如何为以其他行值为条件的值差异创建一个新列答案

【问题标题】：How to create a new column for difference of value conditionated to other row values如何为以其他行值为条件的值差异创建一个新列
【发布时间】：2021-12-26 21:59:57
【问题描述】：

我有这个数据集

> dput(head(data_negctr1, 50))
structure(list(ID = c("01", "01", "01", "01", "01", "01", "02", 
"02", "02", "02", "04", "04", "04", "04", "04", "04", "06", "06", 
"06", "06", "06", "06", "07", "07", "07", "07", "07", "07", "08", 
"08", "08", "08", "08", "08", "09", "09", "09", "09", "10", "10", 
"10", "10", "10", "10", "11", "11", "11", "11", "11", "11"), 
    COND = c("NEG-CTR", "NEG-NOC", "NEG-CTR", "NEG-NOC", "NEG-CTR", 
    "NEG-NOC", "NEG-CTR", "NEG-NOC", "NEG-CTR", "NEG-NOC", "NEG-CTR", 
    "NEG-NOC", "NEG-CTR", "NEG-NOC", "NEG-CTR", "NEG-NOC", "NEG-CTR", 
    "NEG-NOC", "NEG-CTR", "NEG-NOC", "NEG-CTR", "NEG-NOC", "NEG-CTR", 
    "NEG-NOC", "NEG-CTR", "NEG-NOC", "NEG-CTR", "NEG-NOC", "NEG-CTR", 
    "NEG-NOC", "NEG-CTR", "NEG-NOC", "NEG-CTR", "NEG-NOC", "NEG-CTR", 
    "NEG-NOC", "NEG-CTR", "NEG-NOC", "NEG-CTR", "NEG-NOC", "NEG-CTR", 
    "NEG-NOC", "NEG-CTR", "NEG-NOC", "NEG-CTR", "NEG-NOC", "NEG-CTR", 
    "NEG-NOC", "NEG-CTR", "NEG-NOC"), SES = c("L", "L", "R", 
    "R", "V", "V", "L", "L", "R", "R", "L", "L", "R", "R", "V", 
    "V", "L", "L", "R", "R", "V", "V", "L", "L", "R", "R", "V", 
    "V", "L", "L", "R", "R", "V", "V", "L", "L", "V", "V", "L", 
    "L", "R", "R", "V", "V", "L", "L", "R", "R", "V", "V"), `LPP2(1000-1500).FCz` = c(-0.397250087672501, 
    1.01519755373062, 0.663115370014327, -1.72160165493474, -0.198736254963744, 
    -3.16101041766438, 9.21248347391488, 4.2029104966206, 4.43538787409554, 
    1.28346027317076, 3.21881665778703, 5.92604591551597, 2.18006640865321, 
    -0.225533814334514, 3.11042068112836, 2.27800090558473, 2.11741401205978, 
    2.41324857857743, 1.84146702695206, 2.1796529384155, 0.70169995101623, 
    1.75290674648806, 5.54833202313032, 6.28414577615156, -0.243931314877871, 
    4.36296654230765, 6.69246804853767, 4.76363910849669, 2.51614439050594, 
    0.544519979811557, 7.01169149718639, -0.0444060583672976, 
    1.37744267413711, 0.973396686787801, 4.96421236672541, 7.15336406198256, 
    4.70492544977759, 12.1173859570715, 3.2659641826697, 2.36748010754166, 
    -1.3357207368765, -0.0920228462827166, -3.27424811388479, 
    3.17906097432267, 5.63727876507353, 1.12124233903887, 5.60016620262074, 
    -0.948471360524649, -0.548079192716476, 4.61956745266038), 
    `LPP2(1000-1500).Cz` = c(2.91913936481813, 1.85489998202138, 
    3.95694572160701, 2.36649557381131, 2.96437294922766, -2.12913230708907, 
    10.0673617448764, 5.28032844958354, 4.54790063971537, 2.16569592764593, 
    2.58426832963409, 5.077230852852, 6.61872855398538, -1.21696526578647, 
