我有这样的数据
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))
我试图获得几个数字的平均值,然后从该数据中的每个数字中减去它,然后获得特定数字的平均值
这就是我做的事情
我首先尝试获得&#34; G05&#34;,&#34; G06&#34;,&#34; G07&#34;的平均值。每套(X1) 然后我从每个值减去它
df2 <- df1 %>%
filter(X2 %in% paste0(paste0("G0", 5:7)) %>%
group_by(X1) %>%
summarise_at(vars(-X2), funs(mean(.)))
哪个应该给我两个数字用于第1组和第2组(基于X1)
平均值(C(0.005931266,0.005936195,0.005895191)) [1] 0.005920884
平均值(C(0.005803604,0.006235161,0.005808725)) [1] 0.005949163
然后我想根据组
从组1和组2中的每个数字中删除此值 例如,0.005648642- 0.005920884 。 。 。 。 0.005840959- 0.005949163
简单来说
1-我们得到两组的G05,G06和G07的平均值,其中X1是1或2
例如
mean(c(0.005931266,0.005936195,0.005895191)) [1] 0.005920884
mean(c(0.005803604,0.006235161,0.005808725)) [1] 0.005949163
2-我们从每个数字中删除这些平均值 例如
0.005648642- 0.005920884
.
.
.
.
0.005840959- 0.005949163
3-在此更正之后然后我想对两个组的特定行进行过多的处理
例如
两组的B02和B03
average(c(0.005648642- 0.005920884,0.005876389- 0.005920884))
和
average(c(0.005808851- 0.005949163,0.005833586 - 0.005949163))
答案 0 :(得分:1)
我这是你之后的事情?
步骤1和2:
在X1
上拆分(即按X1
分组)并根据X3
,G05
,{{1}的平均值将值放在G06
中心}}:
G07
第3步
对于每个组,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
和X3
的平均居中B02
值。
B03