1

我有一个最多包含 5 个测量值 (x) 及其相应时间的数据框：

df = structure(list(x1 = c(92.9595722286402, 54.2085219673818, 
46.3227062573019, 
NA, 65.1501442134141, 49.736451235317), time1 = c(43.2715277777778, 
336.625, 483.975694444444, NA, 988.10625, 510.072916666667), 
x2 = c(82.8368681534474, 53.7981639701784, 12.9993531230419, 
NA, 64.5678816290574, 55.331442940348), time2 = c(47.8166666666667, 
732, 506.747222222222, NA, 1455.25486111111, 958.976388888889
), x3 = c(83.5433119686794, 65.723072881366, 19.0147593408309, 
NA, 65.1989838202356, 36.7000828457705), time3 = c(86.5888888888889, 
1069.02083333333, 510.275, NA, 1644.21527777778, 1154.95694444444
), x4 = c(NA, 66.008102917677, 40.6243513885846, NA, 62.1694420909955, 
29.0078249523063), time4 = c(NA, 1379.22986111111, 520.726388888889, 
NA, 2057.20833333333, 1179.86805555556), x5 = c(NA, 61.0047472617535, 
45.324715258421, NA, 59.862110645527, 45.883161439362), time5 = c(NA, 
1825.33055555556, 523.163888888889, NA, 3352.26944444444, 
1364.99513888889)), class = c("tbl_df", "tbl", "data.frame"
), row.names = c(NA, -6L))

“NA”表示该人（行）没有测量值。

我想计算最后一个现有测量值和第一个测量值之间的差异。

所以第一个是 x3 减去 x1 (6.4)，第二个是 -6.8，依此类推。

我尝试了这样的事情，但没有奏效：

df$diff = apply(df %>% select(., contains("x")), 1, function(x) head(x, 
na.rm = T) - tail(x, na.rm=T))

有什么建议么？另外，apply/rowwise 是最有效的方式，还是有一个矢量化函数可以做到这一点？

4

2 回答 2

1

一种矢量化的方法是使用max.col我们得到的位置"first"和使用参数"last"的非 NA 值ties.method

#Get column number of first and last col
first_col <- max.col(!is.na(df[x_cols]), ties.method = "first")
last_col <- max.col(!is.na(df[x_cols]), ties.method = "last")

#subset the dataframe to include only `"x"` cols
new_df <- as.data.frame(df[grep("^x", names(df))])

#Subtract last non-NA value with the first one
df$new_calc <- new_df[cbind(1:nrow(df), last_col)] - 
               new_df[cbind(1:nrow(df), first_col)]

使用apply你可以做

x_cols <- grep("^x", names(df))

df$new_calc <- apply(df[x_cols], 1, function(x) {
    new_x <- x[!is.na(x)]
    if (length(new_x) > 0)
      new_x[length(new_x)] - new_x[1L]
    else NA
})
于 2019-06-30T12:38:27.520 回答
1

我们可以tidyverse在tbl_df. 创建rownames_to_column行名列（gather_ na.rm = TRUE_ diff_ df'firstlast

library(tidyverse)
rownames_to_column(df, 'rn') %>% 
    select(rn, starts_with('x')) %>% 
    gather(key, val, -rn, na.rm = TRUE) %>%
    group_by(rn) %>%
    summarise(Diff = diff(c(first(val), last(val)))) %>% 
    mutate(rn = as.numeric(rn)) %>%
    complete(rn = min(rn):max(rn)) %>% 
    pull(Diff) %>%
    bind_cols(df, new_col = .)
# A tibble: 6 x 11
#     x1 time1    x2  time2    x3  time3    x4 time4    x5 time5 new_col
#  <dbl> <dbl> <dbl>  <dbl> <dbl>  <dbl> <dbl> <dbl> <dbl> <dbl>   <dbl>
#1  93.0  43.3  82.8   47.8  83.5   86.6  NA     NA   NA     NA   -9.42 
#2  54.2 337.   53.8  732    65.7 1069.   66.0 1379.  61.0 1825.   6.80 
#3  46.3 484.   13.0  507.   19.0  510.   40.6  521.  45.3  523.  -0.998
#4  NA    NA    NA     NA    NA     NA    NA     NA   NA     NA   NA    
#5  65.2 988.   64.6 1455.   65.2 1644.   62.2 2057.  59.9 3352.  -5.29 
#6  49.7 510.   55.3  959.   36.7 1155.   29.0 1180.  45.9 1365.  -3.85
于 2019-06-30T16:07:45.347 回答