我的简单问题是:如何对两个数据框的每一列进行
例如,我们有两个数据框:
注意:这只是一个例子 - 实际情况会包括更多列,它们包含特定位置上某个元素的浓度。
现在我想在这两个数据框之间运行 ks.test:
ks.test
测试?例如,我们有两个数据框:
D1 <- data.frame(D$Ag, D$Al, D$As, D$Ba, D$Be, D$Ca, D$Cd, D$Co, D$Cu, D$Cr)
D2 <- data.frame(S$Ag, S$Al, S$As, S$Ba, S$Be, S$Ca, S$Cd, S$Co, S$Cu, S$Cr)
注意:这只是一个例子 - 实际情况会包括更多列,它们包含特定位置上某个元素的浓度。
现在我想在这两个数据框之间运行 ks.test:
ks.test(D$Ag, S$Ag)
ks.test(D$Al, S$Al)
ks.test(D$As, S$As)
如何在不进行苦力工作的情况下完成etc.?
当我在一个数据框上使用shapiro.test时,我只需要简单地使用:
lshap1 <- lapply(D1, shapiro.test)
lres1 <- sapply(lshap1, `[`, c("statistic","p.value"))
我读了一些关于循环,聚合和mapply的内容 - 尝试了不同的东西,例如:
apply(D1, 2, function(D2) ks.test(D2,D1[,1])$p.value)
但是当我手动操作时,我没有得到很多的p值为0的情况。
编辑:2017年9月10日 我将数据导入为两个数据框,然后从中提取一些数据到“较小”的数据框进行分析 - 例如,在这种情况下查看有毒元素并排除其他元素。
示例数据:dput(head(D1))
和 dput(head(D2))
。
## Output dput(head(D1)):
structure(list(DF.As = c(-0.154868225169351, -0.291459578010276,
0.0355227595866723, 0.0892191549433623, 0.189115121672669,
-0.365222418641706
), DF.Cd = c(1.28810277421719, 1.45844987179892, 0.642331353138319,
0.673164023466527, 0.131548822144598, 0.146964746525726), DF.Cu
c(8.01131080231879,
6.52606822875086, 2.93449454196807, 4.08720148249298, 1.55494291704341,
1.73663851851503), DF.Cr = c(0.164849379809527, 0.196759436988158,
0.307645386162046, 0.302917612808149, 0.187202322026229, 0.25358922601195
), DF.Ni = c(0.362592459542858, 0.527078409257359, 0.477116357433909,
0.469287608844157, 0.225865184678244, 0.355321456594576), DF.Pb
c(0.414448963979605,
0.616598678960665, -0.0531899082482045, 0.47477978516042,
0.422106471495816,
0.0326241032568164), DF.Zn = c(74.7657982668, 74.2978919524635,
36.6575117549406, 47.8440365300156, 21.4962811912273, 23.3823413091772
)), .Names = c("DF.As", "DF.Cd", "DF.Cu", "DF.Cr", "DF.Ni", "DF.Pb",
"DF.Zn"), row.names = c(NA, 6L), class = "data.frame")
## Output dput(head(D2)):
structure(list(DO.As = c(0.0150158517208966, -0.0477743050574027,
-0.121541780066373, -0.0376195600535572, 0.115393920133327,
0.265450918075612), DO.Cd = c(0.367936811743133, 0.445545318262818,
0.350071986298948,
0.331513644782201, 0.603874629105229, 0.598527030667747), DO.Cu
c(1.65127139067621,
1.90306634226191, 1.08280240161368, 1.12130376047927, 1.23137174481965,
1.16618813144813), DO.Cr = c(0.162996340978278, 0.493799568371693,
0.18441814919492, 0.179883906525139, 0.128058190333676, 0.030406737049484
), DO.Ni = c(0.290717040452464, 0.331891307317008, 0.387987078391917,
0.36147470695146, 0.774910299821917, 0.323259411199816), DO.Pb
c(-0.0584055598838365,
0.377799120780818, -0.0741768575020139, 0.511278669452117,
0.320822577941608, 0.250377389869303), DO.Zn = c(16.5625482436821,
14.5084409384572, 16.571001044493, 18.4509635406253, 15.6876446591721,
12.7649440587945)), .Names = c("DO.As", "DO.Cd", "DO.Cu", "DO.Cr", "DO.Ni",
"DO.Pb", "DO.Zn"), row.names = c(NA, 6L), class = "data.frame")
我发布这篇文章是因为我仍然遇到一个错误:
## This is code for execution:
col.names = colnames(D1)
lapply(col.names, function(t, d1, d2){ks.test(d1[, t], d2[, t])}, D1, D2)
## Output:
Error in `[.data.frame`(d2, , t) : undefined columns chosen
(“traceback”按钮显示):
6.stop("undefined columns selected")
5.`[.data.frame`(d2, , t)
4.d2[, t]
3.ks.test(d1[, t], d2[, t])
2.FUN(X[[i]], ...)
1.lapply(col.names, function(t, d1, d2) {ks.test(d1[, t], d2[, t])}, D1, D2)