R中的0到1归一化，同时保留列1和标题[重复]

uujelgoq 于 2023-07-31 发布在其他

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Standardize data columns in R（16个回答）
5天前关闭。
我正在尝试将我的数据集在0到1之间进行归一化。每列应独立归一化。我想输出一个新的dataframe，它保留了第一列（未规范化）和所有原始列标题。
这是我的数据的一个子集：

SEC <- structure(list(ml = c(0, 0.03, 0.06, 0.09, 0.12, 0.15, 0.18, 
0.21, 0.24, 0.27), A1_280 = c(0.542, 0.322, 0.286, 0.261, 0.19, 
-0.258, -0.272, -0.046, -0.005, 0.138), A1_420 = c(-0.06, -0.303, 
-0.192, -0.381, 0.15, -0.268, -0.576, -0.016, -0.541, -0.41), 
    A2_280 = c(9.877, 27.637, 3.513, -0.882, -1.92, -1.251, -2.284, 
    -2.129, -3.131, -2.913), A2_420 = c(-0.445, 13.337, 1.075, 
    -1.402, -2.156, -2.263, -1.988, -2.105, -2.082, -2.61), A3_280 = c(8.782, 
    59.775, 56.769, 22.842, 9.086, 3.466, 2.256, 1.341, 0.946, 
    0.754), A3_420 = c(0.54, 30.736, 29.073, 12.277, 4.413, 1.77, 
    1.123, 0.488, 0.634, -0.011), B1_280 = c(14.95, 61.441, 37.189, 
    10.928, 4.316, 2.292, 0.757, 0.995, 0.997, -0.07), B1_420 = c(2.455, 
    30.966, 18.61, 4.779, 1.511, 0.74, 0.267, 0.533, 0.149, -0.551
    ), B2_280 = c(-0.288, -0.304, -0.006, -0.158, -0.284, -0.131, 
    -0.443, -0.081, -0.387, -0.04), B2_420 = c(-0.074, -0.256, 
    0.022, 0.104, -0.287, -0.139, -0.015, 0.1, -0.021, -0.146
    ), B3_280 = c(0.084, 0.043, 0.061, 0.032, 0.038, 0.072, 0.03, 
    0.128, 0.077, 0.098), B3_420 = c(-0.056, 0.095, 0.05, -0.015, 
    -0.106, 0.106, -0.017, -0.001, 0.036, 0.139), AB1_280 = c(1.599, 
    1.908, 0.735, 0.49, 0.708, 0.109, 0.702, -0.487, -0.009, 
    -0.196), AB1_420 = c(0.199, 1.218, 0.469, 0.564, 0.498, -0.2, 
    -0.322, 0.294, 0.367, -0.281), AB2_280 = c(-1.46, -1.2, -1.977, 
    -2.736, -2.087, -2.144, -2.246, -2.84, -2.304, -3.106), AB2_420 = c(-1, 
    -0.468, -0.459, -0.345, -1.145, -0.924, -1.622, -0.869, -1.028, 
    -1.183), AB3_280 = c(0.306, 1.392, -2.248, -3.247, -3.715, 
    -2.699, -3.896, -2.744, -3.653, -3.387), AB3_420 = c(-0.899, 
    0.817, -1.41, -1.162, -1.258, -1.409, -1.7, -1.309, -1.946, 
    -1.658), AB4_280 = c(6.847, 55.721, 51.163, 21.166, 8.441, 
    3.105, 2.631, 1.265, -0.184, 0.529), AB4_420 = c(-0.861, 
    27.465, 25.185, 10.767, 4.136, 1.414, 0.545, -0.098, 0.242, 
    -0.509)), row.names = c(NA, -10L), spec = structure(list(
    cols = list(ml = structure(list(), class = c("collector_double", 
    "collector")), A1_280 = structure(list(), class = c("collector_double", 
    "collector")), A1_420 = structure(list(), class = c("collector_double", 
    "collector")), A2_280 = structure(list(), class = c("collector_double", 
    "collector")), A2_420 = structure(list(), class = c("collector_double", 
    "collector")), A3_280 = structure(list(), class = c("collector_double", 
    "collector")), A3_420 = structure(list(), class = c("collector_double", 
    "collector")), B1_280 = structure(list(), class = c("collector_double", 
    "collector")), B1_420 = structure(list(), class = c("collector_double", 
    "collector")), B2_280 = structure(list(), class = c("collector_double", 
    "collector")), B2_420 = structure(list(), class = c("collector_double", 
    "collector")), B3_280 = structure(list(), class = c("collector_double", 
    "collector")), B3_420 = structure(list(), class = c("collector_double", 
    "collector")), AB1_280 = structure(list(), class = c("collector_double", 
    "collector")), AB1_420 = structure(list(), class = c("collector_double", 
    "collector")), AB2_280 = structure(list(), class = c("collector_double", 
    "collector")), AB2_420 = structure(list(), class = c("collector_double", 
    "collector")), AB3_280 = structure(list(), class = c("collector_double", 
    "collector")), AB3_420 = structure(list(), class = c("collector_double", 
    "collector")), AB4_280 = structure(list(), class = c("collector_double", 
    "collector")), AB4_420 = structure(list(), class = c("collector_double", 
    "collector"))), default = structure(list(), class = c("collector_guess", 
    "collector")), delim = ","), class = "col_spec"), problems = <pointer: 0x5606ec29c390>, class = c("spec_tbl_df", 
"tbl_df", "tbl", "data.frame"))

