Question

我正在使用一些时间序列（瞳孔扩大）数据，并希望根据不同的因子变量（Time）过滤不同的时间范围（SOA）

示例数据：

library(dplyr)        

Data <- structure(list(Subject = structure(c(1L, 1L, 2L, 2L, 1L, 1L, 
        2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 
        2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L), .Label = c("12", "14", 
        "15", "16", "18", "20", "21", "22", "23", "28", "29", "30", "33", 
        "36", "37", "38", "40", "42", "43", "44"), class = "factor"), 
        SOA = structure(c(2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 
        2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 
        1L, 2L, 1L, 2L, 1L, 2L, 1L), .Label = c("Long SOA", "Short SOA"
        ), class = "factor"), Time = c(-66.68, -66.68, -66.68, -66.68, 
        -33.34, -33.34, -33.34, -33.34, 0, 0, 0, 0, 33.34, 33.34, 
        33.34, 33.34, 66.68, 66.68, 66.68, 66.68, 100.02, 100.02, 
        100.02, 100.02, 133.36, 133.36, 133.36, 133.36, 166.7, 166.7, 
        166.7, 166.7), Pcent_Chng = c(0.14391, 0.076759, -0.022377, 
        0.038111, 0.21093, 0.11448, -0.0047064, 0.078232, 0.27924, 
        0.1527, -0.0085276, 0.12385, 0.38328, 0.21299, 0.01988, 0.15626, 
        0.47471, 0.25357, 0.050318, 0.20517, 0.58012, 0.2888, 0.080629, 
        0.20616, 0.65861, 0.33622, 0.12892, 0.20832, 0.75277, 0.38181, 
        0.17921, 0.21789)), class = "data.frame", row.names = c(NA, 
        -32L), .Names = c("Subject", "SOA", "Time", "Pcent_Chng"))

我想对Time的{{1}}期间的SOA = "Short"期间的平均时间比对SOA = "Long"的平均值要长。

我已经针对Type = "Word"（filter之前和之后的group_by）尝试了这些内容：

Data %>% 
filter(Time[SOA = "Short SOA"] >= 0 & Time[SOA = "Short SOA"] <= 100,
       Time[SOA = "Long SOA"] >= 0 & Time[SOA = "Long SOA"] <= 150) %>%
group_by(Subject,SOA) %>%
summarize(Word_Avg_Pcent = mean(Pcent_Chng,na.rm=TRUE))

和

Data %>% 
group_by(Subject,SOA) %>%
filter(Time[SOA = "Short SOA"] >= 0 & Time[SOA = "Short SOA"] <= 100,
       Time[SOA = "Long SOA"] >= 0 & Time[SOA = "Long SOA"] <= 150) %>%
summarize(Word_Avg_Pcent = mean(Pcent_Chng,na.rm=TRUE))

两者都会产生空数据帧;列在那里，但没有数据。如果我不使用第二个过滤器，我会得到一个完整的数据框。

有没有办法在dplyr链中使用管道和过滤器来实现我想要的东西？

Answer 1

如评论中所述，您需要OR |您正在寻找的两个AND &条件。

您的过滤器是：

filter(Time[SOA = "Short SOA"] >= 0 & Time[SOA = "Short SOA"] <= 1200, 
       Time[SOA = "Long SOA"] >= 0 & Time[SOA = "Long SOA"] <= 3000)

哪些语句不能评估为逻辑（例如SOA =＆＃34; Short SOA＆＃34;）。你需要做的是更明确。

您希望过滤到SOA等同于＆＃34; Short SOA＆＃34;并且时间值介于0到1200之间，或者SOA等于＆＃34; Long SOA＆＃34;并且时间值介于0到3000之间。

SOA == "Short SOA" and 0 <= Time <= 1200 OR SOA == "Long SOA" and 0 <= Time <= 1200

您可以使用between中的dplyr作为时间条件。

实施的是：

library(tidyverse)

Data <- eval(parse(file("http://pastebin.com/raw.php?i=VTWCVgCA")))

Data %>% 
  gather(Sample, Prop_Chng, X.8:X100) %>%
  mutate(Sample = gsub("[.]","-", Sample)) %>%
  mutate(Sample = as.numeric(gsub("X","", Sample))) %>%
  mutate(Time = Sample*33.34) %>%
  mutate(Pcent_Chng = Prop_Chng*100) %>%
  filter(Type == "Word") %>% 
  filter((SOA == "Short SOA" & between(Time, 0, 1200)) |  (SOA == "Long SOA" & between(Time, 0, 3000))) %>% 
  group_by(Subject, NsCond,Close,SOA) %>%
  summarize(Word_Avg_Pcent = mean(Pcent_Chng,na.rm=TRUE))

使用dplyr过滤器

1 个答案: