如何根据仅复制一次的行减去 dataframe 中的多列值？
[英]How to substract multiple columns of values in dataframe based on the row being duplicated just once?
问题描述
Imagine I have a dataframe as such:
想象一下，我有一个 dataframe 如下：
#Stack example
df <- data.frame(DATE = c("2022-08-29", "2022-08-29", "2022-08-29", "2022-08-29", "2022-08-29", "2022-08-29",
"2022-09-05", "2022-09-05", "2022-09-05", "2022-09-05", "2022-09-05", "2022-09-05",
"2022-08-29", "2022-08-29", "2022-08-29", "2022-08-29", "2022-08-29", "2022-08-29",
"2022-09-05", "2022-09-05", "2022-09-05", "2022-09-05", "2022-09-05", "2022-09-05",
"2022-08-29", "2022-08-29", "2022-08-29", "2022-08-29", "2022-08-29", "2022-08-29",
"2022-09-05", "2022-09-05", "2022-09-05", "2022-09-05", "2022-09-05", "2022-09-05",
"2022-08-29", "2022-08-29", "2022-08-29", "2022-08-29", "2022-08-29", "2022-08-29",
"2022-09-05", "2022-09-05", "2022-09-05", "2022-09-05", "2022-09-05", "2022-09-05",
"2022-09-05", "2022-08-29", "2022-09-05", "2022-08-29", "2022-09-05", "2022-08-29"),
ID = c("1", "2", "3", "4", "5", "6",
"1", "2", "3", "4", "5", "6",
"1", "2", "3", "4", "5", "6",
"1", "2", "3", "4", "5", "6",
"1", "2", "3", "4", "5", "6",
"1", "2", "3", "4", "5", "6",
"1", "2", "3", "4", "5", "6",
"1", "2", "3", "4", "5", "6",
"7", "8", "9", "10", "9", "10"),
CV = c("SV", "SV", "SV", "SV", "SV", "SV",
"SV", "SV", "SV", "SV", "SV", "SV",
"PD", "PD", "PD", "PD", "PD", "PD",
"PD", "PD", "PD", "PD", "PD", "PD",
"SV", "SV", "SV", "SV", "SV", "SV",
"SV", "SV", "SV", "SV", "SV", "SV",
"PD", "PD", "PD", "PD", "PD", "PD",
"PD", "PD", "PD", "PD", "PD", "PD",
"PD", "PD", "PD", "SV", "PD", "SV"),
TR= c("T1", "T1", "T1", "T1", "T1", "T1",
"T1", "T1", "T1", "T1", "T1", "T1",
"T1", "T1", "T1", "T1", "T1", "T1",
"T1", "T1", "T1", "T1", "T1", "T1",
"T2", "T2", "T2", "T2", "T2", "T2",
"T2", "T2", "T2", "T2", "T2", "T2",
"T2", "T2", "T2", "T2", "T2", "T2",
"T2", "T2", "T2", "T2", "T2", "T2",
"T2", "T2", "T2", "T2", "T2", "T2"),
Values_1 = c(34.9003695, 34.9003695, 28.2389394, 28.2389394, 26.0821875, 26.0821875,
30.5515533, 30.5515533, 22.6469958, 22.6469958, 34.5662974, 34.5662974,
35.2881883, 35.2881883, 41.3885176, 41.3885176, 19.9440042, 19.9440042,
5.6987524, 5.6987524, 37.4641052, 37.4641052, 2.4808126, 2.4808126,
1.7883822, 21.1799057, 21.1799057, 21.1799057, 2.7334442, 2.7334442,
2.7334442, 11.3880187, 11.3880187, 11.3880187, 7.8442267, 7.8442267,
7.8442267, 5.2445510, 5.2445510, 5.2445510, 20.4706600, 20.4706600,
15.7275634, 15.7275634, 4.4575814, 4.4575814, 17.0854186, 17.0854186,
5.6987524, 5.6987524, 37.4641052, 37.4641052, 2.4808126, 2.4808126),
Values_2 = c(76.24359, 76.24359, 58.52421, 58.52421, 80.14131, 80.14131,
59.05000, 102.19699, 102.19699, 72.39848, 72.39848, 58.15000,
68.31217, 68.31217, 53.67941, 53.67941, 56.88980, 56.88980,
108.98399, 96.64207, 96.64207, 38.88542, 38.88542, 54.60000,
52.12500, 52.12500, 17.20875, 17.20875, 47.26923, 47.26923,
67.80738, 60.41250, 60.41250, 83.93404, 83.93404, 37.20336,
50.02500, 50.02500, 94.73309, 94.73309, 41.31748, 41.31748,
56.88344, 59.74702, 59.74702, 48.23750, 48.23750, 95.14831,
108.98399, 96.64207, 96.64207, 38.88542, 38.88542, 54.60000))
For every ID, CV, TR combination that has two timepoints (being "2022-08-29" & "2022-09-05") I would like to subtract the first timepoint values (Values_1 & Values 2) from the last timepoint and return the substracted difference, generating the following output:对于具有两个时间点（即“2022-08-29”和“2022-09-05”）的每个 ID、CV、TR 组合，我想从最后一个时间点中减去第一个时间点值（Values_1 和 Values 2），然后返回减去的差，生成以下 output：