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以下是我到目前为止的代码：

normalize_0_to_1_columnwise <- function(SEC) {
  normalized_SEC <- data.frame(ml = SEC$ml)  # Copy the first column as it is
  
  # Apply normalization for each column (excluding the first column 'ml')
  for (col in names(SEC)[-1]) {
    normalized_SEC[[col]] <- (SEC[[col]] - min(SEC[[col]])) / (max(SEC[[col]]) - min(SEC[[col]]))
  }
  
  # Preserve the original column headers
  colnames(normalized_SEC)[-1] <- colnames(SEC)[-1]

}

# Output normalized dataframe
normalized_SEC

型
这用于保留第一列“ml”和所有列标题，但 Dataframe 中的所有“值”都是NA。我哪里做错了？
我知道还有其他类似的问题，但我不能让他们为我的数据和所需的输出工作。

来源：https://stackoverflow.com/questions/76764194/0-to-1-normalisation-in-r-whilst-preserving-column-1-and-headers

3条答案

按热度按时间

np8igboo1#

在tidyverse中，您可以：

library(tidyverse)
sec_scaled <- mutate(SEC, across(-ml, scales::rescale))

sec_scaled

# A tibble: 10 × 21
      ml A1_280 A1_420  A2_280 A2_420  A3_280 A3_420 B1_280 B1_420 B2_280 B2_420 B3_280
   <dbl>  <dbl>  <dbl>   <dbl>  <dbl>   <dbl>  <dbl>  <dbl>  <dbl>  <dbl>  <dbl>  <dbl>
 1  0    1      0.711  0.423   0.136  0.136   0.0179 0.244  0.0954  0.355 0.545  0.551 
 2  0.03 0.730  0.376  1       1      1       1      1      1       0.318 0.0793 0.133 
 3  0.06 0.686  0.529  0.216   0.231  0.949   0.946  0.606  0.608   1     0.790  0.316 
 4  0.09 0.655  0.269  0.0731  0.0758 0.374   0.400  0.179  0.169   0.652 1      0.0204
 5  0.12 0.568  1      0.0394  0.0285 0.141   0.144  0.0713 0.0654  0.364 0      0.0816
 6  0.15 0.0172 0.424  0.0611  0.0218 0.0459  0.0579 0.0384 0.0410  0.714 0.379  0.429 
 7  0.18 0      0      0.0275  0.0390 0.0254  0.0369 0.0134 0.0260  0     0.696  0     
 8  0.21 0.278  0.771  0.0326  0.0317 0.00995 0.0162 0.0173 0.0344  0.828 0.990  1     
 9  0.24 0.328  0.0482 0       0.0331 0.00325 0.0210 0.0173 0.0222  0.128 0.680  0.480 
10  0.27 0.504  0.229  0.00709 0      0       0      0      0       0.922 0.361  0.694 
# ℹ 9 more variables: B3_420 <dbl>, AB1_280 <dbl>, AB1_420 <dbl>, AB2_280 <dbl>,
#   AB2_420 <dbl>, AB3_280 <dbl>, AB3_420 <dbl>, AB4_280 <dbl>, AB4_420 <dbl>