    3.44844607014521, 3.02403433835637, 1.31359485387859, 3.04464581710327, 
    -1.66589658134171, 2.27922842547595, -0.119440173019054, 
    0.668265457194833, 7.62396496656744, 8.44945750972713, -0.147829603799427, 
    4.04101914581768, 7.32640154038155, 6.75063445570149, 3.24583129375449, 
    -0.71967320732169, 6.0531518403553, -0.610046927519558, 0.716894061961362, 
    1.15587576192688, 6.44199941531222, 8.83328750324128, 6.44896958389742, 
    10.5178519351064, 5.72982692246142, 5.81594496182059, 1.83049907582528, 
    3.33758848411935, -0.881527211868382, 4.99936090838488, 8.6569554790885, 
    -1.00196034198877, 6.68708434240297, -1.05436002526955, 0.914180881345654, 
    4.57223746317305), `LPP2(1000-1500).Pz` = c(8.38141924882334, 
    4.88842858528025, 3.88831364571248, 5.83201076558185, 6.28027312932027, 
    5.24535230966772, 8.55827491190151, 4.37390045274906, 6.04702803069286, 
    5.29458007289059, 6.30887598671621, 5.9614279900414, 3.66646157996528, 
    0.240956222277802, 1.08242973465635, 2.99896314000211, 2.79229514887897, 
    5.44503756798021, 4.18316343024083, 7.57635217480856, 1.22060069796986, 
    0.43007709818575, 10.1988511186129, 6.91943531570565, 4.85299225301444, 
    4.65269577391896, 8.73650967214853, 7.20593389743811, 0.0275313390411946, 
    -2.25099759243389, 4.84396306101255, 0.490818836727051, -1.33620384654018, 
    -1.28885901608471, 10.2129994727413, 10.8178803270846, 8.11715753014206, 
    13.3060153720352, 7.12058567533851, 7.32756919594656, 4.44482623240119, 
    3.96749443325504, 2.88799699943079, 5.57921718107867, 5.26994720706095, 
    0.150423100714413, 6.7884358866985, 1.43104145556053, -1.35508717090199, 
    5.24004334238142), `LPP2(1000-1500).POz` = c(9.61455213192824, 
    7.21299011910717, 6.46723813995257, 7.25962635499353, 7.91468942320841, 
    9.94838815736199, 5.76901446016799, 1.63411653734436, 4.09091078261671, 
    5.32593378182311, 6.86490468078958, 5.26280996552585, 5.0384350436334, 
    0.361670119961531, 1.07455889922813, 1.65917850515029, 4.15964501337974, 
    5.35297505953096, 4.11643541344308, 6.89982254066399, 0.742475871599007, 
    3.46485267529148, 7.78087935610006, 5.56652323496618, 4.7450495762156, 
    6.10110636998382, 4.48645178533554, 6.25868881853194, -0.823830529987789, 
    -2.33362938716716, 2.65023738859414, -1.22376317353831, -2.42743430604504, 
    -1.97143573568321, 6.51666682131685, 9.0739717272552, 0.649335906601939, 
    10.4599728475769, 3.117902050049, 5.32174579379253, -0.0968694240545827, 
    1.23896069355975, 1.10176599828887, 4.64632804145351, 4.53436293179661, 
    -0.736223349923721, 5.48908879966524, 3.07114450173319, -2.06826278615017, 
    0.745452283454849)), row.names = c(NA, -50L), class = c("tbl_df", 
"tbl", "data.frame"))
>

