今天我未能完成一项简单的任务:
我想用我之前计算过的特定单个值替换数据帧每行中的最大值。我不知道我怎么知道
DF <- structure(list(age_n = c(2, 3, 3, 3, 3, 3), female_noe_n = c(0.674555812063115,
0.681331945419623, 0.632659475334461, 0.649517172953091, 0.64918662986253,
0.667531771388686), male_noe_n = c(0.734797290317925, 0.742565052946117,
0.715791062610646, 0.729425965096302, 0.710171830071103, 0.734962561863205
), female_v_n = c(0.745209397670608, 0.86387436718214, 0.915191181991792,
1.01146185711779, 1.04658206048994, 1.11211222819373), male_v_n = c(0.783635031948367,
0.860486300503885, 0.909571949452249, 0.995595788770845, 1.03071599214299,
1.12996155508404)), row.names = c(NA, 6L), class = "data.frame")
ID <- apply(DF, 1, function(x) which.max(x))
# Now I need to overwrite the rowwise maximum by ceiling
overwrite_value <-
sapply(1:nrow(DF), function(x)
DF[x,ID[[x]]] %>% ceiling())
我想有简单的解决方案，但我陷入了过于复杂的问题
最佳答案
这是一个 dplyr 解决方案:
library(dplyr)
individual_value <- 5
DF %>%
rowwise() %>%
mutate( x = max(c_across(2:5))) %>%
mutate(across(2:5, ~case_when(. == x ~ individual_value,
TRUE ~ .)), .keep="unused"
)
age_n female_noe_n male_noe_n female_v_n male_v_n
<dbl> <dbl> <dbl> <dbl> <dbl>
1 2 0.675 0.735 0.745 5
2 3 0.681 0.743 5 0.860
3 3 0.633 0.716 5 0.910
4 3 0.650 0.729 5 0.996
5 3 0.649 0.710 5 1.03
6 3 0.668 0.735 1.11 5
关于r - 如何用单个值替换数据框中每行的最大值，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/73899338/