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赞(0）回复(0）举报 2023-07-31

i34xakig2#

函数返回最后一行。函数内部的最后一行是colnames(SEC)[-1]。在函数关闭}之后有normalized_SEC，但它需要作为函数定义中的最后一行才能正确返回。如果我做了这个修正，你的函数就可以很好地处理你的样本数据了。
但是如果它仍然不能处理完整的数据，并且您正在获得NA输出，则输入中可能有NA值，并且您需要将na.rm = TRUE参数添加到min()和max()调用中。
也就是说，我可以从您的tbl_df类中看到您正在使用一些tidyverse函数。一个不错的dplyr方法是这样的：

library(dplyr)
result = SEC |> mutate(across(-ml, \(x) {
  min = min(x, na.rm = TRUE);
  max = max(x, na.rm = TRUE);
  (x - min) / (max - min)
}))

# # A tibble: 10 × 21
#       ml A1_280 A1_420  A2_280 A2_420  A3_280 A3_420 B1_280 B1_420 B2_280 B2_420 B3_280 B3_420 AB1_280 AB1_420
#    <dbl>  <dbl>  <dbl>   <dbl>  <dbl>   <dbl>  <dbl>  <dbl>  <dbl>  <dbl>  <dbl>  <dbl>  <dbl>   <dbl>   <dbl>
#  1  0    1      0.711  0.423   0.136  0.136   0.0179 0.244  0.0954  0.355 0.545  0.551   0.204   0.871  0.338 
#  2  0.03 0.730  0.376  1       1      1       1      1      1       0.318 0.0793 0.133   0.820   1      1     
#  3  0.06 0.686  0.529  0.216   0.231  0.949   0.946  0.606  0.608   1     0.790  0.316   0.637   0.510  0.514 
#  4  0.09 0.655  0.269  0.0731  0.0758 0.374   0.400  0.179  0.169   0.652 1      0.0204  0.371   0.408  0.575 
#  5  0.12 0.568  1      0.0394  0.0285 0.141   0.144  0.0713 0.0654  0.364 0      0.0816  0       0.499  0.532 
#  6  0.15 0.0172 0.424  0.0611  0.0218 0.0459  0.0579 0.0384 0.0410  0.714 0.379  0.429   0.865   0.249  0.0792
#  7  0.18 0      0      0.0275  0.0390 0.0254  0.0369 0.0134 0.0260  0     0.696  0       0.363   0.496  0     
#  8  0.21 0.278  0.771  0.0326  0.0317 0.00995 0.0162 0.0173 0.0344  0.828 0.990  1       0.429   0      0.4   
#  9  0.24 0.328  0.0482 0       0.0331 0.00325 0.0210 0.0173 0.0222  0.128 0.680  0.480   0.580   0.200  0.447 
# 10  0.27 0.504  0.229  0.00709 0      0       0      0      0       0.922 0.361  0.694   1       0.122  0.0266
# # ℹ 6 more variables: AB2_280 <dbl>, AB2_420 <dbl>, AB3_280 <dbl>, AB3_420 <dbl>, AB4_280 <dbl>, AB4_420 <dbl>

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使用此示例数据（dput()来自问题，删除了pointer和spec_*内容）：

