R 将数据缩放回其原始值
R descale data back to their original values
我有以下可重现的数据：
MyScaledData 包含 6 个变量的 0 到 1 之间的缩放值。 minvec 和 maxvec 是命名向量，包含用于创建缩放数据框 MyScaledData 的原始数据集中的最大值和最小值。 minvec 和 maxvec 包含原始数据集的所有 22 个变量的值，包括我现在在 MyScaledData.
中的 6 个变量
X14863 X15066 X15067 X15068 X15069 X15070
0.6014784 0.6975109 0.5043208 0.15284648 0.9416364 0.7860731
0.2495215 0.7801444 0.6683925 0.13768245 0.4277954 0.2058412
0.6167705 0.3344044 0.9254125 0.12777565 0.3826231 0.2590457
0.1227380 0.4448501 0.3961802 0.19117246 0.7789835 0.7587897
0.7299760 0.6375931 0.5760061 0.44746838 0.3634903 0.1079679
0.1988647 0.7814712 0.6572054 0.71409305 0.6715690 0.4029459
0.5041371 0.6374958 0.9333635 0.89057831 0.5716711 0.7219823
0.5774327 0.7677038 0.7622717 0.45288270 0.2817869 0.2572325
0.6809509 0.6089656 0.8191862 0.01151454 0.2780449 0.4655353
0.5754383 0.5662045 0.7003630 0.62559642 0.2865510 0.1847980
MyScaledData<-structure(list(X14863=c(0.601478444979532,0.249521497274968,0.616770466379489,0.122737966507165,0.729975993009922,0.198864661389536,0.504137054265617,0.577432671357089,0.680950947164095,0.575438259547452),X15066=c(0.697510926657699,0.780144354632397,0.334404422875259,0.444850091405716,0.637593061483412,0.781471212351781,0.637495834667556,0.7677038048039,0.608965550162107,0.566204459603197),X15067=c(0.50432083998529,0.668392530333367,0.925412484830622,0.396180214305286,0.576006062451239,0.657205387087382,0.933363470346907,0.762271729415789,0.819186151914183,0.700362991098644),X15068=c(0.152846483002917,0.137682446305942,0.127775652495726,0.191172455317975,0.447468375530484,0.714093046059637,0.890578310935752,0.452882699805154,0.011514536383708,0.625596417031532),X15069=c(0.94163636689763,0.427795395079331,0.38262308941233,0.77898345642139,0.363490265569212,0.671568951210917,0.571671115989958,0.281786881885636,0.278044876559552,0.286551022600823),X15070=c(0.786073059382553,0.205841229942702,0.259045736299276,
0.758789694211416,0.107967864736275,0.402945912782515,0.721982268066207,0.257232456508833,0.46553533255268,0.184798001614338)),row.names=c(NA,10L),class="data.frame"); minvec<-c(X14861=22.95,X14862=29.95,X14863= 39.95,X15066=59.95,X15067=79.95,X15068=14.99,X15069=24.99,X15070=33.45,X15071=36.95,X15072=44.95,X15073=54.95,X15074=74.95,X15132=12.95,X15548=12.95,X15549=22.95,X15550=29.95,X15551=39.95,X15552=59.95,X15553=79.95,X15956=49.95,X15957=49.95,X16364=3.5);maxvec<-c(X14861=29.99,X14862=39.99,X14863=49.99,X15066=79.99,X15067=99.99,X15068=19.99,X15069=29.99,X15070=39.99,X15071=49.99,X15072=59.99,X15073=79.99,X15074=99.99,X15132=19.99,X15548=19.99,X15549=29.99,X15550=39.99,X15551=49.99,X15552=79.99,X15553=99.99,X15956=59.99,X15957=59.99,X16364=9.99)
我想通过将 min/max 值与基于名称的每个对应列相匹配，将 MyScaledData 重新缩放到它们的原始比例。我尝试了以下方法：
descale <- function(x,minval,maxval) {x*(maxval-minval) + minval}
as.data.frame(Map(descale,MyScaledData,minvec,maxvec))
