英文:
convert large to long
问题
我想将数据转换为长格式。这是一个重复测量设计，有3个条件。
这是我目前的数据：
| 参与者编号 | 测量1（条件1） | 测量1（条件2） | 测量1（条件3） | 测量2（条件1） | 测量2（条件2） | 测量2（条件3） | 测量3（条件1） | 测量3（条件2） | 测量3（条件3） | 年龄 | 性别 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| 1 | |||||||||||
| 2 | |||||||||||
| 3 |
我希望得到的结果是：
| 参与者编号 | 条件 | 测量1 | 测量2 | 测量3 | 年龄 | 性别 |
|---|---|---|---|---|---|---|
| 1 | 1 | |||||
| 1 | 2 | |||||
| 1 | 3 |
我尝试了以下代码：data_long <- gather(df, condition_measure1, measure1, measure1_cond1, measure1_cond2, measure1_cond3, factor_key=TRUE)。
这个方法可以工作，但是我不知道如何将所有3个条件都转换成长格式。我尝试重复相同的代码来处理测量2，但是它没有起作用，它为每个参与者添加了另外3行。
你能帮助我吗？我对R非常陌生，所以请原谅我，我只是想转换我的数据并回到Jamovi ^^
谢谢！
编辑：这是数据的结构：
structure(list(num_pp = c(1, 2, 3, 4, 5, 6), nombre_dp1 = c(24,14, 2, 6, 6, 21), nombre_dp05 = c(20, 28, 2, 9, 8, 21), nombre_dp0 = c(24,20, 4, 11, 8, 20), jugement_causal_dp1 = c("Oui", "Oui", "Oui","Oui", "Oui", "Oui"), jugement_causal_dp05 = c("Non", "Oui","Non", "Non", "Oui", "Non"), jugement_causal_dp0 = c("Non", "Non","Oui", "Non", "Non", "Non"), confiance_dp1 = c(90, 80, 63, 80,90, 80), confiance_dp05 = c(60, 50, 86, 65, 50, 90), confiance_dp0 = c(65,60, 55, 43, 50, 80), age = c(33, 22, 20, 20, 18, 18), genre = c("Masculin","Feminin", "Feminin", "Feminin", "Feminin", "Feminin"), etude = c("L1","L1", "L1", "L1", "L1", "L1"), ordre = c("dp_05|dp_1|dp_0", "dp_0|dp_1|dp_05","dp_0|dp_1|dp_05", "dp_0|dp_05|dp_1", "dp_1|dp_05|dp_0", "dp_1|dp_05|dp_0"), wdif_dp1dp05 = c(-4, 14, 0, 3, 2, 0)), row.names = c(NA, -6L), class = c("tbl_df", "tbl", "data.frame"))
英文:
I want to convert data to a long format. It's a repeated measure design, with 3 conditions.
This is what I have :
| participant id | measure 1 (cond1) | measure 1 (cond2) | measure 1 (cond3) | measure2 (cond1) | measure2 (cond2) | measure2 (cond3) | measure3 (cond1) | measure3(cond2) | measure3 (cond3) | age | gender |
|---|---|---|---|---|---|---|---|---|---|---|---|
| 1 | |||||||||||
| 2 | |||||||||||
| 3 |
And this is what I would like:
| participant id | condition | measure1 | measure2 | measure3 | age | gender |
|---|---|---|---|---|---|---|
| 1 | 1 | |||||
| 1 | 2 | |||||
| 1 | 3 |
I Tried data_long <- gather(df, condition_measure1, measure1, measure1_cond1, measure1_cond2, measure1_cond3, factor_key=TRUE)
It works, but if I don't know how to put all 3 conditions in long format. I tried repeating the same code but for measure2, it did not work, it added another 3 rows for each participant.