SEC = structure(list(ml = c(0, 0.03, 0.06, 0.09, 0.12, 0.15, 0.18, 
0.21, 0.24, 0.27), A1_280 = c(0.542, 0.322, 0.286, 0.261, 0.19, 
-0.258, -0.272, -0.046, -0.005, 0.138), A1_420 = c(-0.06, -0.303, 
-0.192, -0.381, 0.15, -0.268, -0.576, -0.016, -0.541, -0.41), 
    A2_280 = c(9.877, 27.637, 3.513, -0.882, -1.92, -1.251, -2.284, 
    -2.129, -3.131, -2.913), A2_420 = c(-0.445, 13.337, 1.075, 
    -1.402, -2.156, -2.263, -1.988, -2.105, -2.082, -2.61), A3_280 = c(8.782, 
    59.775, 56.769, 22.842, 9.086, 3.466, 2.256, 1.341, 0.946, 
    0.754), A3_420 = c(0.54, 30.736, 29.073, 12.277, 4.413, 1.77, 
    1.123, 0.488, 0.634, -0.011), B1_280 = c(14.95, 61.441, 37.189, 
    10.928, 4.316, 2.292, 0.757, 0.995, 0.997, -0.07), B1_420 = c(2.455, 
    30.966, 18.61, 4.779, 1.511, 0.74, 0.267, 0.533, 0.149, -0.551
    ), B2_280 = c(-0.288, -0.304, -0.006, -0.158, -0.284, -0.131, 
    -0.443, -0.081, -0.387, -0.04), B2_420 = c(-0.074, -0.256, 
    0.022, 0.104, -0.287, -0.139, -0.015, 0.1, -0.021, -0.146
    ), B3_280 = c(0.084, 0.043, 0.061, 0.032, 0.038, 0.072, 0.03, 
    0.128, 0.077, 0.098), B3_420 = c(-0.056, 0.095, 0.05, -0.015, 
    -0.106, 0.106, -0.017, -0.001, 0.036, 0.139), AB1_280 = c(1.599, 
    1.908, 0.735, 0.49, 0.708, 0.109, 0.702, -0.487, -0.009, 
    -0.196), AB1_420 = c(0.199, 1.218, 0.469, 0.564, 0.498, -0.2, 
    -0.322, 0.294, 0.367, -0.281), AB2_280 = c(-1.46, -1.2, -1.977, 
    -2.736, -2.087, -2.144, -2.246, -2.84, -2.304, -3.106), AB2_420 = c(-1, 
    -0.468, -0.459, -0.345, -1.145, -0.924, -1.622, -0.869, -1.028, 
    -1.183), AB3_280 = c(0.306, 1.392, -2.248, -3.247, -3.715, 
    -2.699, -3.896, -2.744, -3.653, -3.387), AB3_420 = c(-0.899, 
    0.817, -1.41, -1.162, -1.258, -1.409, -1.7, -1.309, -1.946, 
    -1.658), AB4_280 = c(6.847, 55.721, 51.163, 21.166, 8.441, 
    3.105, 2.631, 1.265, -0.184, 0.529), AB4_420 = c(-0.861, 
    27.465, 25.185, 10.767, 4.136, 1.414, 0.545, -0.098, 0.242, 
    -0.509)), row.names = c(NA, -10L), class = c("tbl_df", "tbl", 
"data.frame"))

型

赞(0）回复(0）举报 2023-07-31

wn9m85ua3#

请原谅我将您的数据从tibble转换为data.frame，因为我在重新创建结构时遇到了问题。下面是基R，所以它也适用于tibles，但会返回data.frame。
首先，数据：

SECDF <- data.frame(ml = c(0, 0.03, 0.06, 0.09, 0.12, 0.15, 0.18, 0.21, 0.24, 0.27),
                    A1_280 = c(0.542, 0.322, 0.286, 0.261, 0.19, -0.258, -0.272, -0.046, -0.005, 0.138),
                    A1_420 = c(-0.06, -0.303, -0.192, -0.381, 0.15, -0.268, -0.576, -0.016, -0.541, -0.41),
                    A2_280 = c(9.877, 27.637, 3.513, -0.882, -1.92, -1.251, -2.284, -2.129, -3.131, -2.913),
                    A2_420 = c(-0.445, 13.337, 1.075,  -1.402, -2.156, -2.263, -1.988, -2.105, -2.082, -2.61),
                    A3_280 = c(8.782,  59.775, 56.769, 22.842, 9.086, 3.466, 2.256, 1.341, 0.946, 0.754),
                    A3_420 = c(0.54, 30.736, 29.073, 12.277, 4.413, 1.77, 1.123, 0.488, 0.634, -0.011),
                    B1_280 = c(14.95, 61.441, 37.189, 10.928, 4.316, 2.292, 0.757, 0.995, 0.997, -0.07),
                    B1_420 = c(2.455, 30.966, 18.61, 4.779, 1.511, 0.74, 0.267, 0.533, 0.149, -0.551),
                    B2_280 = c(-0.288, -0.304, -0.006, -0.158, -0.284, -0.131, -0.443, -0.081, -0.387, -0.04),
                    B2_420 = c(-0.074, -0.256, 0.022, 0.104, -0.287, -0.139, -0.015, 0.1, -0.021, -0.146),
                    B3_280 = c(0.084, 0.043, 0.061, 0.032, 0.038, 0.072, 0.03, 0.128, 0.077, 0.098),
                    B3_420 = c(-0.056, 0.095, 0.05, -0.015, -0.106, 0.106, -0.017, -0.001, 0.036, 0.139),
                    AB1_280 = c(1.599, 1.908, 0.735, 0.49, 0.708, 0.109, 0.702, -0.487, -0.009, -0.196),
                    AB1_420 = c(0.199, 1.218, 0.469, 0.564, 0.498, -0.2,  -0.322, 0.294, 0.367, -0.281),
                    AB2_280 = c(-1.46, -1.2, -1.977, -2.736, -2.087, -2.144, -2.246, -2.84, -2.304, -3.106),
                    AB2_420 = c(-1, -0.468, -0.459, -0.345, -1.145, -0.924, -1.622, -0.869, -1.028, -1.183),
                    AB3_280 = c(0.306, 1.392, -2.248, -3.247, -3.715, -2.699, -3.896, -2.744, -3.653, -3.387),
                    AB3_420 = c(-0.899, 0.817, -1.41, -1.162, -1.258, -1.409, -1.7, -1.309, -1.946, -1.658),
                    AB4_280 = c(6.847, 55.721, 51.163, 21.166, 8.441, 3.105, 2.631, 1.265, -0.184, 0.529),
                    AB4_420 = c(-0.861, 27.465, 25.185, 10.767, 4.136, 1.414, 0.545, -0.098, 0.242, -0.509))