TR CV ID Values_1 Values_2
1 T1 PD 1 -29.5894359 40.67182
2 T1 PD 2 -29.5894359 28.32990
3 T1 PD 3 -3.9244124 42.96266
4 T1 PD 4 -3.9244124 -14.79399
5 T1 PD 5 -17.4631916 -18.00438
6 T1 PD 6 -17.4631916 -2.28980
7 T1 SV 1 -4.3488162 -17.19359
8 T1 SV 2 -4.3488162 25.95340
9 T1 SV 3 -5.5919436 43.67278
10 T1 SV 4 -5.5919436 13.87427
11 T1 SV 5 8.4841099 -7.74283
12 T1 SV 6 8.4841099 -21.99131
13 T2 PD 1 7.8833367 6.85844
14 T2 PD 2 10.4830124 9.72202
15 T2 PD 3 -0.7869696 -34.98607
16 T2 PD 4 -0.7869696 -46.49559
17 T2 PD 5 -3.3852414 6.92002
18 T2 PD 6 -3.3852414 53.83083
19 T2 SV 1 0.9450620 15.68238
20 T2 SV 2 -9.7918870 8.28750
21 T2 SV 3 -9.7918870 43.20375
22 T2 SV 4 -9.7918870 66.72529
23 T2 SV 5 5.1107825 36.66481
24 T2 SV 6 5.1107825 -10.06587
I found the following solution, but it does not return the desired result, especially not when there are stray rows or additional duplications because my solution is based on sorting the dataframe and assuming I have a correct sorted dataframe:我找到了以下解决方案，但它没有返回所需的结果，尤其是在有杂散行或其他重复项时，因为我的解决方案基于对 dataframe 进行排序并假设我有正确排序的 dataframe：
#Remove rows that are not duplicated (do not have two timepoints)
dupe = df[,c("ID", "CV", "TR")] # select columns to check duplicates
ONLY_DUPES=df[duplicated(dupe) | duplicated(dupe, fromLast=TRUE),]
#Now we substract last timepoint with first timepoint
#First we melt the data
ONLY_DUPES <- as.data.frame(ONLY_DUPES)
melted<-melt(ONLY_DUPES, id=c("DATE", "ID", "CV", "TR"))
#Then we cast with diff
#First we order the dataframe
melted= melted[with(melted, order(TR, CV, ID, variable, DATE)),]
casted=cast(melted, TR+CV+ID ~variable, fun.aggregate = diff)
#Returns:
TR CV ID Values_1 Values_2
1 T1 PD 1 -29.5894359 40.67182
2 T1 PD 2 -29.5894359 28.32990
3 T1 PD 3 -3.9244124 42.96266
4 T1 PD 4 -3.9244124 -14.79399
5 T1 PD 5 -17.4631916 -18.00438
6 T1 PD 6 -17.4631916 -2.28980
7 T1 SV 1 -4.3488162 -17.19359
8 T1 SV 2 -4.3488162 25.95340
9 T1 SV 3 -5.5919436 43.67278
10 T1 SV 4 -5.5919436 13.87427
11 T1 SV 5 8.4841099 -7.74283
12 T1 SV 6 8.4841099 -21.99131
13 T2 PD 1 7.8833367 6.85844
14 T2 PD 2 10.4830124 9.72202
15 T2 PD 3 -0.7869696 -34.98607
16 T2 PD 4 -0.7869696 -46.49559
17 T2 PD 5 -3.3852414 6.92002
18 T2 PD 6 -3.3852414 53.83083
19 T2 PD 9 -34.9832926 -57.75665 !
20 T2 SV 1 0.9450620 15.68238
21 T2 SV 10 -34.9832926 15.71458 !
22 T2 SV 2 -9.7918870 8.28750
23 T2 SV 3 -9.7918870 43.20375
24 T2 SV 4 -9.7918870 66.72529
25 T2 SV 5 5.1107825 36.66481
26 T2 SV 6 5.1107825 -10.06587
What method can I apply to perform this operation in a fail-proof way?我可以应用什么方法以防故障的方式执行此操作？
1 个解决方案
解决方案1
1 2022-09-06 08:49:32
Update for >2 value columns: >2 值列的更新：
If you have multiple columns you can use across() to define the columns you want to apply a function.如果您有多个列，您可以使用 cross across()来定义要应用 function 的列。 With this you only have to enter the start and end column you want to manipulate:有了这个，您只需输入要操作的开始和结束列：
For this example I just copied the Values_1 and Values_2 and stored them in new variables Values_3 and Values_4 respectively.对于这个例子，我只是复制了Values_1和Values_2并将它们分别存储在新变量Values_3和Values_4中。
library(tidyverse)
df %>%
group_by(ID, CV, TR) %>%
summarise(across(Values_1:Values_4, ~ .x[DATE == "2022-09-05"] - .x[DATE == "2022-08-29"])) %>%
arrange(TR, CV) %>%
print(., n = 24)
#> `summarise()` has grouped output by 'ID', 'CV', 'TR'. You can override using
#> the `.groups` argument.