Can you hep me ? I a very new to R, so forgive me, I just want to convert my data and go back to Jamovi ^^
Thank you!
edit: here is the data
structure(list(num_pp = c(1, 2, 3, 4, 5, 6), nombre_dp1 = c(24,14, 2, 6, 6, 21), nombre_dp05 = c(20, 28, 2, 9, 8, 21), nombre_dp0 = c(24,20, 4, 11, 8, 20), jugement_causal_dp1 = c("Oui", "Oui", "Oui","Oui", "Oui", "Oui"), jugement_causal_dp05 = c("Non", "Oui","Non", "Non", "Oui", "Non"), jugement_causal_dp0 = c("Non", "Non","Oui", "Non", "Non", "Non"), confiance_dp1 = c(90, 80, 63, 80,90, 80), confiance_dp05 = c(60, 50, 86, 65, 50, 90), confiance_dp0 = c(65,60, 55, 43, 50, 80), age = c(33, 22, 20, 20, 18, 18), genre = c("Masculin","Feminin", "Feminin", "Feminin", "Feminin", "Feminin"), etude = c("L1","L1", "L1", "L1", "L1", "L1"), ordre = c("dp_05|dp_1|dp_0", "dp_0|dp_1|dp_05","dp_0|dp_1|dp_05", "dp_0|dp_05|dp_1", "dp_1|dp_05|dp_0", "dp_1|dp_05|dp_0"), wdif_dp1dp05 = c(-4, 14, 0, 3, 2, 0)), row.names = c(NA, -6L), class = c("tbl_df", "tbl", "data.frame"))
答案1
得分: 2
使用OP提供的数据，您可以直接使用pivot_longer函数，如下所示：
df %>%pivot_longer(matches('_dp\\d+$'), names_to = c('.value', 'dp'),names_pattern = '(.*)_(\\w+)')# A tibble: 18 × 10num_pp age genre etude ordre wdif_dp1dp05 dp nombre jugement_causal confiance<dbl> <dbl> <chr> <chr> <chr> <dbl> <chr> <dbl> <chr> <dbl>1 1 33 Masculin L1 dp_05… -4 dp1 24 Oui 902 1 33 Masculin L1 dp_05… -4 dp05 20 Non 603 1 33 Masculin L1 dp_05… -4 dp0 24 Non 654 2 22 Feminin L1 dp_0|… 14 dp1 14 Oui 805 2 22 Feminin L1 dp_0|… 14 dp05 28 Oui 506 2 22 Feminin L1 dp_0|… 14 dp0 20 Non 607 3 20 Feminin L1 dp_0|… 0 dp1 2 Oui 638 3 20 Feminin L1 dp_0|… 0 dp05 2 Non 869 3 20 Feminin L1 dp_0|… 0 dp0 4 Oui 5510 4 20 Feminin L1 dp_0|… 3 dp1 6 Oui 8011 4 20 Feminin L1 dp_0|… 3 dp05 9 Non 6512 4 20 Feminin L1 dp_0|… 3 dp0 11 Non 4313 5 18 Feminin L1 dp_1|… 2 dp1 6 Oui 9014 5 18 Feminin L1 dp_1|… 2 dp05 8 Oui 5015 5 18 Feminin L1 dp_1|… 2 dp0 8 Non 5016 6 18 Feminin L1 dp_1|… 0 dp1 21 Oui 8017 6 18 Feminin L1 dp_1|… 0 dp05 21 Non 9018 6 18 Feminin L1 dp_1|… 0 dp0 20 Non 80
我们在names_to中使用.value，以确保这3个测量值分布在不同的列中。
英文:
Using the data OP provided, you could directly use pivot_longer as shown below:
df %>%pivot_longer(matches('_dp\\d+$'), names_to = c('.value', 'dp'),names_pattern = '(.*)_(\\w+)')# A tibble: 18 × 10num_pp age genre etude ordre wdif_dp1dp05 dp nombre jugement_causal confiance<dbl> <dbl> <chr> <chr> <chr> <dbl> <chr> <dbl> <chr> <dbl>1 1 33 Masculin L1 dp_05… -4 dp1 24 Oui 902 1 33 Masculin L1 dp_05… -4 dp05 20 Non 603 1 33 Masculin L1 dp_05… -4 dp0 24 Non 654 2 22 Feminin L1 dp_0|… 14 dp1 14 Oui 805 2 22 Feminin L1 dp_0|… 14 dp05 28 Oui 506 2 22 Feminin L1 dp_0|… 14 dp0 20 Non 607 3 20 Feminin L1 dp_0|… 0 dp1 2 Oui 638 3 20 Feminin L1 dp_0|… 0 dp05 2 Non 869 3 20 Feminin L1 dp_0|… 0 dp0 4 Oui 5510 4 20 Feminin L1 dp_0|… 3 dp1 6 Oui 8011 4 20 Feminin L1 dp_0|… 3 dp05 9 Non 6512 4 20 Feminin L1 dp_0|… 3 dp0 11 Non 4313 5 18 Feminin L1 dp_1|… 2 dp1 6 Oui 9014 5 18 Feminin L1 dp_1|… 2 dp05 8 Oui 5015 5 18 Feminin L1 dp_1|… 2 dp0 8 Non 5016 6 18 Feminin L1 dp_1|… 0 dp1 21 Oui 8017 6 18 Feminin L1 dp_1|… 0 dp05 21 Non 9018 6 18 Feminin L1 dp_1|… 0 dp0 20 Non 80
We use .value within the names_to to ensure that the 3 measure values are spread across different columns.
