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这是一个由两部分组成的问题，非常复杂。

第一的。我想计算数据帧“ind_scores”中每个样本之间的“n”维欧几里德距离，它是数据帧“质心”中的相应组质心。

然后，我想选择最接近各自组质心的 3 个单独样本。我想将这些保存到一个新的数据框中，其中包含“单个样本名称”、“组”和“到质心的距离”的信息。

以下是数据示例：

ind_scores <- data.frame(row.names = c("OP2413iiiaMOU","OP2413iiibMOU","OP2413iiicMOU","OP2645ii_aPOR","OP2645ii_bPOR","OP2645ii_cPOR","OP2645ii_dPOR","OP2645ii_ePOR","OP3088i__aPOR","OP5043___aWAT","OP5043___bWAT","OP5044___aMOU","OP5044___bMOU","OP5044___cMOU","OP5046___aWAT","OP5046___bWAT","OP5046___cWAT","OP5046___dWAT","OP5046___eWAT","OP5047___aPHA","OP5047___bPHA","OP5048___bPHA","OP5048___cPHA","OP5048___dPHA","OP5048___ePHA","OP5048___fPHA","OP5048___gPHA","OP5048___hPHA","OP5049___aWAT","OP5049___bWAT","OP5051DNAaCOM","OP5051DNAbCOM","OP5051DNAcCOM","OP5052DNAaWAT","OP5053DNAaPHA","OP5053DNAbPHA","OP5053DNAcPHA","OP5054DNAaMOU","OP5054DNAbMOU","OP5054DNAcMOU"),
                        group = c("4","4","4","1","1","1","1","1","3","3","3","5","5","5","2","5","2","2","5","3","3","3","3","1","3","3","3","3","2","2","4","5","4","2","3","3","3","5","5","5"),
                        CV.1 = c(-13.3864612433581,-12.8079930877268,-12.8078461023615,11.609290941109,10.5489837203281,10.8802079446603,11.7559827821839,10.769027609963,2.93788199576291,5.14343682437333,1.1768471575429,-3.59878541566711,-3.69656648297924,-3.13205394000296,-1.88190759998412,-3.50181277277038,0.563858206656491,-1.38629942623866,-3.73771209413208,3.40039117982473,2.86962877144321,3.80869463338469,4.15722705333298,9.08529455175736,3.15497802125988,2.42193314853044,0.600699372070624,4.14515087614032,-3.3599436881205,-1.8893406509868,-13.355031250023,-4.10118631444206,-11.4911993949333,-1.55841778422586,2.91834267480086,1.58762181687645,3.08125993208779,-3.84248479288043,-3.60800082570682,-3.47369634755007),
                        CV.2 = c(-5.98931418061097,-6.48685652483353,-6.48781938591041,-5.4121748521578,-4.56051914391762,-5.14772881585026,-4.3883054106957,-3.06298578319138,0.25688954313487,1.01459325674394,1.47381593062751,5.11285501685872,6.32219277017476,4.93757903863915,-1.98974199849122,6.8029453586845,-4.47482073821288,-2.89353901685366,6.19654462202962,1.44791941276988,2.01950206487354,3.29347544821835,1.70411388918498,-3.36842394773708,0.843537649290457,1.53904192617335,-0.0653393231022099,2.43481086719558,-2.28081054006986,-1.12101221091068,-5.74678650527647,2.81164429296665,-4.7739502651084,-0.836323550526183,1.21550795042252,1.3943021883996,1.4814166592311,5.83324212843683,5.74898742272061,5.20153475667944),
                        CV.3 = c(-1.98030009996666,-0.130982057250324,-0.13182806033636,4.66419380929057,5.76073945060135,4.68132496125842,4.76343610149589,4.14550671815003,-4.32639082067268,-4.24665489024982,-4.41960026466873,3.48306980151309,3.33978102573513,5.7630709271421,1.72213262278476,3.4138699327986,-0.214011687254588,-1.35717946591182,3.99742433050098,-4.11899265115508,-4.850265219848,-4.56241597162798,-5.1673124571133,3.88620294769555,-7.55945071289283,-5.18624310325486,-2.64740221288213,-3.34585676732483,-0.146912983782168,0.183282683148834,0.341803164827804,3.08878325423758,0.402559648490399,-0.589462854225432,-4.66295564242554,-4.70902036477095,-3.15037329091412,4.46721009678144,4.19323467451728,5.20598542755799),
                        CV.4 = c(-1.85773720384766,-3.29816018270707,-3.29805035723744,-1.0463680864694,-0.164642808251456,-1.88434766843655,-2.76184052196793,-1.69491772471098,0.0194432918943446,0.900426089523736,-0.581953934607345,-0.230042890025999,-1.79667524325622,-2.45893275735924,6.71016957191989,1.8888359729478,5.48587185602468,7.45260127587355,-0.447573770298677,-1.61748546155154,-2.01415972868345,-1.50135791552696,-0.439840157186184,-1.26569596255966,-1.04297110114946,-1.59978271452128,-0.471298592990895,-0.466524983137062,6.36590517153234,6.62852590954231,-3.04695209017556,-0.936146169909344,-2.4145719914164,5.10804058988218,-0.0744344020096521,-1.17738342385673,-1.67635978290671,-1.05954691377259,-0.0467102661118772,1.81264507750015))