SECDF
     ml A1_280 A1_420 A2_280 A2_420 A3_280 A3_420 B1_280 B1_420 B2_280 B2_420 B3_280 B3_420 AB1_280 AB1_420 AB2_280 AB2_420 AB3_280 AB3_420 AB4_280 AB4_420
1  0.00  0.542 -0.060  9.877 -0.445  8.782  0.540 14.950  2.455 -0.288 -0.074  0.084 -0.056   1.599   0.199  -1.460  -1.000   0.306  -0.899   6.847  -0.861
2  0.03  0.322 -0.303 27.637 13.337 59.775 30.736 61.441 30.966 -0.304 -0.256  0.043  0.095   1.908   1.218  -1.200  -0.468   1.392   0.817  55.721  27.465
3  0.06  0.286 -0.192  3.513  1.075 56.769 29.073 37.189 18.610 -0.006  0.022  0.061  0.050   0.735   0.469  -1.977  -0.459  -2.248  -1.410  51.163  25.185
4  0.09  0.261 -0.381 -0.882 -1.402 22.842 12.277 10.928  4.779 -0.158  0.104  0.032 -0.015   0.490   0.564  -2.736  -0.345  -3.247  -1.162  21.166  10.767
5  0.12  0.190  0.150 -1.920 -2.156  9.086  4.413  4.316  1.511 -0.284 -0.287  0.038 -0.106   0.708   0.498  -2.087  -1.145  -3.715  -1.258   8.441   4.136
6  0.15 -0.258 -0.268 -1.251 -2.263  3.466  1.770  2.292  0.740 -0.131 -0.139  0.072  0.106   0.109  -0.200  -2.144  -0.924  -2.699  -1.409   3.105   1.414
7  0.18 -0.272 -0.576 -2.284 -1.988  2.256  1.123  0.757  0.267 -0.443 -0.015  0.030 -0.017   0.702  -0.322  -2.246  -1.622  -3.896  -1.700   2.631   0.545
8  0.21 -0.046 -0.016 -2.129 -2.105  1.341  0.488  0.995  0.533 -0.081  0.100  0.128 -0.001  -0.487   0.294  -2.840  -0.869  -2.744  -1.309   1.265  -0.098
9  0.24 -0.005 -0.541 -3.131 -2.082  0.946  0.634  0.997  0.149 -0.387 -0.021  0.077  0.036  -0.009   0.367  -2.304  -1.028  -3.653  -1.946  -0.184   0.242
10 0.27  0.138 -0.410 -2.913 -2.610  0.754 -0.011 -0.070 -0.551 -0.040 -0.146  0.098  0.139  -0.196  -0.281  -3.106  -1.183  -3.387  -1.658   0.529  -0.509

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现在，为缩放创建一个（矢量化）函数，并将其apply到数据框列。如果传递正确的center和scale值，也可以使用scale的基本版本，但它足够简单，我们可以滚动自己的：

scale01 <- function(x) (x - min(x)) / (max(x) - min(x))