我得到的输出比 MyScaledData 多了 6 列。我感觉到该函数甚至没有按名称匹配列，因此输出计算不正确。如何按列名匹配函数，以便它为每一列采用相应的 minvec 和 maxvec 元素，并且 return 只有我拥有的 6 列？
所需的输出应为：
MyDeScaledData <- structure(list(X14863 = c(45.9888435875945, 42.4551958326407,46.1423754824501, 41.1822891837319, 47.2789589698196, 41.9466012003509,45.0115360248268, 45.7474240204252, 46.7867475095275, 45.7274001258564), X15066 = c(73.9281189702203, 75.5840928668332, 66.6514646344202,68.8647958317706, 72.7273649521276, 75.6106830955297, 72.7254165267378,75.3347842482702, 72.1536696252486, 71.2967373704481), X15067 = c(90.0565896333052,93.3445863078807, 98.4952661960057, 87.8894514946779, 91.4931614915228,93.1203959572311, 98.654603945752, 95.2259254574924, 96.3664904843602,93.9852743416168), X15068 = c(15.7542324150146, 15.6784122315297,15.6288782624786, 15.9458622765899, 17.2273418776524, 18.5604652302982,19.4428915546788, 17.2544134990258, 15.0475726819185, 18.1179820851577), X15069 = c(29.6981818344881, 27.1289769753967, 26.9031154470616,28.8849172821069, 26.8074513278461, 28.3478447560546, 27.8483555799498,26.3989344094282, 26.3802243827978, 26.4227551130041), X15070 = c(38.5909178083619,34.7962016438253, 35.1441591153973, 38.4124846001427, 34.1561098353752,36.0852662695976, 38.171764033153, 35.1323002655678, 36.4946010748945,34.6585789305578)), row.names = c(NA, 10L), class = "data.frame")
感谢@Shirin Yavari 提供解决方案：
MyDeScaledData<-as.data.frame(Map(descale,MyScaledData,minvec[names(MyScaledData)],maxvec[names(MyScaledData)]))
我有以下可重现的数据：
MyScaledData 包含 6 个变量的 0 到 1 之间的缩放值。 minvec 和 maxvec 是命名向量，包含用于创建缩放数据框 MyScaledData 的原始数据集中的最大值和最小值。 minvec 和 maxvec 包含原始数据集的所有 22 个变量的值，包括我现在在 MyScaledData.
X14863 X15066 X15067 X15068 X15069 X15070
0.6014784 0.6975109 0.5043208 0.15284648 0.9416364 0.7860731
0.2495215 0.7801444 0.6683925 0.13768245 0.4277954 0.2058412
0.6167705 0.3344044 0.9254125 0.12777565 0.3826231 0.2590457
0.1227380 0.4448501 0.3961802 0.19117246 0.7789835 0.7587897
0.7299760 0.6375931 0.5760061 0.44746838 0.3634903 0.1079679
0.1988647 0.7814712 0.6572054 0.71409305 0.6715690 0.4029459
0.5041371 0.6374958 0.9333635 0.89057831 0.5716711 0.7219823
0.5774327 0.7677038 0.7622717 0.45288270 0.2817869 0.2572325
0.6809509 0.6089656 0.8191862 0.01151454 0.2780449 0.4655353
0.5754383 0.5662045 0.7003630 0.62559642 0.2865510 0.1847980
MyScaledData<-structure(list(X14863=c(0.601478444979532,0.249521497274968,0.616770466379489,0.122737966507165,0.729975993009922,0.198864661389536,0.504137054265617,0.577432671357089,0.680950947164095,0.575438259547452),X15066=c(0.697510926657699,0.780144354632397,0.334404422875259,0.444850091405716,0.637593061483412,0.781471212351781,0.637495834667556,0.7677038048039,0.608965550162107,0.566204459603197),X15067=c(0.50432083998529,0.668392530333367,0.925412484830622,0.396180214305286,0.576006062451239,0.657205387087382,0.933363470346907,0.762271729415789,0.819186151914183,0.700362