centroids <- data.frame(group = c("1","2","3","4","5"),
                        CV.1 = c(10.7747979250003,-1.58534182381657,2.95743524695937,-12.7697062156805,-3.63247766512568),
                        CV.2 = c(-4.32335632559164,-2.26604134251075,1.43239910451168,-5.89694537234795,5.44083615635448),
                        CV.3 = c(4.65023399808197,-0.0670252808734024,-4.49663816927149,-0.299749480847027,4.1058254967538),
                        CV.4 = c(-1.469635462066,6.29185239579583,-0.838834486907799,-2.78309436507683,-0.363794106698444))

提前谢谢了！干杯。迪翁。

4

2 回答 2

2

就个人而言，我喜欢使用整洁的小标题（没有更多的行名，并且是长格式），所以我会首先将你的数据框转换为那个。

library(tidyverse)

ind_scores <- ind_scores %>% 
  as_tibble(rownames = "name") %>% 
  pivot_longer(cols = starts_with("CV"),
               names_to = "CV")

centroids <- centroids %>% 
  pivot_longer(cols = starts_with("CV"),
               names_to = "CV")

现在，很容易将正确的质心加入到个体中，逐个分组，并计算其欧几里得距离。生成的 tibble 包含列name和distance. 排序依据distance在顶部给出最接近的示例。

ind_scores %>% 
  left_join(centroids, by = c("group", "CV"), suffix = c("", "_centroid")) %>% 
  group_by(group, name) %>% 
  summarise(distance = sqrt(sum((value - value_centroid)^2))) %>% # euclidean distance
  top_n(-3, distance) %>% # bottom 3 sorted by distance
  arrange(group, distance) # sort them
于 2019-12-30T12:09:12.663 回答
1

Bas 的版本更漂亮且更易于理解，但如果您想将其转换为 Rcpp（我个人在集群方面工作很多，而对于大数据，R 有点太慢了），这也可能会有所帮助。

groups <- ind_scores[,1] %>% as.character() %>% as.numeric()
ind_scores[,1] <- NULL
centroids <- list()
j <- 1
for(i in unique(groups)){
  centroids[[j]] <- (ind_scores[groups==i,] %>% apply(.,2,mean))
  names(centroids)[j] <- i
  ind_scores <- rbind(ind_scores,centroids[[j]])
  j <- j +1
}
## the ast j of ind scores will be centroids


dist_mat <-  dist(ind_scores %>% as.matrix()) %>% as.matrix() ## get the distance matrix
# > dist_mat[1:5,1:5]
# OP2413iiiaMOU OP2413iiibMOU OP2413iiicMOU OP2645ii_aPOR OP2645ii_bPOR
# OP2413iiiaMOU      0.000000   2.465150984   2.464681274     25.882974     25.253460
# OP2413iiibMOU      2.465151   0.000000000   0.001294793     25.008458     24.367816
# OP2413iiicMOU      2.464681   0.001294793   0.000000000     25.008508     24.367942
# OP2645ii_aPOR     25.882974  25.008458155  25.008508380      0.000000      1.956891
# OP2645ii_bPOR     25.253460  24.367815962  24.367941651      1.956891      0.000000


## do not touch j
thresh <- 3
new_data_frame <- data.frame(sample_name=NA,group=NA,centr_distance=NA)
for(i in nrow(dist_mat):(nrow(dist_mat)-j+2)){
  distances_to_cluster <- dist_mat[i,-i]
  indexes <- order(distances_to_cluster,decreasing = F)[1:thresh]
  ## collect thresh minimum distances
  ##[1]  2  3 31
  for(z in indexes){
    ## get indexes name, which group and distance to centorid
    tmp <- c(rownames(dist_mat)[z],groups[z],distances_to_cluster[z])
    new_data_frame <- rbind(new_data_frame,tmp)
  }
}
new_data_frame[order(new_data_frame$group),] %>% na.omit()
# 11 OP2645ii_cPOR     1 0.929329939818952
# 12 OP2645ii_ePOR     1    1.376251766813
# 13 OP2645ii_aPOR     1   1.4357069775206
# 2  OP5049___bWAT     2  1.25678563440431
# 3  OP5049___aWAT     2  1.77800330839339
# 4  OP5046___dWAT     2  1.85612687904496
# 8  OP5053DNAaPHA     3 0.812735500386649
# 9  OP5047___aPHA     3 0.972298470684858
# 10 OP5048___fPHA     3  1.16307022957174
# 14 OP2413iiibMOU     4 0.802020132014482
# 15 OP2413iiicMOU     4 0.802473693143821
# 16 OP5051DNAaCOM     4 0.919980313623531
# 5  OP5054DNAbMOU     5 0.451374395540337
# 6  OP5044___aMOU     5 0.717231370935914
# 7  OP5046___eWAT     5 0.775202821859753
于 2019-12-30T12:13:29.517 回答