答案2
得分: 1
假设数据如下所示：
df <- structure(list(`participant id` = c(1, 2, 3), `measure 1 (cond1)` = c(10,12, 8), `measure 1 (cond2)` = c(15, 14, 9), `measure 1 (cond3)` = c(20,18, 10), `measure2 (cond1)` = c(25, 22, 15), `measure2 (cond2)` = c(30,28, 19), `measure2 (cond3)` = c(35, 30, 22), `measure3 (cond1)` = c(40,38, 25), `measure3 (cond2)` = c(45, 42, 28), `measure3 (cond3)` = c(50,48, 32), age = c(25, 30, 27), gender = c("Male", "Female", "Male")), class = "data.frame", row.names = c(NA, -3L))
你可以这样操作：
library(dplyr)library(tidyr)df <- pivot_longer(df,cols = starts_with("measure"),names_pattern = "measure ?(\\d+) \\(cond(\\d+)\\)",names_to = c("measure", "condition")) %>%mutate(condition = as.integer(condition),measure = as.integer(measure))# 输出结果：`participant id` age gender measure condition value<dbl> <dbl> <chr> <int> <int> <dbl>1 1 25 Male 1 1 102 1 25 Male 1 2 153 1 25 Male 1 3 204 1 25 Male 2 1 255 1 25 Male 2 2 306 1 25 Male 2 3 357 1 25 Male 3 1 408 1 25 Male 3 2 459 1 25 Male 3 3 5010 2 30 Female 1 1 12# ℹ 还有17行数据
要恢复为宽格式：
df %>% pivot_wider(names_from = measure, values_from = value, names_prefix = "measure")
英文:
Assuming the data is something like this:
df <- structure(list(`participant id` = c(1, 2, 3), `measure 1 (cond1)` = c(10,12, 8), `measure 1 (cond2)` = c(15, 14, 9), `measure 1 (cond3)` = c(20,18, 10), `measure2 (cond1)` = c(25, 22, 15), `measure2 (cond2)` = c(30,28, 19), `measure2 (cond3)` = c(35, 30, 22), `measure3 (cond1)` = c(40,38, 25), `measure3 (cond2)` = c(45, 42, 28), `measure3 (cond3)` = c(50,48, 32), age = c(25, 30, 27), gender = c("Male", "Female", "Male")), class = "data.frame", row.names = c(NA, -3L))
You can do something like this:
library(dplyr)library(tidyr)df <- pivot_longer(df,cols = starts_with("measure"),names_pattern = "measure ?(\\d+) \\(cond(\\d+)\\)",names_to = c("measure", "condition")) |>mutate(condition = as.integer(condition),measure = as.integer(measure))# Output:`participant id` age gender measure condition value<dbl> <dbl> <chr> <int> <int> <dbl>1 1 25 Male 1 1 102 1 25 Male 1 2 153 1 25 Male 1 3 204 1 25 Male 2 1 255 1 25 Male 2 2 306 1 25 Male 2 3 357 1 25 Male 3 1 408 1 25 Male 3 2 459 1 25 Male 3 3 5010 2 30 Female 1 1 12# ℹ 17 more rows
To make it wide again:
df |> pivot_wider(names_from = measure, values_from = value, names_prefix = "measure")
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