991098644),X15068=c(0.152846483002917,0.137682446305942,0.127775652495726,0.191172455317975,0.447468375530484,0.714093046059637,0.890578310935752,0.452882699805154,0.011514536383708,0.625596417031532),X15069=c(0.94163636689763,0.427795395079331,0.38262308941233,0.77898345642139,0.363490265569212,0.671568951210917,0.571671115989958,0.281786881885636,0.278044876559552,0.286551022600823),X15070=c(0.786073059382553,0.205841229942702,0.259045736299276,
0.758789694211416,0.107967864736275,0.402945912782515,0.721982268066207,0.257232456508833,0.46553533255268,0.184798001614338)),row.names=c(NA,10L),class="data.frame"); minvec<-c(X14861=22.95,X14862=29.95,X14863= 39.95,X15066=59.95,X15067=79.95,X15068=14.99,X15069=24.99,X15070=33.45,X15071=36.95,X15072=44.95,X15073=54.95,X15074=74.95,X15132=12.95,X15548=12.95,X15549=22.95,X15550=29.95,X15551=39.95,X15552=59.95,X15553=79.95,X15956=49.95,X15957=49.95,X16364=3.5);maxvec<-c(X14861=29.99,X14862=39.99,X14863=49.99,X15066=79.99,X15067=99.99,X15068=19.99,X15069=29.99,X15070=39.99,X15071=49.99,X15072=59.99,X15073=79.99,X15074=99.99,X15132=19.99,X15548=19.99,X15549=29.99,X15550=39.99,X15551=49.99,X15552=79.99,X15553=99.99,X15956=59.99,X15957=59.99,X16364=9.99)
我想通过将 min/max 值与基于名称的每个对应列相匹配，将 MyScaledData 重新缩放到它们的原始比例。我尝试了以下方法：
descale <- function(x,minval,maxval) {x*(maxval-minval) + minval}
as.data.frame(Map(descale,MyScaledData,minvec,maxvec))
我得到的输出比 MyScaledData 多了 6 列。我感觉到该函数甚至没有按名称匹配列，因此输出计算不正确。如何按列名匹配函数，以便它为每一列采用相应的 minvec 和 maxvec 元素，并且 return 只有我拥有的 6 列？
所需的输出应为：
MyDeScaledData <- structure(list(X14863 = c(45.9888435875945, 42.4551958326407,46.1423754824501, 41.1822891837319, 47.2789589698196, 41.9466012003509,45.0115360248268, 45.7474240204252, 46.7867475095275, 45.7274001258564), X15066 = c(73.9281189702203, 75.5840928668332, 66.6514646344202,68.8647958317706, 72.7273649521276, 75.6106830955297, 72.7254165267378,75.3347842482702, 72.1536696252486, 71.2967373704481), X15067 = c(90.0565896333052,93.3445863078807, 98.4952661960057, 87.8894514946779, 91.4931614915228,93.1203959572311, 98.654603945752, 95.2259254574924, 96.3664904843602,93.9852743416168), X15068 = c(15.7542324150146, 15.6784122315297,15.6288782624786, 15.9458622765899, 17.2273418776524, 18.5604652302982,19.4428915546788, 17.2544134990258, 15.0475726819185, 18.1179820851577), X15069 = c(29.6981818344881, 27.1289769753967, 26.9031154470616,28.8849172821069, 26.8074513278461, 28.3478447560546, 27.8483555799498,26.3989344094282, 26.3802243827978, 26.4227551130041), X15070 = c(38.5909178083619,34.7962016438253, 35.1441591153973, 38.4124846001427, 34.1561098353752,36.0852662695976, 38.171764033153, 35.1323002655678, 36.4946010748945,34.6585789305578)), row.names = c(NA, 10L), class = "data.frame")
感谢@Shirin Yavari 提供解决方案：
MyDeScaledData<-as.data.frame(Map(descale,MyScaledData,minvec[names(MyScaledData)],maxvec[names(MyScaledData)]))