Question

我正在处理随时间变化的数据集，需要计算峰值变化发生的时间。我遇到了一个问题，因为有些科目缺少数据（NA＆＃39;）。

示例：

library(dplyr)

Data <- structure(list(Subject = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 6L, 6L, 
6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 
6L, 6L), .Label = c("1", "10", "11", "12", "13", "14", "16", 
"17", "18", "19", "2", "20", "21", "22", "23", "24", "25", "26", 
"27", "28", "29", "3", "31", "32", "4", "5", "7", "8", "9"), class = "factor"), 
Close = 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, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L
), .Label = c("High Predictability", "Low Predictability"
), 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, 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, -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), Pcent_Chng = c(0.12314, 
0.048254, -0.098007, 0.023216, 0.20327, 0.08338, -0.15157, 
0.030008, 0.26442, 0.12019, -0.22878, 0.035547, 0.31849, 
0.15488, -0.26887, 0.038992, 0.39489, 0.15112, -0.31185, 
0.02144, NA, 0.046474, NA, 0.17541, NA, 0.14975, NA, 0.3555, 
NA, -0.1736, NA, 0.72211, NA, -0.32201, NA, 1.0926, NA, -0.39551, 
0.72211, 1.4406)), class = "data.frame", row.names = c(NA, -40L
), .Names = c("Subject", "Close", "SOA", "Time", "Pcent_Chng"
))

我在以下尝试时遇到错误：

Data %>%
group_by(Subject,Close,SOA) %>%
summarize(Peak_Pcent = max(Pcent_Chng), 
                    Peak_Latency = Time[which.max(Pcent_Chng)])

错误是：

Error in summarise_impl(.data, dots) : 
  Column `Peak_Latency` must be length 1 (a summary value), not 0

这似乎是由于NA，仅在某些SOA条件下。将complete.cases()与我的实际数据一起使用过于激进并删除过多数据。

是否有一种解决方法可以忽略NA？

Answer 1

这应该可以解决问题：

Data %>% 
  group_by(Subject, Close, SOA) %>% 
  mutate(Peak_Pcent = max(Pcent_Chng)) %>% 
  arrange(Subject, Close, SOA) %>% 
  filter(Peak_Pcent == Pcent_Chng)

输出：

# A tibble: 6 x 6
# Groups:   Subject, Close, SOA [6]
  Subject               Close       SOA   Time Pcent_Chng Peak_Pcent
   <fctr>              <fctr>    <fctr>  <dbl>      <dbl>      <dbl>
1       1 High Predictability  Long SOA  33.34   0.154880   0.154880
2       1 High Predictability Short SOA  66.68   0.394890   0.394890
3       1  Low Predictability  Long SOA  33.34   0.038992   0.038992
4       1  Low Predictability Short SOA -66.68  -0.098007  -0.098007
5      14 High Predictability  Long SOA -33.34   0.149750   0.149750
6      14  Low Predictability  Long SOA  66.68   1.440600   1.440600

Answer 2

您有一个小组Peak_Pcent全部为NA，另一个小组只有一个Peak_Pcent。我认为最好过滤掉Peak_Pcent所有NA的群组，并在使用na.rm = TRUE功能时设置max。

Data %>%
  group_by(Subject,Close,SOA) %>%
  filter(!all(is.na(Pcent_Chng))) %>% # Filter out groups with Pcent_Chng all is NA
  summarize(Peak_Pcent = max(Pcent_Chng, na.rm = TRUE),  # Set na.rm = TRUE
            Peak_Latency = Time[which.max(Pcent_Chng)]) 

# # A tibble: 7 x 5
# # Groups:   Subject, Close [?]
# Subject               Close       SOA Peak_Pcent Peak_Latency
# <fctr>              <fctr>    <fctr>      <dbl>        <dbl>
# 1       1 High Predictability  Long SOA   0.154880        33.34
# 2       1 High Predictability Short SOA   0.394890        66.68
# 3       1  Low Predictability  Long SOA   0.038992        33.34
# 4       1  Low Predictability Short SOA  -0.098007       -66.68
# 5      14 High Predictability  Long SOA   0.149750       -33.34
# 6      14  Low Predictability  Long SOA   1.440600        66.68
# 7      14  Low Predictability Short SOA   0.722110        66.68

Dplyr总结了哪个.max和数据与NA＆＃39; s

2 个答案: