data.frame列的子集,以最大化"完成"意见

时间:2014-06-03 13:35:09

标签: r optimization mathematical-optimization subquery

我的数据框大约有20个数字列,每个数据列包含大量的NA值。我想选择这些列的一个子集,它将为我提供包含零NA值的最多行。详尽的搜索需要花费大量的计算时间 - 是否有更好的方法来获得近似值?

以下是一个较小数据框(完全任意)的示例:

set.seed(2)
foo = as.data.frame(matrix(rnorm(200), nr = 20))
foo[sapply(foo, function(x) x > abs(x[1]))] = NA
foo = foo[-1, ]

round(foo, 3)

       V1     V2     V3     V4     V5     V6     V7     V8     V9    V10
2   0.185 -1.200 -1.959     NA -1.696  0.261  0.139  0.410 -0.638 -1.262
3      NA  1.590 -0.842 -0.703 -0.533 -0.314     NA -0.807 -0.268  0.392
4  -1.130  1.955     NA  0.158 -1.372 -0.750 -0.431  0.086  0.360 -1.131
5  -0.080  0.005     NA  0.506 -2.208 -0.862 -1.044     NA -1.313  0.544
6   0.132 -2.452     NA -0.820     NA     NA  0.538 -0.654 -0.884     NA
7   0.708  0.477 -0.305 -1.999 -0.653  0.940 -0.670     NA     NA  0.025
8  -0.240 -0.597 -0.091 -0.479 -0.285     NA  0.639  0.550 -2.099  0.515
9      NA  0.792 -0.184  0.084 -0.387 -0.421 -1.724 -0.807 -1.239 -0.654
10 -0.139  0.290 -1.199 -0.895  0.387 -0.351 -1.742 -0.997     NA  0.504
11  0.418  0.739 -0.838 -0.921     NA -1.027  0.690     NA     NA -1.272
12     NA  0.319     NA  0.330     NA -0.251  0.331 -0.169     NA -0.077
13 -0.393  1.076 -0.562 -0.142 -1.184  0.472  0.871     NA  0.057 -1.345
14 -1.040 -0.284     NA  0.435 -1.358     NA -2.016 -0.844  0.324 -0.266
15     NA -0.777 -1.048 -0.054 -1.513  0.564  1.213     NA -0.905     NA
16 -2.311 -0.596 -1.966 -0.907 -1.253  0.456  1.200 -1.343 -0.652  0.701
17  0.879 -1.726 -0.323  1.304     NA     NA  1.032     NA -0.262 -0.443
18  0.036 -0.903     NA  0.772  0.008     NA  0.786  0.464 -0.935 -0.789
19     NA -0.559     NA  1.053 -0.843  0.107     NA  0.268     NA -0.857
20  0.432 -0.247     NA -1.410 -0.601 -0.783 -1.454     NA -1.624 -0.746

dim(na.omit(foo))
[1]  1 10

以下是我如何制定详尽的搜索:

best.list = list()
for (i in 5:ncol(foo)) {
    # get best subset for each size
    collist = combn(ncol(foo), i)
    numobs = apply(collist, 2, function(x) nrow(na.omit(foo[, x])))
    cat("for subset size", i, "most complete obs is", max(numobs), "\n")
    best = which(numobs == max(numobs))[1]
    best.list = c(best.list, list(collist[, best]))
}

例如,best.list[[1]]告诉我,如果我保留5列,我可以有12个完整的观察值(行数为零的NAs),第1,2,4,7和10列是我应该的选择。

虽然这适用于非常小的数据帧,但对于较大的数据帧,它很快就会变得过高。 R中有没有办法有效地估计给定大小的最佳子集?我唯一能找到的是subselect包,但我无法弄清楚如何为手头的问题实现其方法。

2 个答案:

答案 0 :(得分:3)

不确定这是否是完整的解决方案,但如果您想要快速的结果,data.table和阴影矩阵是最可能的成分。

library(data.table)
df = data.table(foo) # your foo dataframe, converted to data.table

y = sort(df[,lapply(.SD, function(x) sum(is.na(x)))]) # nr of NA in columns, increasing
setcolorder(df, names(y)) # now the columns are ordered - less NA first

df[, idx := rowSums(is.na(df))] # count nr of NA in rows
df = df[order(idx),] # sort by nr of NA in rows
df[, idx := NULL] # idx not needed anymore
# now your data.table is sorted: columns with least NA to the left,  
# rows with with least NA on top

# shadow matrix
x= data.table(abs(!is.na(df)))  # 0 = NA value
y = as.data.table(t(x))
y = y[,lapply(.SD, cumprod)]
y = as.data.table(t(y))
y[,lapply(.SD, sum)] 

# nr of complete cases from column selections:
# V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
# 1: 19 18 16 14 11 10  7  5  2   1

答案 1 :(得分:0)

旧帖子,但有一个内置函数可以做到这一点。我打赌它效率很高:

df_noNAs <- df[complete.cases(df[,1:20]),]