r:在for()循环中部署NbClust()调用时发生错误-"if((res [ncP-min_nc + 1，15]< = resCritical [ncP-min_nc +:''的错误 [英] r: error for NbClust() call when deploying it within for() loop - "Error in if ((res[ncP - min_nc + 1, 15] <= resCritical[ncP - min_nc + :"
问题描述
我想为几个数据框调用NbClust()函数.我这样做是通过包含NbClust()函数调用的for循环全部"发送它们.
代码如下:
I want to call the NbClust() function for a couple of dataframes. I do so by "sending" them all through a for loop that contains the NbClust() function call.
The code looks like this:
#combos of just all columns from df
variations = unlist(lapply(seq_along(df), function(x) combn(df, x, simplify=FALSE)), recursive=FALSE)
for(i in 1:length(variations)){
df = data.frame(variations[i])
nc = NbClust(scale(df), distance="euclidean", min.nc=2, max.nc=10, method="complete")
}
不幸的是，它总是产生以下错误.奇怪的是，如果我在不使用循环的情况下应用相同的函数调用(即，仅对一个数据帧)，它将完美工作……那是怎么回事?
Unfortunately it always generates the below error. Strangely enough, if I am applying the same function call without the loop (i.e. to only one data frame) it works perfectly... so what is wrong?
我看过NbClust的源代码，确实有一行包含错误消息的代码，但是我无法相应地更改代码.您知道可能是什么问题吗?
I have had a look at the source code of NbClust and indeed there is a line that contains the code of the error message but I am unable to change the code accordingly. Do you have any idea what the problem might be?
if((res [ncP-min_nc + 1，15]< = resCritical [ncP-min_nc + :缺少需要TRUE/FALSE的值
Error in if ((res[ncP - min_nc + 1, 15] <= resCritical[ncP - min_nc + : missing value where TRUE/FALSE needed
此外，它还会产生以下警告:
Additionally it produces the following warnings:
In addition: Warning messages:
1: In max(DiffLev[, 5], na.rm = TRUE) :
no non-missing arguments to max; returning -Inf
2: In matrix(c(results), nrow = 2, ncol = 26) :
data length [51] is not a sub-multiple or multiple of the number of rows [2]
3: In matrix(c(results), nrow = 2, ncol = 26, dimnames = list(c("Number_clusters", :
data length [51] is not a sub-multiple or multiple of the number of rows [2]
数据如下:
df = structure(list(GDP = c(18.2, 8.5, 54.1, 1.4, 2.1, 83.6, 17, 4.9,
7.9, 2, 14.2, 48.2, 17.1, 10.4, 37.5, 1.6, 49.5, 10.8, 6.2, 7.1,
7.8, 3, 3.7, 4.2, 8.7, 2), Population = c(1.22, 0.06, 0, 0.54,
2.34, 0.74, 1.03, 1.405095932, 0.791124402, 2.746318326, 0.026149254,
11.1252, 0.05183432, 2.992952671, 0.705447655, 0, 0.900246028,
1.15476828, 0, 1.150673397, 1.441975309, 0, 0.713777778, 1.205504587,
1.449230769, 0.820985507), Birth.rate = c(11.56, 146.75, 167.23,
7, 7, 7, 10.07, 47.42900998, 20.42464115, 7.520608751, 7, 7,
15.97633136, 15.1531143, 20.41686405, 7, 22.60379293, 7, 7, 18.55225902,
7, 7.7, 7, 7, 7, 7), Income = c(54L, 94L, 37L, 95L, 98L, 31L,
78L, 74L, 81L, 95L, 16L, 44L, 63L, 95L, 20L, 95L, 83L, 98L, 98L,
84L, 62L, 98L, 98L, 97L, 98L, 57L), Savings = c(56.73, 56.49,
42.81, 70.98, 88.24, 35.16, 46.18, 35.043, 46.521, 58.024, 22.738,
60.244, 77.807, 80.972, 13.08, 40.985, 46.608, 63.32, 51.45,
74.803, 73.211, 50.692, 65.532, 83.898, 60.857, 40.745)), .Names = c("GDP", "Population", "Birth.rate", "Income", "Savings"), class = "data.frame", row.names = c(NA, -26L))
推荐答案
某些聚类方法无法直接适应您的数据集或数据类型.您可以选择最佳方法，也可以全部使用.当使用所有它们时，经常会产生一条错误消息(不是错误).通过禁用使循环停止的ERROR消息，可以采用以下替代方法:
Some of the Clustering methods are not directly adapted to your datasets or type of data. You can select the best methods, or use all of them. When using all of them, it often happens that this produces an ERROR message (which is not a bug). By disabling the ERROR message that stops the loop, the below could be an alternative:
vc.method <- c("kl","ch", "hartigan","ccc", "scott","marriot","trcovw", "tracew","friedman", "rubin", "cindex", "db", "silhouette", "duda", "beale", "ratkowsky", "ball", "ptbiserial", "pseudot2", "gap", "frey", "mcclain", "gamma", "gplus", "tau", "dunn", "hubert", "sdindex", "dindex", "sdbw", "alllong")
val.nb <- c()
for(method in 1:length(vc.method)){
tryCatch({
en.nb <- NbClust(na.omit(sum.sn), distance = "euclidean", min.nc = 2,
max.nc = vc.K.max, method = "kmeans",
index = vc.method[method])
val.nb <- c(val.nb, as.numeric(en.nb$Best.nc[1]))
}, error=function(e){cat("ERROR :",conditionMessage(e), "\n")})
}
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