型
现在简单地说：

normSECDF <- cbind(SECDF[, 1L], apply(SECDF[, -1L], 2L, scale01))
normSECDF
               A1_280     A1_420      A2_280     A2_420      A3_280     A3_420     B1_280     B1_420    B2_280     B2_420     B3_280    B3_420   AB1_280    AB1_420
 [1,] 0.00 1.00000000 0.71074380 0.422776911 0.13576221 0.136019383 0.01792045 0.24418397 0.09537710 0.3546911 0.54475703 0.55102041 0.2040816 0.8709812 0.33831169
 [2,] 0.03 0.72972973 0.37603306 1.000000000 1.00000000 1.000000000 1.00000000 1.00000000 1.00000000 0.3180778 0.07928389 0.13265306 0.8204082 1.0000000 1.00000000
 [3,] 0.06 0.68550369 0.52892562 0.215938638 0.23107795 0.949068975 0.94591342 0.60572906 0.60795761 1.0000000 0.79028133 0.31632653 0.6367347 0.5102296 0.51363636
 [4,] 0.09 0.65479115 0.26859504 0.073095424 0.07575092 0.374239677 0.39964875 0.17879729 0.16911508 0.6521739 1.00000000 0.02040816 0.3714286 0.4079332 0.57532468
 [5,] 0.12 0.56756757 1.00000000 0.039359074 0.02846930 0.141170092 0.14388396 0.07130432 0.06542501 0.3638444 0.00000000 0.08163265 0.0000000 0.4989562 0.53246753
 [6,] 0.15 0.01719902 0.42424242 0.061102444 0.02175958 0.045949747 0.05792435 0.03839964 0.04096202 0.7139588 0.37851662 0.42857143 0.8653061 0.2488518 0.07922078
 [7,] 0.18 0.00000000 0.00000000 0.027528601 0.03900420 0.025448569 0.03688165 0.01344475 0.02595425 0.0000000 0.69565217 0.00000000 0.3632653 0.4964509 0.00000000
 [8,] 0.21 0.27764128 0.77134986 0.032566303 0.03166740 0.009945613 0.01622923 0.01731398 0.03439414 0.8283753 0.98976982 1.00000000 0.4285714 0.0000000 0.40000000
 [9,] 0.24 0.32800983 0.04820937 0.000000000 0.03310968 0.003253079 0.02097766 0.01734649 0.02221024 0.1281465 0.68030691 0.47959184 0.5795918 0.1995825 0.44740260
[10,] 0.27 0.50368550 0.22865014 0.007085283 0.00000000 0.000000000 0.00000000 0.00000000 0.00000000 0.9221968 0.36061381 0.69387755 1.0000000 0.1215031 0.02662338
        AB2_280   AB2_420    AB3_280    AB3_420    AB4_280    AB4_420
 [1,] 0.8635887 0.4870791 0.79462935 0.37893594 0.12576693 0.00000000
 [2,] 1.0000000 0.9036805 1.00000000 1.00000000 1.00000000 1.00000000
 [3,] 0.5923400 0.9107283 0.31164902 0.19399204 0.91846883 0.91950858
 [4,] 0.1941238 1.0000000 0.12273071 0.28374955 0.38189786 0.41050625
 [5,] 0.5346275 0.3735317 0.03422844 0.24900471 0.15427958 0.17641037
 [6,] 0.5047219 0.5465936 0.22636157 0.19435396 0.05883195 0.08031491
 [7,] 0.4512067 0.0000000 0.00000000 0.08903366 0.05035328 0.04963638
 [8,] 0.1395593 0.5896633 0.21785174 0.23054651 0.02591897 0.02693638
 [9,] 0.4207765 0.4651527 0.04595310 0.00000000 0.00000000 0.03893949
[10,] 0.0000000 0.3437745 0.09625567 0.10423453 0.01275378 0.01242675

> apply(normSECDF, 2L, min)
         A1_280  A1_420  A2_280  A2_420  A3_280  A3_420  B1_280  B1_420  B2_280  B2_420  B3_280  B3_420 AB1_280 AB1_420 AB2_280 AB2_420 AB3_280 AB3_420 AB4_280 AB4_420 
      0       0       0       0       0       0       0       0       0       0       0       0       0       0       0       0       0       0       0       0       0 
> apply(normSECDF, 2L, max)
         A1_280  A1_420  A2_280  A2_420  A3_280  A3_420  B1_280  B1_420  B2_280  B2_420  B3_280  B3_420 AB1_280 AB1_420 AB2_280 AB2_420 AB3_280 AB3_420 AB4_280 AB4_420 
   0.27    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00

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赞(0）回复(0）举报 2023-07-31

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R中的0到1归一化，同时保留列1和标题[重复]

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