

動作する回避策は次のとおりです。

よろしくお願いします！

mm=lapply(1:length(maxz),function(x) paste(diffz[[x]]+1,"",minz[[x]],"-",maxz[[x]],"",names(maxz)[[x]]))

names(mm)=names(minz)

hb=lapply(names(mm),function(x) rep(mm[[x]],length(data[data$Name==x,2])))

どこ

head(names(minz))
[1] ">>Primus<< sbk"                    "A & B:s skofabriks arbetares sbk." "A. B. Elevators arberates sbk"    
[4] "Abk. N.K.B. (Nya Klöfverbladet)"   "absolutisternas sbk"               "Aftonbladets personals sbk"
4

1 に答える 1

3

Rにデータを格納するための標準のデータ構造を使用すると、作業がはるかに楽になりますthe data.frame。次のコードは、入力をデータフレームに変換し、サブセット化を使用して行を繰り返します。

mm <- list("8  1901 - 1908  >>Primus<< sbk", "12  1901 - 1912  A & B:s skofabriks arbetares sbk.", 
    "5  1907 - 1911  A. B. Elevators arberates sbk", "5  1901 - 1905  Abk. N.K.B. (Nya Klöfverbladet)", 
    "2  1904 - 1905  absolutisternas sbk", "12  1901 - 1912  Aftonbladets personals sbk")

# Convert to a character vector
m <- unlist(mm)

# Convert multiple character separator to single
m2 <- gsub(" {2, }", ",", m)

# Parse with read.csv
df <- read.csv(text = m2, header = false)
names(df) <- c("n", "years", "company")

# Finally, duplicate each row
df[rep(1:nrow(df), df$n), -1]
于 2013-02-05T13:57:25.427 に答える