基于参数的位置评级
Rating of locations based on a parameter
我正在寻找一种利用流量计数的方法，并根据该方法给出特定位置的评分，评分应介于 0 到 10 之间。我可以简单地给出 1 到 30 的排名，其中排名 1 的大部分交通发生的位置。我可能需要将它缩放到 0 到 10 之间的东西。有什么建议吗？
> dput(Data)
structure(list(Traffic.Views = c(3175760L, 2949940L, 2685756L,
2535156L, 2437236L, 2210328L, 2085276L, 1974840L, 1961424L, 1923308L,
1844408L, 1781592L, 1761252L, 1675820L, 1582748L, 1475928L, 1399336L,
1311940L, 1309980L, 1305544L, 1160140L, 1144348L, 1137584L, 1106904L,
946304L, 931992L, 891176L, 815812L, 789788L, 662652L), Names = c("df01",
"df02", "df03", "df04", "df05", "df06", "df07", "df08", "df09",
"df10", "df11", "df12", "df13", "df14", "df15", "df16", "df17",
"df18", "df19", "df20", "df21", "df22", "df23", "df24", "df25",
"df26", "df27", "df28", "df29", "df30")), row.names = c(NA, -30L
), class = "data.frame")
您可以使用：
rating_0_10 <- function(x) (x-min(x))/(max(x)-min(x)) * 10
Data$scaled <- rating_0_10(Data$Traffic.Views)
Data
# Traffic.Views Names scaled
#1 3175760 df01 10.0000000
#2 2949940 df02 9.1014314
#3 2685756 df03 8.0502072
#4 2535156 df04 7.4509492
#5 2437236 df05 7.0613121
#6 2210328 df06 6.1584142
#7 2085276 df07 5.6608152
#8 1974840 df08 5.2213753
#9 1961424 df09 5.1679912
#10 1923308 df10 5.0163224
#11 1844408 df11 4.7023685
#12 1781592 df12 4.4524151
#13 1761252 df13 4.3714795
#14 1675820 df14 4.0315339
#15 1582748 df15 3.6611877
#16 1475928 df16 3.2361363
#17 1399336 df17 2.9313663
#18 1311940 df18 2.5836056
#19 1309980 df19 2.5758065
#20 1305544 df20 2.5581551
#21 1160140 df21 1.9795727
#22 1144348 df22 1.9167342
#23 1137584 df23 1.8898193
#24 1106904 df24 1.7677394
#25 946304 df25 1.1286901
#26 931992 df26 1.0717406
#27 891176 df27 0.9093282
#28 815812 df28 0.6094446
#29 789788 df29 0.5058915
#30 662652 df30 0.0000000
我正在寻找一种利用流量计数的方法，并根据该方法给出特定位置的评分，评分应介于 0 到 10 之间。我可以简单地给出 1 到 30 的排名，其中排名 1 的大部分交通发生的位置。我可能需要将它缩放到 0 到 10 之间的东西。有什么建议吗？
> dput(Data)
structure(list(Traffic.Views = c(3175760L, 2949940L, 2685756L,
2535156L, 2437236L, 2210328L, 2085276L, 1974840L, 1961424L, 1923308L,
1844408L, 1781592L, 1761252L, 1675820L, 1582748L, 1475928L, 1399336L,
1311940L, 1309980L, 1305544L, 1160140L, 1144348L, 1137584L, 1106904L,
946304L, 931992L, 891176L, 815812L, 789788L, 662652L), Names = c("df01",
"df02", "df03", "df04", "df05", "df06", "df07", "df08", "df09",
"df10", "df11", "df12", "df13", "df14", "df15", "df16", "df17",
"df18", "df19", "df20", "df21", "df22", "df23", "df24", "df25",
"df26", "df27", "df28", "df29", "df30")), row.names = c(NA, -30L
), class = "data.frame")
您可以使用：
rating_0_10 <- function(x) (x-min(x))/(max(x)-min(x)) * 10
Data$scaled <- rating_0_10(Data$Traffic.Views)
Data
# Traffic.Views Names scaled
#1 3175760 df01 10.0000000
#2 2949940 df02 9.1014314
#3 2685756 df03 8.0502072
#4 2535156 df04 7.4509492
#5 2437236 df05 7.0613121
#6 2210328 df06 6.1584142
#7 2085276 df07 5.6608152
#8 1974840 df08 5.2213753
#9 1961424 df09 5.1679912
#10 1923308 df10 5.0163224
#11 1844408 df11 4.7023685
#12 1781592 df12 4.4524151
#13 1761252 df13 4.3714795
#14 1675820 df14 4.0315339
#15 1582748 df15 3.6611877
#16 1475928 df16 3.2361363
#17 1399336 df17 2.9313663
#18 1311940 df18 2.5836056
#19 1309980 df19 2.5758065
#20 1305544 df20 2.5581551
#21 1160140 df21 1.9795727
#22 1144348 df22 1.9167342
#23 1137584 df23 1.8898193
#24 1106904 df24 1.7677394
#25 946304 df25 1.1286901
#26 931992 df26 1.0717406
#27 891176 df27 0.9093282
#28 815812 df28 0.6094446
#29 789788 df29 0.5058915
#30 662652 df30 0.0000000