Aid*_*dis -1 json nested r matrix
我有不同球员的统计数据.我想知道如何操纵我的数据.我怎么能找到:
如何在三个矩阵的列中比较"胜利"的值?
library(jsonlite)
iris2 <- fromJSON("C:\\Users\\Saonkfas\\Desktop\\WOWPAPI\\wowpfinaljson.json")
print(iris2)
for (x in iris2){print (x)}
Run Code Online (Sandbox Code Playgroud)这是我打印后的内容:
wins losses max_killed battles plane_id max_ground_object_destroyed
1 118 40 7 158 4401 3
2 100 58 7 158 2401 3
3 120 38 7 158 2403 3
wins losses max_killed battles plane_id max_ground_object_destroyed
1 12 450 7 158 4401 3
2 150 8 7 158 2401 3
3 120 328 7 158 2403 3
wins losses max_killed battles plane_id max_ground_object_destroyed
1 158 40 7 158 4401 3
2 1010 584 7 158 2401 3
3 120 348 7 158 2403 3
Run Code Online (Sandbox Code Playgroud)
数据框包括最高值wins:
iris2[[which.max(sapply(iris2, function(x) max(x$wins)))]]
wins losses max_killed battles plane_id max_ground_object_destroyed
1 158 40 7 158 4401 3
2 1010 584 7 158 2401 3
3 120 348 7 158 2403 3
Run Code Online (Sandbox Code Playgroud)具有最高的行wins到losses的所有数据帧的比率:
lapply(iris2, function(x) x[which.max(x$wins / x$losses), ])
[[1]]
wins losses max_killed battles plane_id max_ground_object_destroyed
3 120 38 7 158 2403 3
[[2]]
wins losses max_killed battles plane_id max_ground_object_destroyed
2 150 8 7 158 2401 3
[[3]]
wins losses max_killed battles plane_id max_ground_object_destroyed
1 158 40 7 158 4401 3
Run Code Online (Sandbox Code Playgroud)这太不明确了......
如何比较所有三个矩阵中的"胜利"值?