如您所见，数据采用宽格式，我被要求计算两个条件级别之间的差异。如果我需要创建一个进一步的列/数据集报告值之间的耦合差异并以每个条件级别为条件（NEG-CTR - NEU-NOC），我应该怎么做？

【问题讨论】：

标签： r variables dplyr tidyverse

【解决方案1】：

也许我们需要按差异分组，然后创建新列 - 按“ID”、“SES”分组，循环 acrossmutate 中的数字列，通过创建基于 ' 的逻辑向量来子集值COND'值，获取差异（-）并通过修改.names添加新列，即在原始列名中添加后缀_diff

library(dplyr)
data_negctr2 <- data_negctr1 %>%
    group_by(ID, SES) %>% 
    mutate(across(where(is.numeric), ~ .[COND == 'NEG-CTR'] - 
        .[COND == 'NEG-NOC'], .names = "{.col}_diff")) %>%
    ungroup

-输出

> print(head(data_negctr2, 2), Inf)
# A tibble: 2 × 11
  ID    COND    SES   `LPP2(1000-1500).FCz` `LPP2(1000-1500).Cz` `LPP2(1000-1500).Pz` `LPP2(1000-1500).POz` `LPP2(1000-1500).FCz_diff`
  <chr> <chr>   <chr>                 <dbl>                <dbl>                <dbl>                 <dbl>                      <dbl>
1 01    NEG-CTR L                    -0.397                 2.92                 8.38                  9.61                      -1.41
2 01    NEG-NOC L                     1.02                  1.85                 4.89                  7.21                      -1.41
  `LPP2(1000-1500).Cz_diff` `LPP2(1000-1500).Pz_diff` `LPP2(1000-1500).POz_diff`
                      <dbl>                     <dbl>                      <dbl>
1                      1.06                      3.49                       2.40
2                      1.06                      3.49                       2.40

如果我们需要为每个组创建一个观察，请使用summarise

library(stringr)
data_negctr3 <- data_negctr1 %>% 
        group_by(ID, SES) %>%  
        summarise(across(where(is.numeric),
       ~ .[COND == 'NEG-CTR'] -  .[COND == 'NEG-NOC'],
        .names = "{.col}_diff"), 
      COND = str_c(COND, collapse = ' - '), .groups = 'drop')

-输出

> data_negctr3
# A tibble: 25 × 7
   ID    SES   `LPP2(1000-1500).FCz_diff` `LPP2(1000-1500).Cz_diff` `LPP2(1000-1500).Pz_diff` `LPP2(1000-1500).POz_diff` COND           
   <chr> <chr>                      <dbl>                     <dbl>                     <dbl>                      <dbl> <chr>          
 1 01    L                         -1.41                      1.06                      3.49                       2.40  NEG-CTR - NEG-…
 2 01    R                          2.38                      1.59                     -1.94                      -0.792 NEG-CTR - NEG-…
 3 01    V                          2.96                      5.09                      1.03                      -2.03  NEG-CTR - NEG-…
 4 02    L                          5.01                      4.79                      4.18                       4.13  NEG-CTR - NEG-…
 5 02    R                          3.15                      2.38                      0.752                     -1.24  NEG-CTR - NEG-…
 6 04    L                         -2.71                     -2.49                      0.347                      1.60  NEG-CTR - NEG-…
 7 04    R                          2.41                      7.84                      3.43                       4.68  NEG-CTR - NEG-…
 8 04    V                          0.832                     0.424                    -1.92                      -0.585 NEG-CTR - NEG-…
 9 06    L                         -0.296                    -1.73                     -2.65                      -1.19  NEG-CTR - NEG-…
10 06    R                         -0.338                    -3.95                     -3.39                      -2.78  NEG-CTR - NEG-…
# … with 15 more rows

【讨论】：

@mały_statystyczny 如果您需要单行，您可能需要summarise 而不是mutate。我不确定模型是否适合重复值
我认为 summarise() 数据集应该是正确的。即使 COND 列丢失。差异似乎是我寻找的那个
一如既往的感谢 ;)
@mały_statystyczny 这很简单，我们可以做到data_negctr1 %>% group_by(ID, SES) %>% summarise(across(where(is.numeric), ~ .[COND == 'NEG-CTR'] - .[COND == 'NEG-NOC'], .names = "{.col}_diff"), COND = str_c(COND, collapse = ' - '), .groups = 'drop')
非常感谢;)