我在R中有以下数据框:
> str(df)
'data.frame': 545227 obs. of 15 variables:
$ ykod : int 93 93 93 93 93 93 93 93 93 93 ...
$ yad : Factor w/ 42 levels "BAKUGAN","BARBIE",..: 30 30 30 30 30 30 30 30 30 30 ...
$ per : Factor w/ 3 levels "2 AYLIK","3 AYLIK",..: 3 3 3 3 3 3 3 3 3 3 ...
$ donem: int 201101 201101 201101 201101 201101 201101 201101 201101 201101 201101 ...
$ sayi : int 201101 201101 201101 201101 201101 201101 201101 201101 201101 201101 ...
$ mkod : int 4 5 9 11 12 18 20 22 25 26 ...
$ mad : Factor w/ 10464 levels " Defne Market ",..: 405 8075 9710 10145 9297 7973 2542 3892 2759 5769 ...
$ mtip : Factor w/ 29 levels "Abone Bürosu ",..: 2 20 20 2 2 2 2 2 2 2 ...
$ kanal: Factor w/ 2 levels "OB","SS": 2 2 2 2 2 2 2 2 2 2 ...
$ bkod : int 110565 110565 110565 110565 110565 110565 110565 110565 110565 110565 ...
$ bad : Factor w/ 212 levels "4. Levent","500 Evler",..: 167 167 167 167 167 167 167 167 167 167 ...
$ bolge: Factor w/ 12 levels "Adana ?ehiriçi",..: 7 7 7 7 7 7 7 7 7 7 ...
$ sevk : int 2 3 3 3 2 2 2 6 2 2 ...
$ iade : int 2 1 0 2 0 2 1 0 0 2 ...
$ satis: int 0 2 3 1 2 0 1 6 2 0 ...
Run Code Online (Sandbox Code Playgroud)
我想列出所选多个变量的唯一(如SQL的DISTINCT)值.例如,unique(yad)
给我每个42个元素的名称,但我需要提取两列(yad
并且per
一起使用所有唯一组合):
yad per
--- ---
BARBIE AYLIK
BAKUGAN 2 AYLIK
MICKEY MOUSE 2 AYLIK
TINKERBELL 3 AYLIK
... ...
Run Code Online (Sandbox Code Playgroud)
我怎样才能做到这一点?
Jos*_*ien 119
如何使用unique()
自己?
df <- data.frame(yad = c("BARBIE", "BARBIE", "BAKUGAN", "BAKUGAN"),
per = c("AYLIK", "AYLIK", "2 AYLIK", "2 AYLIK"),
hmm = 1:4)
df
# yad per hmm
# 1 BARBIE AYLIK 1
# 2 BARBIE AYLIK 2
# 3 BAKUGAN 2 AYLIK 3
# 4 BAKUGAN 2 AYLIK 4
unique(df[c("yad", "per")])
# yad per
# 1 BARBIE AYLIK
# 3 BAKUGAN 2 AYLIK
Run Code Online (Sandbox Code Playgroud)
mic*_*cah 13
基于任何列都是唯一的,并使用 保留所有其他列dplyr
。
df <- df %>%
distinct(col1, col2, .keep_all = TRUE)
Run Code Online (Sandbox Code Playgroud)
raf*_*ira 12
这是Josh答案的补充.
您还可以在data.table中过滤掉重复的行时保留其他变量的值
例:
library(data.table)
#create data table
dt <- data.table(
V1=LETTERS[c(1,1,1,1,2,3,3,5,7,1)],
V2=LETTERS[c(2,3,4,2,1,4,4,6,7,2)],
V3=c(1),
V4=c(2) )
> dt
# V1 V2 V3 V4
# A B 1 2
# A C 1 2
# A D 1 2
# A B 1 2
# B A 1 2
# C D 1 2
# C D 1 2
# E F 1 2
# G G 1 2
# A B 1 2
# set the key to all columns
setkey(dt)
# Get Unique lines in the data table
unique( dt[list(V1, V2), nomatch = 0] )
# V1 V2 V3 V4
# A B 1 2
# A C 1 2
# A D 1 2
# B A 1 2
# C D 1 2
# E F 1 2
# G G 1 2
Run Code Online (Sandbox Code Playgroud)
警报:如果其他变量中有不同的值组合,那么结果将是
V1和V2的独特组合
有几种方法可以获得一系列因素的所有独特组合.
with(df, interaction(yad, per, drop=TRUE)) # gives labels
with(df, yad:per) # ditto
aggregate(numeric(nrow(df)), df[c("yad", "per")], length) # gives a data frame
Run Code Online (Sandbox Code Playgroud)
归档时间: |
|
查看次数: |
137738 次 |
最近记录: |