2

次のようなデータセットがあります。

partyid            coninc
Ind,Near Dem       25926
Not Str Democrat   33333
Not Str Democrat   41667
Strong Democrat    69444
Ind,Near Dem       60185
Ind,Near Dem       50926
Ind,Near Dem       18519
Strong Democrat    3704
Strong Democrat    25926
Strong Democrat    18519
Not Str Republican 18519
Strong Democrat    18519
Not Str Democrat   18519

私がやりたいのは、データセットを次のような形式にフォーマットすることです。

partyid             0-50,000   50,000-100,000   100,000-150,000   >150,000
Strong Democrat     2344       3423             4342              54
Not Str Democrat    2643       934              ..
Ind, Near Dem       7656       343              ..
Ind, Near Rep       7655       833              .. 
Not Str Republican  2443       343
Strong Republican   3444       773

つまり、行を partyid 変数のレベルで並べ替え、列を coninc 変数の範囲のカウントで並べ替えます。

dput私のデータのA ：

structure(list(partyid = structure(c(3L, 2L, 2L, 1L, 3L, 3L, 3L, 1L, 1L, 1L, 5L, 1L, 2L, 1L, 1L, 4L, 4L, 3L, 4L, 3L), .Label = c("Strong Democrat", "Not Str Democrat", "Ind,Near Dem", "Ind,Near Rep", "Not Str Republican", "Strong Republican"), class = "factor"), coninc = c(25926L, 33333L, 41667L, 69444L, 60185L, 50926L, 18519L, 3704L, 25926L, 18519L, 18519L, 18519L, 18519L, 25926L, 18519L, 33333L, 25926L, 60185L, 69444L, 50926L)), .Names = c("partyid", "coninc"), row.names = c(1L, 2L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 21L), class = "data.frame")
4

2 に答える 2

2

パッケージを使用すると、これを非常に簡単に行うことができplyrます (サンプル データは少し読みにくいため、 のコンマとスペースを削除しましたpartyid)。

# creating sample data
dat <- structure(list(partyid = structure(c(3L, 2L, 2L, 1L, 3L, 3L, 3L, 1L, 1L, 1L, 5L, 1L, 2L, 1L, 1L, 4L, 4L, 3L, 4L, 3L), .Label = c("Strong Democrat", "Not Str Democrat", "Ind,Near Dem", "Ind,Near Rep", "Not Str Republican", "Strong Republican"), class = "factor"), coninc = c(25926L, 33333L, 41667L, 69444L, 60185L, 50926L, 18519L, 3704L, 25926L, 18519L, 18519L, 18519L, 18519L, 25926L, 18519L, 33333L, 25926L, 60185L, 69444L, 50926L)), .Names = c("partyid", "coninc"), row.names = c(1L, 2L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 21L), class = "data.frame")

# summarising the data with plyr
require(plyr)
dat2 <- ddply(dat, .(partyid), summarise,
              zero = sum(coninc < 50001),
              fifty = sum(coninc > 50000 & coninc < 100001),
              hundred = sum(coninc > 100000 & coninc < 150001),
              hfifty = sum(coninc > 150000))

これにより、次の出力が得られます。

dat2 <- structure(list(partyid = structure(1:5, .Label = c("Strong Democrat", "Not Str Democrat", "Ind,Near Dem", "Ind,Near Rep", "Not Str Republican", "Strong Republican"), class = "factor"), zero = c(6L, 3L, 2L, 2L, 1L), fifty = c(1L, 0L, 4L, 1L, 0L), hundred = c(0L, 0L, 0L, 0L, 0L), hfifty = c(0L, 0L, 0L, 0L, 0L)), .Names = c("partyid", "zero", "fifty", "hundred", "hfifty"), row.names = c(NA, -5L), class = "data.frame")
于 2014-04-05T08:58:15.627 に答える
2

ベースRでcutandを使用できます：table

dat$cat <- cut(dat$coninc, breaks = c(0, 50000, 100000, 150000, Inf),
               labels = c("< 50K", "50K - 100K", "100K - 150K", "> 150K"))
table(dat$partyid, dat$cat)
#                     
#                      < 50K 50K - 100K 100K - 150K > 150K
#   Strong Democrat        6          1           0      0
#   Not Str Democrat       3          0           0      0
#   Ind,Near Dem           2          4           0      0
#   Ind,Near Rep           2          1           0      0
#   Not Str Republican     1          0           0      0
#   Strong Republican      0          0           0      0
于 2014-04-05T13:37:26.967 に答える