CODEWITHSUNDEEP
r
nik
nik

Reputation: 2584

how can I remove an average from each number and get average from certain region

I have a data like this 

df<- structure(list(X1 = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), X2 = structure(c(1L, 2L, 3L, 
4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 
18L, 19L, 20L, 21L, 22L, 23L, 24L, 7L, 8L, 1L, 2L, 3L, 4L, 5L, 
6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 
19L, 20L, 21L, 22L, 23L, 24L), .Label = c("B02", "B03", "B04", 
"B05", "B06", "B07", "C02", "C03", "C04", "C05", "C06", "C07", 
"D02", "D03", "D04", "D05", "D06", "D07", "G02", "G03", "G04", 
"G05", "G06", "G07"), class = "factor"), X3 = c(0.005648642, 
0.005876389, 0.00592532, 0.006244456, 0.005987075, 0.006075874, 
0.006198667, 0.006003758, 0.006041885, 0.006186987, 0.006041323, 
0.006071594, 0.005902391, 0.005976096, 0.00593805, 0.005866524, 
0.0059831, 0.005902586, 0.005914309, 0.005887304, 0.006054509, 
0.005931266, 0.005936195, 0.005895191, 0.005840959, 0.005849247, 
0.005808851, 0.005833586, 0.005825153, 0.00584873, 0.005983976, 
0.00598669, 0.006011548, 0.005997747, 0.005851022, 0.005919044, 
0.005854566, 0.0058226, 0.00578052, 0.005784874, 0.005933198, 
0.005996407, 0.005898848, 0.00595775, 0.005918857, 0.005882898, 
0.005877808, 0.005803604, 0.006235161, 0.005808725)), .Names = c("X1", 
"X2", "X3"), class = "data.frame", row.names = c(NA, -50L))

I am trying to get the average of several numbers and then minus it from every single number in that data and then get average of specific numbers

Here is what I did

I first try to get the average of "G05", "G06", "G07" for each set (X1) Then I minus it from each value 

df2 <- df1 %>%
  filter(X2 %in% paste0(paste0("G0", 5:7)) %>%
  group_by(X1) %>%
  summarise_at(vars(-X2), funs(mean(.)))

which should give me two numbers for group 1 and group 2( based on X1)

mean(c(0.005931266,0.005936195,0.005895191)) [1] 0.005920884

mean(c(0.005803604,0.006235161,0.005808725)) [1] 0.005949163

Then I want to remove this value from each number in group 1 and group 2 based on their group

for example 0.005648642- 0.005920884 . . . . 0.005840959- 0.005949163

In simple words

1- We get the mean of G05 , G06 and G07 for two groups where X1 is 1 or 2

for example 

mean(c(0.005931266,0.005936195,0.005895191)) [1] 0.005920884

mean(c(0.005803604,0.006235161,0.005808725)) [1] 0.005949163

2- We remove these mean values from every single number for example 

0.005648642- 0.005920884 
.
.
.
.
0.005840959- 0.005949163

3- After this correction Then I want to take avererge of specific rows which will be for both groups

For example

B02 and B03 for both group

average(c(0.005648642- 0.005920884,0.005876389- 0.005920884))

and

average(c(0.005808851- 0.005949163,0.005833586 - 0.005949163))

Upvotes: 0

Views: 103

Answers (1)

Maurits Evers
Maurits Evers

Reputation: 50678

I this what you're after?

  1. Steps 1 and 2:

    Split on X1 (i.e. group by X1) and center the values in X3 based on the mean across G05, G06, G07: 

    lst <- lapply(split(df, df$X1), function(w) {
    w.G0567 <- subset(w, grepl("G0[567]", w$X2));
    print(mean(w.G0567$X3));
    w$X3 <- w$X3 - mean(w.G0567$X3);
    return(w);
})
#[1] 0.005920884
#[1] 0.005949163
lst;
#$`1`
#   X1  X2           X3
#1   1 B02 -0.000272242
#2   1 B03 -0.000044495
#3   1 B04  0.000004436
#4   1 B05  0.000323572
#5   1 B06  0.000066191
#6   1 B07  0.000154990
#7   1 C02  0.000277783
#8   1 C03  0.000082874
#9   1 C04  0.000121001
#10  1 C05  0.000266103
#11  1 C06  0.000120439
#12  1 C07  0.000150710
#13  1 D02 -0.000018493
#14  1 D03  0.000055212
#15  1 D04  0.000017166
#16  1 D05 -0.000054360
#17  1 D06  0.000062216
#18  1 D07 -0.000018298
#19  1 G02 -0.000006575
#20  1 G03 -0.000033580
#21  1 G04  0.000133625
#22  1 G05  0.000010382
#23  1 G06  0.000015311
#24  1 G07 -0.000025693
#
#$`2`
#   X1  X2            X3
#25  2 C02 -1.082043e-04
#26  2 C03 -9.991633e-05
#27  2 B02 -1.403123e-04
#28  2 B03 -1.155773e-04
#29  2 B04 -1.240103e-04
#30  2 B05 -1.004333e-04
#31  2 B06  3.481267e-05
#32  2 B07  3.752667e-05
#33  2 C02  6.238467e-05
#34  2 C03  4.858367e-05
#35  2 C04 -9.814133e-05
#36  2 C05 -3.011933e-05
#37  2 C06 -9.459733e-05
#38  2 C07 -1.265633e-04
#39  2 D02 -1.686433e-04
#40  2 D03 -1.642893e-04
#41  2 D04 -1.596533e-05
#42  2 D05  4.724367e-05
#43  2 D06 -5.031533e-05
#44  2 D07  8.586667e-06
#45  2 G02 -3.030633e-05
#46  2 G03 -6.626533e-05
#47  2 G04 -7.135533e-05
#48  2 G05 -1.455593e-04
#49  2 G06  2.859977e-04
#50  2 G07 -1.404383e-04

  2. Step 3

    For every group, average centred X3 values for B02 and B03.

    lapply(lst, function(w) mean(subset(w, X2 %in% c("B03", "B03"))$X3))
#$`1`
#[1] -4.4495e-05
#
#$`2`
#[1] -0.0001155773

Upvotes: 1

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