R语言 计算两个或多个时间点的评分变化

u1ehiz5o  于 2023-01-10  发布在  其他
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我有一个很简单的问题:想象一个认知测试有6个项目,每个项目都在0-10量表上单独评分,现在假设5名患者在一天中的3个(或更多)不同时间进行了该测试。

df <- tibble(
  id = rep(1:5, each = 3),
  time = rep(c("morning", "evening", "night"), each = 1, times = 5),
  i1 = round(runif(15, 0, 10)),
  i2 = round(runif(15, 0, 10)),
  i3 = round(runif(15, 0, 10)),
  i4 = round(runif(15, 0, 10)),
  i5 = round(runif(15, 0, 10)),
  i6 = round(runif(15, 0, 10))
)

现在,我如何计算每位患者在一天中不同时间的每个单项评分变化?tidyverse方法非常受欢迎。
P.S:输出最好是这样的:

df <- tibble(
  id = rep(1:5, each = 3),
  time = rep(c("morning", "evening", "night"), each = 1, times = 5),
  i1 = round(runif(15, 0, 10)),
  i2 = round(runif(15, 0, 10)),
  i3 = round(runif(15, 0, 10)),
  i4 = round(runif(15, 0, 10)),
  i5 = round(runif(15, 0, 10)),
  i6 = round(runif(15, 0, 10)),
  diff_morning_evening = rep(NA, 15),
  diff_morning_night = rep(NA, 15),
  diff_evening_night = rep(NA, 15),
)
nnsrf1az

nnsrf1az1#

以简洁方式计算变化的方法:

  • 第1行:* 上午 * - * 晚上 *
  • 第2行:* 上午 * - * 晚上 *
  • 第3行:* 晚上 * - * 晚上 *
df %>% 
  group_by(id) %>% 
  mutate(across(i1:i6, ~ ifelse(is.na(.x - lag(.x)), 
                           lead(.x, 2) - .x, .x - lag(.x)), 
           .names="{.col}_chg")) %>% 
  ungroup()
# A tibble: 15 × 14
      id time       i1    i2    i3    i4    i5    i6 i1_chg i2_chg i3_chg i4_chg
   <int> <chr>   <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>  <dbl>  <dbl>  <dbl>  <dbl>
 1     1 morning     1     2     4     9     5     6      7     -1      1     -8
 2     1 evening     5     0     4     8     9     4      4     -2      0     -1
 3     1 night       8     1     5     1     4     5      3      1      1     -7
 4     2 morning     7     7     4     3     2     9     -5     -1      4      5
 5     2 evening     8     9     1     2     4     5      1      2     -3     -1
 6     2 night       2     6     8     8    10     2     -6     -3      7      6
 7     3 morning     9     6     6     1     5     5     -8     -1      2      0
 8     3 evening     3     2     8     1     3    10     -6     -4      2      0
 9     3 night       1     5     8     1     3     3     -2      3      0      0
10     4 morning     7     2     9     1     5     8      1      1     -6      0
11     4 evening     3     5     9     5     6     3     -4      3      0      4
12     4 night       8     3     3     1     3     2      5     -2     -6     -4
13     5 morning     4     1     3     7     1     0      4      6      4      2
14     5 evening     7     2     7     7     5     2      3      1      4      0
15     5 night       8     7     7     9     8     4      1      5      0      2
# … with 2 more variables: i5_chg <dbl>, i6_chg <dbl>

特殊行求和法

df %>% 
  group_by(id) %>% 
  summarize(across(i1:i6, ~ 
    .x[which(time == "morning")] + .x[which(time == "night")]))
# A tibble: 5 × 7
     id    i1    i2    i3    i4    i5    i6
  <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1     1     9     3     9    10     9    11
2     2     9    13    12    11    12    11
3     3    10    11    14     2     8     8
4     4    15     5    12     2     8    10
5     5    12     8    10    16     9     4
gxwragnw

gxwragnw2#

您可以使用combn生成2次的所有组合,并通过设置FUN = diffdiff应用于每个组合以计算变化。

library(tidyverse)

df %>% 
  group_by(id) %>% 
  summarise(DIFF = combn(time, 2, paste, collapse = '_'),
            across(i1:i6, ~ combn(.x, 2, diff)), .groups = 'drop')

# # A tibble: 15 × 8
#       id DIFF                   i1        i2        i3        i4        i5        i6
#    <int> <chr[1d]>       <dbl[1d]> <dbl[1d]> <dbl[1d]> <dbl[1d]> <dbl[1d]> <dbl[1d]>
#  1     1 morning_evening         1        -2         0         5         1         3
#  2     1 morning_night          -2         5         1         3        -2        -1
#  3     1 evening_night          -3         7         1        -2        -3        -4
#  4     2 morning_evening         1         5        -2        -3         5         7
#  5     2 morning_night           2        -1         6         2        -2         1
#  6     2 evening_night           1        -6         8         5        -7        -6
# ...

如果需要宽格式输出,请将上面的代码传递给tidyr::pivot_wider()

... %>%
  pivot_wider(names_from = DIFF, values_from = i1:i6, names_prefix = 'diff_')

# # A tibble: 5 × 19
#      id i1_diff_morning_evening i1_diff_morning_night i1_diff_evening_night i2_diff_morning_evening i2_diff_morning_night i2_diff_evening_night
#   <int>                   <dbl>                 <dbl>                 <dbl>                   <dbl>                 <dbl>                 <dbl>
# 1     1                       1                    -2                    -3                      -2                     5                     7
# 2     2                       1                     2                     1                       5                    -1                    -6
# 3     3                       2                    -1                    -3                       5                    -1                    -6
# 4     4                       1                     2                     1                      -2                    -4                    -2
# 5     5                       6                     7                     1                       1                     0                    -1
#
# … with 12 more variables:
# i3_diff_morning_evening <dbl>, i3_diff_morning_night <dbl>, i3_diff_evening_night <dbl>,
# i4_diff_morning_evening <dbl>, i4_diff_morning_night <dbl>, i4_diff_evening_night <dbl>,
# i5_diff_morning_evening <dbl>, i5_diff_morning_night <dbl>, i5_diff_evening_night <dbl>,
# i6_diff_morning_evening <dbl>, i6_diff_morning_night <dbl>, i6_diff_evening_night <dbl>

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