问题描述
df <- data.frame(PatientID = c("0002","0002","0005","0009","0018","0039","0043","0046","0048","0048"),Timepoint= c("A","B","A","B"),sex= c("F","F","M","F"),country= c("I","I","S","I"),A = c(NA,977.146,NA,964.315,952.311,950.797,947.465,902.852,985.124,930.141,1007.790,1027.110,999.414),B = c(998.988,998.680,1020.560,955.540,911.606,964.039,988.087,902.367,959.338,1029.050,987.374,1066.400,957.512,917.597),C = c( 987.140,961.810,929.466,978.166,969.469,943.398,936.034,965.292,996.404,920.610,967.047,913.517,893.428,921.606,929.590,950.493),D = c( 961.810,1005.820,925.752,976.192,E = c(1006.330,1028.070,954.274,1005.910,949.969,992.820,934.407,948.913,961.375,955.296,961.128,998.119,1009.110,994.891,1000.170,982.763),G= c(NA,958.990,924.680,955.927,949.384,973.348,984.392,943.894,961.468,995.368,994.997,979.454,952.605,956.507),stringsAsFactors = F)
我有这个代码来对有 3 个或更多列超出范围的人进行分类,阈值是 1015:
cols <- 5:10
df$Myo <- ifelse(rowSums(df[cols] > 1015,na.rm = TRUE) >= 3,'Yes','No')
a) 第一个阈值是居住在瑞典 (df$sex==M
) 的男性 (df$country==S
),这个阈值是 900
b) 第二个阈值是生活在瑞典 (df$sex==F
) 的女性 (df$country==S
),这个阈值是 1016
c) 第一个阈值是居住在冰岛 (df$sex==M
) 的男性 (df$country==I
),这个阈值是 800
d) 第二个阈值是生活在冰岛 (df$sex==F
) 的女性 (df$country==I
),这个阈值是 1000。
谢谢!!
解决方法
我更喜欢使用 data.table
来解决这个问题。函数fcase
用于处理多阈值分支。我不确定 Myo
的输入是否是您想要的。不管怎样,我想你可以纠正它,我稍后会编辑它。
df <- data.frame(PatientID = c("0002","0002","0005","0009","0018","0039","0043","0046","0048","0048"),Timepoint= c("A","B","A","B"),sex= c("F","F","M","F"),country= c("I","I","S","I"),A = c(NA,977.146,NA,964.315,952.311,950.797,947.465,902.852,985.124,930.141,1007.790,1027.110,999.414),B = c(998.988,998.680,1020.560,955.540,911.606,964.039,988.087,902.367,959.338,1029.050,987.374,1066.400,957.512,917.597),C = c( 987.140,961.810,929.466,978.166,969.469,943.398,936.034,965.292,996.404,920.610,967.047,913.517,893.428,921.606,929.590,950.493),D = c( 961.810,1005.820,925.752,976.192,E = c(1006.330,1028.070,954.274,1005.910,949.969,992.820,934.407,948.913,961.375,955.296,961.128,998.119,1009.110,994.891,1000.170,982.763),G= c(NA,958.990,924.680,955.927,949.384,973.348,984.392,943.894,961.468,995.368,994.997,979.454,952.605,956.507),stringsAsFactors = F)
library(data.table)
setDT(df)
cols <- 5:10
df[,Myo := fcase(sex == "F" & rowSums(.SD > 1004,na.rm = T) >= 3,"Yes",sex == "M" & rowSums(.SD > 986,default = "No"),.SDcols = cols]
df[,Myo2 := fcase(sex == "M" & country == "S" & rowSums(.SD > 900,sex == "F" & country == "S" & rowSums(.SD > 1016,sex == "M" & country == "I" & rowSums(.SD > 800,sex == "F" & country == "I" & rowSums(.SD > 1000,.SDcols = cols]
df
#> PatientID Timepoint sex country A B C D E
#> 1: 0002 A F I NA 998.988 987.140 961.810 1006.330
#> 2: 0002 B F I 977.146 NA 961.810 929.466 1028.070
#> 3: 0005 A M S NA 998.680 929.466 978.166 954.274
#> 4: 0005 B M S 964.315 NA 978.166 1005.820 1005.910
#> 5: 0009 A F S NA 1020.560 969.469 925.752 949.969
#> 6: 0009 B F S 952.311 955.540 943.398 969.469 992.820
#> 7: 0018 A M S NA 911.606 936.034 943.398 934.407
#> 8: 0018 B M S 950.797 964.039 965.292 965.292 948.913
#> 9: 0039 A F S 947.465 988.087 996.404 996.404 961.375
#> 10: 0039 B F S 902.852 902.367 920.610 967.047 955.296
#> 11: 0043 A M I 985.124 959.338 967.047 NA 961.128
#> 12: 0043 B M I NA 1029.050 913.517 893.428 998.119
#> 13: 0046 A M I 930.141 987.374 893.428 921.606 1009.110
#> 14: 0046 B M I 1007.790 1066.400 921.606 976.192 994.891
#> 15: 0048 A F I 1027.110 957.512 929.590 929.590 1000.170
#> 16: 0048 B F I 999.414 917.597 950.493 950.493 982.763
#> G Myo Myo2
#> 1: NA No No
#> 2: 958.990 No No
#> 3: 924.680 No Yes
#> 4: 955.927 No Yes
#> 5: NA No No
#> 6: 949.384 No No
#> 7: 973.348 No Yes
#> 8: 984.392 No Yes
#> 9: 943.894 No No
#> 10: 961.468 No No
#> 11: 995.368 No Yes
#> 12: 994.997 Yes Yes
#> 13: 979.454 No Yes
#> 14: 952.605 Yes Yes
#> 15: NA No No
#> 16: 956.507 No No
由 reprex package (v2.0.0) 于 2021 年 8 月 3 日创建
,我发现这段代码也适用于第一种情况:
cols <- 5:10
df$sex= as.factor(df$sex)
df %>% mutate(Myo=ifelse(sex == "F" & (rowSums(df[cols] > 1004,na.rm = TRUE) >=3),'Yes',ifelse(sex == "M" & (rowSums(df[cols] > 986,'No')))-> df