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基于滤波准则的突变在DPLYR中的应用

mutate dplyr filter r

Lmm · 技术社区 · 7 年前

好的,我有一个时间序列数据集(见虚构数据),当值B满足某个标准(>100),当B是lt;100时,我想改变值A,并生成一个新的变量(Xi,XII,Xiii)。to=值A。

我已经设法做到了这一点,但是这让我在B<100时在新值中使用了NAS—是否有办法更有效地将其连接在一起:

df2 <- df %>%
left_join( #this will grab the  value when B > 100 of each day
df %>% filter(B >100) %>% select(B, X = A),
by = 'B'
 ) %>% 
  mutate(
  Xi = ( X*0.1 ) + A
  )%>% 
mutate(
  Xii = ( X*0.5 ) + A
)%>% 
mutate(
   Xiii = ( X*0.9 ) + A
)

***编辑输出应如下所示:

output <- structure(list(datetime = structure(c(1467354545, 1467358145, 
1467361745, 1467365345, 1467368945, 1467372545), class = c("POSIXct", 
"POSIXt"), tzone = "Canada/Mountain"), A = c(0.3896284155, 0.5047586155, 
0.4988, 0.5978083335, 0.610775, 0.665325), B = c(74.41898788, 
58.563069355, 43.227076165, 26.66528807, 13.70157303, 6.8983046415
), Xi = c(0.42859125705, 0.55523447705, NA, NA, NA, NA), Xii = 
c(0.58444262325, 
0.75713792325, NA, NA, NA, NA), Xiii = c(0.74029398945, 0.95904136945, 
NA, NA, NA, NA)), row.names = c(NA, 6L), class = "data.frame")

但是,如果NAS是这些值,则应该=A。

### dummy data
df<-structure(list(datetime = structure(c(1467354545, 1467358145, 
1467361745, 1467365345, 1467368945, 1467372545, 1467376145, 1467379745, 
1467383345, 1467386945, 1467390545, 1467394145, 1467397745, 1467401345, 
1467404945, 1467408545, 1467412145, 1467415745, 1467419345, 1467422945, 
1467426545, 1467430145, 1467433745, 1467437345, 1467440945, 1467444545, 
1467448145, 1467451745, 1467455345, 1467458945, 1467462545, 1467466145, 
1467469745, 1467473345, 1467476945, 1467480545, 1467484145, 1467487745, 
1467491345, 1467494945, 1467498545, 1467502145, 1467505745, 1467509345, 
1467512945, 1467516545, 1467520145, 1467523745), class = c("POSIXct", 
"POSIXt"), tzone = "Canada/Mountain"), A = c(0.3896284155, 0.5047586155, 
 0.4988, 0.5978083335, 0.610775, 0.665325, 0.61445, 0.6224416665, 
 0.6195833335, 0.6156666665, 0.628616667, 0.594925, 0.592366667, 
 0.448325, 0.4093166665, 0.4076166665, 0.4199666665, 0.3724, 0.3550416665, 
 0.4245833335, 0.406518785, 0.2448416665, 0.2305666665, 0.325375, 
 0.3413166665, 0.409642232, 0.400575, 0.4452833335, 0.48665, 0.5067833335, 
 0.563375, 0.5323416665, 0.499816667, 0.49065, 0.480566667, 0.4535833335, 
 0.491475, 0.370733333, 0.2932916665, 0.307991667, 0.375575, 0.363816667, 
 0.220041667, 0.2227583335, 0.2400166665, 0.226133333, 0.25015, 
 0.286516667), B = c(74.41898788, 58.563069355, 43.227076165, 
 26.66528807, 13.70157303, 6.8983046415, 4.051383457, 3.9928000525, 
 6.6264775935, 11.56529787, 19.381280495, 39.294139195, 63.738904315, 
 87.15298014, 99.92137692, 123.905882, 132.52431235, 122.1168814, 
 112.8298307, 70.39181064, 68.67000666, 181.4195973, 179.78868045, 
 95.658796805, 70.15210006, 53.19819125, 57.79353269, 21.815294445, 
 17.00485432, 8.294689507, 6.159616564, 6.447905295, 11.06234652, 
 18.723238995, 34.42328062, 61.348978945, 71.989398775, 113.3879366, 
 124.0605001, 122.3875736, 101.9667852, 91.15826362, 192.1395999, 
 192.19658895, 180.7257523, 171.3751495, 157.8631188, 126.9319141
  )), class = "data.frame", row.names = c(NA, -48L))

2 回复 | 直到 7 年前

Kerry Jackson 7 年前

library(dplyr)

df2 <- 
  df %>%
  mutate(x = if_else(B > 100, A, A),
         Xi = if_else(B > 100,  x*0.1 + A, A),
         Xii = if_else(B > 100,  x*0.5 + A, A),
         Xiii = if_else(B > 100,  x*0.9 + A, A))

DuckPyjamas 7 年前

除了可复制的示例外,还应提供预期的结果。

为什么要创建新的变量,以后是否要将它们折叠成一列?如果是这样,您现在就可以创建单列了。你可以使用 ifelse() 两种结果,或 dplyr::case_when() 超过两个。

library(dplyr)

iris %>% 
    mutate(new_petal_length = case_when(Petal.Length <= 2 ~ "A",
                                        Petal.Length <= 5 ~ "B",
                                                     TRUE ~ "C")) %>% 
    select(Petal.Length, new_petal_length) %>% 
    sample_n(10)

#> Petal.Length new_petal_length
#>          6.6                C
#>          3.9                B
#>          1.4                A
#>          4.5                B
#>          3.0                B
#>          1.4                A
#>          4.5                B
#>          5.0                B
#>          1.5                A
#>          5.5                C

注意到 case_when() 一旦达到真正的结果就停止评估,这就是为什么 Petal.Length <= 5 不覆盖 Petal.Length <= 2 . 决赛 TRUE 当前面的所有事例都没有计算为时的catch all值 真的 .