#> # A tibble: 24 × 7
#> # Groups: ID, CV, TR [24]
#> ID CV TR Values_1 Values_2 Values_3 Values_4
#> <chr> <chr> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 1 PD T1 -29.6 40.7 -29.6 40.7
#> 2 2 PD T1 -29.6 28.3 -29.6 28.3
#> 3 3 PD T1 -3.92 43.0 -3.92 43.0
#> 4 4 PD T1 -3.92 -14.8 -3.92 -14.8
#> 5 5 PD T1 -17.5 -18.0 -17.5 -18.0
#> 6 6 PD T1 -17.5 -2.29 -17.5 -2.29
#> 7 1 SV T1 -4.35 -17.2 -4.35 -17.2
#> 8 2 SV T1 -4.35 26.0 -4.35 26.0
#> 9 3 SV T1 -5.59 43.7 -5.59 43.7
#> 10 4 SV T1 -5.59 13.9 -5.59 13.9
#> 11 5 SV T1 8.48 -7.74 8.48 -7.74
#> 12 6 SV T1 8.48 -22.0 8.48 -22.0
#> 13 1 PD T2 7.88 6.86 7.88 6.86
#> 14 2 PD T2 10.5 9.72 10.5 9.72
#> 15 3 PD T2 -0.787 -35.0 -0.787 -35.0
#> 16 4 PD T2 -0.787 -46.5 -0.787 -46.5
#> 17 5 PD T2 -3.39 6.92 -3.39 6.92
#> 18 6 PD T2 -3.39 53.8 -3.39 53.8
#> 19 1 SV T2 0.945 15.7 0.945 15.7
#> 20 2 SV T2 -9.79 8.29 -9.79 8.29
#> 21 3 SV T2 -9.79 43.2 -9.79 43.2
#> 22 4 SV T2 -9.79 66.7 -9.79 66.7
#> 23 5 SV T2 5.11 36.7 5.11 36.7
#> 24 6 SV T2 5.11 -10.1 5.11 -10.1
Created on 2022-09-06 with reprex v2.0.2使用reprex v2.0.2创建于 2022-09-06
First Answer:第一个答案：
This should work.这应该有效。 Using the tidyverse package, you can first group by your Variables ID , CV and TR and then use summarise() to calculate the subtractions based on their value on DATE 1 and DATE 2:使用tidyverse package，您可以首先按变量ID 、 CV和TR分组，然后使用summarise()根据 DATE 1 和 DATE 2 的值计算减法：
Mind that the last command print(., n = 24) is only used to show the output here and can be omitted when you copy the code.注意最后一个命令print(., n = 24)在这里只用来显示 output，复制代码时可以省略。
library(tidyverse)
df %>%
group_by(ID, CV, TR) %>%
summarise(Values_1_new = Values_1[DATE == "2022-09-05"] - Values_1[DATE == "2022-08-29"],
Values_2_new = Values_2[DATE == "2022-09-05"] - Values_2[DATE == "2022-08-29"]) %>%
arrange(TR, CV) %>%
print(., n = 24)
#> `summarise()` has grouped output by 'ID', 'CV', 'TR'. You can override using
#> the `.groups` argument.
#> # A tibble: 24 × 5
#> # Groups: ID, CV, TR [24]
#> ID CV TR Values_1_new Values_2_new
#> <chr> <chr> <chr> <dbl> <dbl>
#> 1 1 PD T1 -29.6 40.7
#> 2 2 PD T1 -29.6 28.3
#> 3 3 PD T1 -3.92 43.0
#> 4 4 PD T1 -3.92 -14.8
#> 5 5 PD T1 -17.5 -18.0
#> 6 6 PD T1 -17.5 -2.29
#> 7 1 SV T1 -4.35 -17.2
#> 8 2 SV T1 -4.35 26.0
#> 9 3 SV T1 -5.59 43.7
#> 10 4 SV T1 -5.59 13.9
#> 11 5 SV T1 8.48 -7.74
#> 12 6 SV T1 8.48 -22.0
#> 13 1 PD T2 7.88 6.86
#> 14 2 PD T2 10.5 9.72
#> 15 3 PD T2 -0.787 -35.0
#> 16 4 PD T2 -0.787 -46.5
#> 17 5 PD T2 -3.39 6.92
#> 18 6 PD T2 -3.39 53.8
#> 19 1 SV T2 0.945 15.7
#> 20 2 SV T2 -9.79 8.29
#> 21 3 SV T2 -9.79 43.2
#> 22 4 SV T2 -9.79 66.7
#> 23 5 SV T2 5.11 36.7
#> 24 6 SV T2 5.11 -10.1
Created on 2022-09-06 with reprex v2.0.2使用reprex v2.0.2创建于 2022-09-06
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