Question

我正在重新问一个问题，我试图简化我的数据集并举例说明我想要的输出。如果这仍然很复杂，请随意发表评论，以帮助我澄清这一点。

我有一张表，其中我将具有类似rt和mz的功能分组。

          orig_feat   mz_mid rt_mid similar_feature
1           f_1 685.4350  466.5             f_1
2           f_2 260.1655  245.0             f_2
185         f_2 260.1665  256.5           f_185
408         f_2 260.1670  239.0           f_408
2334        f_2 260.1650  250.0          f_2334
3           f_3 288.1980  276.0             f_3
7           f_3 288.1990  289.0             f_7
414         f_3 288.1970  275.0           f_414
2181        f_3 288.1980  270.0          f_2181
2969        f_3 288.1965  297.5          f_2969
4           f_4 537.3915  454.5             f_4
2271        f_4 537.3965  435.5          f_2271
5           f_5 439.2990  153.5             f_5
6           f_6 325.0690  210.5             f_6
10          f_6 325.0685  227.0            f_10
747         f_6 325.0685  184.5           f_747
2068        f_6 325.0695  225.0          f_2068
2929        f_6 325.0685  218.0          f_2929
2970        f_6 325.0680  237.0          f_2970
31          f_7 288.1980  276.0             f_3
71          f_7 288.1990  289.0             f_7
4141        f_7 288.1970  275.0           f_414
21811       f_7 288.1980  270.0          f_2181
29691       f_7 288.1965  297.5          f_2969

我想列出一个包含每个组的条目的列表。具有相同$ orig_feat的所有行应该“分组”，并且对于这些“组”中的每一个，我想要所有特征的向量。请参阅下面的示例所需输出。

    $grf_1
[1] "f_1"

$grf_2
[1] "f_2"    "f_185"  "f_408"  "f_2334"

$grf_3
[1] "f_3"    "f_7"    "f_414"  "f_2181" "f_2969"

$grf_4
[1] "f_4"    "f_2771"

$grf_5
[1] "f_5"

$grf_6
[1] "f_6"    "f_10"   "f_747"  "f_2068" "f_2929" "f_2970"

但重要的是我希望这是非冗余的（例如gf_3：包含f_7，f_414，f_2181，f_2696，因此当我到达f_7时，我不会为f_7创建一个组，因为f_3组已经包含所有功能在f_7组中）

下面是我的代码。目前，生成的输出在grf_3之后停止。我不确定为什么它似乎过早退出循环。

mkFeatGroupsList<-function(simFeatsTab){
  features_seen<-vector()
  GroupingList<-list()
  counter=1
  for (i in 1:length(unique(simFeatsTab$orig_feat))){
    orig_feat2Grp<-simFeatsTab$orig_feat[i]
    if (orig_feat2Grp%in%features_seen == TRUE) next
    matchingFeats<-subset(simFeatsTab,orig_feat==orig_feat2Grp)$feature
    grFeatNm<-paste("grf_",counter,sep="")
    GroupingList[[grFeatNm]]<-matchingFeats
    features_seen<-c(features_seen,matchingFeats)
    counter=counter+1
  }
  return(GroupingList)
}

如果你需要测试数据。

> dput(simFeatsTab.10.30.test)
structure(list(orig_feat = structure(c(1L, 2L, 2L, 2L, 2L, 3L, 
3L, 3L, 3L, 3L, 4L, 4L, 5L, 6L, 6L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 
7L, 7L), .Label = c("f_1", "f_2", "f_3", "f_4", "f_5", "f_6", 
"f_7"), class = "factor"), mz_mid = c(685.435, 260.1655, 260.1665, 
260.167, 260.165, 288.198, 288.199, 288.197, 288.198, 288.1965, 
537.3915, 537.3965, 439.299, 325.069, 325.0685, 325.0685, 325.0695, 
325.0685, 325.068, 288.198, 288.199, 288.197, 288.198, 288.1965
), rt_mid = c(466.5, 245, 256.5, 239, 250, 276, 289, 275, 270, 
297.5, 454.5, 435.5, 153.5, 210.5, 227, 184.5, 225, 218, 237, 
276, 289, 275, 270, 297.5), similar_feature = c("f_1", "f_2", 
"f_185", "f_408", "f_2334", "f_3", "f_7", "f_414", "f_2181", 
"f_2969", "f_4", "f_2271", "f_5", "f_6", "f_10", "f_747", "f_2068", 
"f_2929", "f_2970", "f_3", "f_7", "f_414", "f_2181", "f_2969"
)), .Names = c("orig_feat", "mz_mid", "rt_mid", "similar_feature"
), class = "data.frame", row.names = c("1", "2", "185", "408", 
"2334", "3", "7", "414", "2181", "2969", "4", "2271", "5", "6", 
"10", "747", "2068", "2929", "2970", "31", "71", "4141", "21811", 
"29691"))

Answer 1

另一种解决方案是使用CREATE TABLE ts.mindesc ( ticker text, time timestamp, close float, high float, low float, numevents int, open float, source text, value float, PRIMARY KEY (ticker, time) ) WITH CLUSTERING ORDER BY (time DESC)包：

igraph

Answer 2

我继续这样做：

将您的数据框（我称之为feat）拆分为orig_feat
使用sapply获取相关功能
循环完成相关功能并消除重复

转化为：

feat.split <- split(feat, my.df$orig_feat)

sim.feat <- sapply(feat.split, function(x){x$similar_feature})

for (i in 2:length(sim.feat))
  {
  # Get all of the previous features
  prev.feat <- do.call("c", sim.feat[1:(i-1)])

  # Remove features already used
  sim.feat[[i]] <- sim.feat[[i]][!sim.feat[[i]] %in% prev.feat]
  }

R - 过早退出循环

2 个答案: