过滤R中的多个列

本文介绍了过滤R中的多个列的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 29岁程序员，3月因学历无情被辞！ 假设一个数据集有几行和一些列，其中一些列为0（我的意思是列中的所有值都是0）。如何筛选这些列？我尝试使用以下代码，但无效。 training_data< - Filer（function（x）{！（all x [，1:99] == 0））}，training_data） 更新： 对不起，在数据集中，NOT不是所有列都是数字的，所以我需要为列指定范围从1:99。 更新V2：添加了一个部分的数据集（使用 dput ） =结构（c（1L，1L，1L，2L，1L，1L），.Label = c（A，B），class =factor），f1 = c（15,24 ，10，9，6，9），f2 = c（6， 14，5，4，2，4），f3 = c（6,7,2,2,1,2），f4 = c（0,0,0,0， 0,0），f5 = c（9，15，6，5，3，5），f6 = c（3,7,2,2,1， 2），f7 = c（1， 0,0,0,0,0），f8 = c（4，11，5，4，2，4），f9 = c（5,3,0 ，0， 0,0），f10 = c（1,3,0,0,0,0），f11 = c（1，4，2，2，1，2）， f12 = c（0,0,0,0,0,0），f13 = c（13，15，7，6，3，6），f14 = c（0， 7,1,1， 1，1），f15 = c（0,0,0,0,0,0），f16 = c（20,30， 11，10，6，10），f17 = c（5,0 ，0,0,0,0），f18 = c（0,0， 0,0,0,0），ft19 = c（28,344,39,28,82,42），f20 = c（2.15， 15.64，49.88，4，20.5,6），f21 = c（0,0,0,0,0,0），f22 = c（0， 0，0，0，0，0），f23 = c（6,7,2,2,1,2），f24 = c（0,0， 0，0，0， 0），f25 = c（19,334,395,23,79,37），f26 = c（0， 26,37,6，16,7），f27 = c（11,64,101 ，5，17，12），f28 = c（0， 0,0,0,0,0），f29 = c（2，37，101，7，26，8），f30 = c 0， 18,32,2,16,4），f31 = c（0,0,0,0,0,0），f32 = c（0， 0，0，0， 0,0），f33 = c（3,0,1,0,1,0），f34 = c（5，44， 32,4,15,5），f35 = c（0,0） 0,0,0,0），f36 = c（0,0,0,0 0,0,0），f37 = c（0,0,0,0,0,0），f38 = c（0,0,0,0， 0,0），f39 = c（6,8,10,3,2,3），f40 = c（4，6，16，4，4， 3），f41 = c（18,36,37,7,5,7），f42 = c（0,18,27,0,14， 1），f43 = c 0,0,0,0,0），f44 = c（54,743,910,65， 184,100），f45 = c（14,133,91,25,18,40）， f46 = c（0,0,0 0,0,0,0），f47 = c（4,25,17,6,6,8），f48 = c（0,0,0,0 0,0,0），f49 = c（0.46,1,1.5,1.14,1.5,1.14），f50 = c（2.67， 1.86,1.83,1.88,1.67， 1.88），f51 = c（3,9,1,2,1,2）， f52 = c（0,1,2,1,1,1），f53 = c（10,12,5 ，4，2，4），f54 = c（0， 0,0,0,0,0），ft55 = c（3，10，3，3，2，3），f56 = c 0.54， 0.07,0.03,0.32,0.07,0.21），f57 = c（0.21,0.04,0.01， 0.14,0.02,0.1），f58 = c（0.21,0.02,0.01,0.07， 0.01,0.05 ），f59 = c（0,0,0,0,0,0），f60 = c（0.32,0.04,0.02,0.18， 0.04,0.12），f61 = c （0.11,0.02,0.01,0.7,0.01,0.05）， f62 = c（0.04,0,0,0,0,0），f63 = c（0.14,0.03,0.01,0.14， = 0.02,0.1），f64 = c（0.18,0.01,0,0,0,0），f65 = c（0.04， 0.01,0,0,0,0），f66 = c（0.04， 0.01,0.01,0.07,0.01， 0.05），f67 = c（0,0,0,0,0,0），f68 = c（0.46,0.04,0.02， 0.21,0.04， 0.14），f69 = c（0,0.02,0,0.04,0.01,0.02）， f70 = c（0,0,0,0,0,0），f71 = c（0.71,0.09,0.03 ，0.36， 0.07,0.24），f72 = c（0.18,0,0,0,0,0），f73 = c（0,0， 0,0,0,0） f74 = c（1,1,1,1,1,1），f75 = c（0.0 8，0.05， 0.12，0.14，0.25，0.14），f76 = c（0,0,0,0,0,0），f77 = c（0， 0,0,0， 0，0），f78 = c（0.21,0.02,0.01,0.07,0.01,0.05 ），f79 = c（0,0,0,0,0,0），f80 = c（0.68,0.97 ，0.99，0.82， 0.96，0.88），f81 = c（0,0.08,0.09,0.21,0.2,0.17），f82 = c（0.39， 0.19,0.25,0.18,0.21,0.29 ），f83 = c（0,0,0,0,0,0）， f84 = c（0.07,0.11,0.25,0.25,0.32,0.19），f85 = c（0,0.05， 0.08,0.07,0.2,0.1），f86 = c（0,0,0,0,0,0），f87 = c（0， 0,0,0,0,0），f88 = c（0.11,0,0,0,0.01,0），f89 = c（0.18， 0.13,0.08,0.14,0.18,0.12），f90 = c（0,0,0,0,0,0 ，0）， f91 = c（0,0,0,0,0,0），f92 = c（0,0,0,0,0,0），f93 = c（0， 0,0,0,0,0），f94 = c（0.21,0.02,0.03,0.11,0.02,0.07 ），f95 = c（0.14,0.02,0.04,0.14,0.05,0.07） ，f96 = c（0.64， 0.1,0.09,0.25,0.06,0.17），f97 = c（0,0.05,0.07,0,0.17， 0.02），f98 = c（0,0 ，0,0,0,0），f99 = c（1.93,2.1 6，2.28， 2.32,2.24,2.38），f100 = c（0.5,0.39,0.23，0.89，0.22， 0.95），f101 = c（0,0,0,0,0， 0），f102 = c（0.14,0.07,0.04， 0.21,0.07,0.19），f103 = c（0,0,0,0,0,0），f104 = c（0.02， 0,0,0.04,0.02,0.03），f105 = c（0.1,0.01,0,07,0.02， 0.04），f106 = c（0.11,0.03,0,07,0.01,0.05）， f107 = c（0， 0,0.01,0.04,0.01,0.02），f108 = c（0.36,0.03,0.01,0.14， 0.02,0.1），f109 = c（0,0， 0,0,0,0），f110 = c（0.11,0.03， 0.01,0.11,0.02,0.07）），Names = c（label，f1，f2，$ b f，f4，f5，f6，f7，f8，f9，f10，f11，f12，f13 f14，f15，f16，f17，f18，ft19，f20，f21，f22，f23，f24 f26，f27，f28，f29，f30，f32，f31， f37，f38，f39，f38，f39，f46，f46，f46， f $，f52，f53，f54，f55，f56，f57，f58 f59，f60，f61，f62，f63，f64，f65，f66，f67 f71，f72，f73，f74，f75，b，f76，f76 f82，f83，f84，f89，f92，f85，f85， f f94，f95，f96，f97，f98，f99，f100，f101，f102 f104，f105，f106，f107，f108，f109，f110），class =data.frame，row.names = c（NA， 6L）） 解决方案 c $ c> all（x == 0）使用 any（x！= 0）稍微更有效，因为任何在元素的第一个实例为！= 0 之后停止，这对于越来越多的行数将是重要的。 使用 plyr 和 colwise 提供不同的解决方案（ dat 是 dput 数据）： library（plyr） f0 colwise（identity，f0）（dat） 要经历dat中的每一列，并返回它（ identity ），但只有当 f0 返回 TRUE ，即该列至少有一个条目！= 0 ，列 is.numeric 编辑：为列表中的每个数据框架执行此操作，例如。 training_data< - list（dat，dat，dat，dat） training_data_clean< - lapply（training_data，function（z）colwise（identity，f0）（z）） sapply（training_data，dim） [，1] [，2] ，3] [，4] [1，] 6 6 6 6 [2，] 111 111 111 111 sapply（training_data_clean，dim） [ ，1] [，2] [，3] [，4] [1，] 6 6 6 6 [2，] 74 74 74 74 EDIT2 ：保留标签列： lapply（training_data，function（z）cbind（label = z $ label，colwise（identity，f0）（z））） / pre> Supposing a data set with several rows and columns with some columns being 0 (I mean all values in the column are 0's). How one can filter out those columns? I have tried with the following code but no avail.training_data <- Filer(function(x) { !(all(x[, 1:99]==0))}, training_data)UPDATE:Sorry. In the data set, NOT all columns are numeric, so I need to specify a range from 1:99 for columns.UPDATE V2:Added a part of my data set (using dput)structure(list(label = structure(c(1L, 1L, 1L, 2L, 1L, 1L), .Label = c("A","B"), class = "factor"), f1 = c(15, 24, 10, 9, 6, 9), f2 = c(6,14, 5, 4, 2, 4), f3 = c(6, 7, 2, 2, 1, 2), f4 = c(0, 0, 0, 0,0, 0), f5 = c(9, 15, 6, 5, 3, 5), f6 = c(3, 7, 2, 2, 1, 2), f7 = c(1,0, 0, 0, 0, 0), f8 = c(4, 11, 5, 4, 2, 4), f9 = c(5, 3, 0, 0,0, 0), f10 = c(1, 3, 0, 0, 0, 0), f11 = c(1, 4, 2, 2, 1, 2), f12 = c(0, 0, 0, 0, 0, 0), f13 = c(13, 15, 7, 6, 3, 6), f14 = c(0, 7, 1, 1, 1, 1), f15 = c(0, 0, 0, 0, 0, 0), f16 = c(20, 30, 11, 10, 6, 10), f17 = c(5, 0, 0, 0, 0, 0), f18 = c(0, 0, 0, 0, 0, 0), ft19 = c(28, 344, 399, 28, 82, 42), f20 = c(2.15, 15.64, 49.88, 4, 20.5, 6), f21 = c(0, 0, 0, 0, 0, 0), f22 = c(0, 0, 0, 0, 0, 0), f23 = c(6, 7, 2, 2, 1, 2), f24 = c(0, 0, 0, 0, 0, 0), f25 = c(19, 334, 395, 23, 79, 37), f26 = c(0, 26, 37, 6, 16, 7), f27 = c(11, 64, 101, 5, 17, 12), f28 = c(0, 0, 0, 0, 0, 0), f29 = c(2, 37, 101, 7, 26, 8), f30 = c(0, 18, 32, 2, 16, 4), f31 = c(0, 0, 0, 0, 0, 0), f32 = c(0, 0, 0, 0, 0, 0), f33 = c(3, 0, 1, 0, 1, 0), f34 = c(5, 44, 32, 4, 15, 5), f35 = c(0, 0, 0, 0, 0, 0), f36 = c(0, 0, 0, 0, 0, 0), f37 = c(0, 0, 0, 0, 0, 0), f38 = c(0, 0, 0, 0, 0, 0), f39 = c(6, 8, 10, 3, 2, 3), f40 = c(4, 6, 16, 4, 4, 3), f41 = c(18, 36, 37, 7, 5, 7), f42 = c(0, 18, 27, 0, 14, 1), f43 = c(0, 0, 0, 0, 0, 0), f44 = c(54, 743, 910, 65, 184, 100), f45 = c(14, 133, 91, 25, 18, 40), f46 = c(0, 0, 0, 0, 0, 0), f47 = c(4, 25, 17, 6, 6, 8), f48 = c(0, 0, 0, 0, 0, 0), f49 = c(0.46, 1, 1.5, 1.14, 1.5, 1.14), f50 = c(2.67, 1.86, 1.83, 1.88, 1.67, 1.88), f51 = c(3, 9, 1, 2, 1, 2), f52 = c(0, 1, 2, 1, 1, 1), f53 = c(10, 12, 5, 4, 2, 4), f54 = c(0, 0, 0, 0, 0, 0), ft55 = c(3, 10, 3, 3, 2, 3), f56 = c(0.54, 0.07, 0.03, 0.32, 0.07, 0.21), f57 = c(0.21, 0.04, 0.01, 0.14, 0.02, 0.1), f58 = c(0.21, 0.02, 0.01, 0.07, 0.01, 0.05 ), f59 = c(0, 0, 0, 0, 0, 0), f60 = c(0.32, 0.04, 0.02, 0.18, 0.04, 0.12), f61 = c(0.11, 0.02, 0.01, 0.07, 0.01, 0.05), f62 = c(0.04, 0, 0, 0, 0, 0), f63 = c(0.14, 0.03, 0.01, 0.14, 0.02, 0.1), f64 = c(0.18, 0.01, 0, 0, 0, 0), f65 = c(0.04, 0.01, 0, 0, 0, 0), f66 = c(0.04, 0.01, 0.01, 0.07, 0.01, 0.05), f67 = c(0, 0, 0, 0, 0, 0), f68 = c(0.46, 0.04, 0.02, 0.21, 0.04, 0.14), f69 = c(0, 0.02, 0, 0.04, 0.01, 0.02), f70 = c(0, 0, 0, 0, 0, 0), f71 = c(0.71, 0.09, 0.03, 0.36, 0.07, 0.24), f72 = c(0.18, 0, 0, 0, 0, 0), f73 = c(0, 0, 0, 0, 0, 0), f74 = c(1, 1, 1, 1, 1, 1), f75 = c(0.08, 0.05, 0.12, 0.14, 0.25, 0.14), f76 = c(0, 0, 0, 0, 0, 0), f77 = c(0, 0, 0, 0, 0, 0), f78 = c(0.21, 0.02, 0.01, 0.07, 0.01, 0.05 ), f79 = c(0, 0, 0, 0, 0, 0), f80 = c(0.68, 0.97, 0.99, 0.82, 0.96, 0.88), f81 = c(0, 0.08, 0.09, 0.21, 0.2, 0.17), f82 = c(0.39, 0.19, 0.25, 0.18, 0.21, 0.29), f83 = c(0, 0, 0, 0, 0, 0), f84 = c(0.07, 0.11, 0.25, 0.25, 0.32, 0.19), f85 = c(0, 0.05, 0.08, 0.07, 0.2, 0.1), f86 = c(0, 0, 0, 0, 0, 0), f87 = c(0, 0, 0, 0, 0, 0), f88 = c(0.11, 0, 0, 0, 0.01, 0), f89 = c(0.18, 0.13, 0.08, 0.14, 0.18, 0.12), f90 = c(0, 0, 0, 0, 0, 0), f91 = c(0, 0, 0, 0, 0, 0), f92 = c(0, 0, 0, 0, 0, 0), f93 = c(0, 0, 0, 0, 0, 0), f94 = c(0.21, 0.02, 0.03, 0.11, 0.02, 0.07 ), f95 = c(0.14, 0.02, 0.04, 0.14, 0.05, 0.07), f96 = c(0.64, 0.1, 0.09, 0.25, 0.06, 0.17), f97 = c(0, 0.05, 0.07, 0, 0.17, 0.02), f98 = c(0, 0, 0, 0, 0, 0), f99 = c(1.93, 2.16, 2.28, 2.32, 2.24, 2.38), f100 = c(0.5, 0.39, 0.23, 0.89, 0.22, 0.95), f101 = c(0, 0, 0, 0, 0, 0), f102 = c(0.14, 0.07, 0.04, 0.21, 0.07, 0.19), f103 = c(0, 0, 0, 0, 0, 0), f104 = c(0.02, 0, 0, 0.04, 0.02, 0.03), f105 = c(0.1, 0.01, 0, 0.07, 0.02, 0.04), f106 = c(0.11, 0.03, 0, 0.07, 0.01, 0.05), f107 = c(0, 0, 0.01, 0.04, 0.01, 0.02), f108 = c(0.36, 0.03, 0.01, 0.14, 0.02, 0.1), f109 = c(0, 0, 0, 0, 0, 0), f110 = c(0.11, 0.03, 0.01, 0.11, 0.02, 0.07)), .Names = c("label", "f1", "f2","f3", "f4", "f5", "f6", "f7", "f8", "f9", "f10", "f11", "f12","f13", "f14", "f15", "f16", "f17", "f18", "ft19", "f20", "f21","f22", "f23", "f24", "f25", "f26", "f27", "f28", "f29", "f30","f31", "f32", "f33", "f34", "f35", "f36", "f37", "f38", "f39","f40", "f41", "f42", "f43", "f44", "f45", "f46", "f47", "f48","f49", "f50", "f51", "f52", "f53", "f54", "ft55", "f56", "f57","f58", "f59", "f60", "f61", "f62", "f63", "f64", "f65", "f66","f67", "f68", "f69", "f70", "f71", "f72", "f73", "f74", "f75","f76", "f77", "f78", "f79", "f80", "f81", "f82", "f83", "f84","f85", "f86", "f87", "f88", "f89", "f90", "f91", "f92", "f93","f94", "f95", "f96", "f97", "f98", "f99", "f100", "f101", "f102","f103", "f104", "f105", "f106", "f107", "f108", "f109", "f110"), class = "data.frame", row.names = c(NA, -6L)) 解决方案 I think in the solutions using all(x == 0) it is slightly more efficient to use any(x!=0), because any stops after the first instance of an element being !=0, which will be important with growing number of rows.To provide a different solution using plyr and colwise (dat being the dputdata): library(plyr)f0 <- function(x) any(x!=0) & is.numeric(x)colwise(identity, f0)(dat)The idea is to go through every column in dat and return it (identity), but only if f0 returns TRUE, i.e. the column has at least one entry !=0 and the column is.numericEDIT:To do this for every data.frame in your list, eg. training_data <- list(dat, dat, dat, dat)training_data_clean <- lapply(training_data, function(z) colwise(identity, f0)(z))sapply(training_data, dim) [,1] [,2] [,3] [,4][1,] 6 6 6 6[2,] 111 111 111 111sapply(training_data_clean, dim) [,1] [,2] [,3] [,4][1,] 6 6 6 6[2,] 74 74 74 74EDIT2:To retain the label column: lapply(training_data, function(z) cbind(label = z$label, colwise(identity, f0)(z))) 这篇关于过滤R中的多个列的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！