数据是
structure(list(country = c("Italy", "Italy", "Italy", "Italy",
"Italy", "Italy", "Italy", "Italy", "Italy", "Austria", "Austria",
"Austria", "Austria", "Austria", "Austria", "Austria", "Austria",
"Austria", "Spain", "Spain", "Spain", "Spain", "Spain", "Spain",
"Spain", "Spain", "Spain", "Portugal", "Portugal", "Portugal",
"Portugal", "Portugal", "Portugal", "Portugal", "Portugal", "Portugal"), date = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 1L,
2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L,
9L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L), levels = c("1000", "1200",
"1300", "1400", "1500", "1600", "1700", "1750", "1800"), class = "factor"), fit.0.025quant = c(0.828858435568409, 2.49016922773755, 2.65828836880409,
2.66581093505707, 2.57548603944065, 1.67885386266028, 0.767937801519583,
0.689014091834041, 0.66513695912892, 0.663699022639656, 0.678689217926065,
0.67888578207667, 0.678889008309296, 0.678889067277956, 0.678889008315121,
0.678885782094281, 0.67868921794039, 0.663699022695475, 0.753024027011781,
1.56176460225556, 1.66925563033888, 1.56663639224176, 0.757961319722877,
0.690760522079772, 0.688181548019768, 0.760878851337501,
1.5613478080279, 0.664563945566996, 0.68508607246818, 0.737286017774939,
1.5119543712026, 0.819687233979314, 1.60507738574301, 1.66679441254118,
1.68306812289583, 1.66919593888341), fit.mean = c(1.13400448293661,
2.87830024613312, 2.98840123437886, 2.99270845416986, 2.93292120022681,
2.00001375693309, 1.06700462995692, 1.00618961591289, 1.0007893132799,
1.00004363941275, 1.00003146052876, 1.00003107297182, 1.00003106386579,
1.00003106366991, 1.00003106386888, 1.00003107297307, 1.00003146052807,
1.00004363941059, 1.06578800415204, 1.92791273808794, 2.00911809484502,
1.92938242492218, 1.05953000948759, 1.00731995863015, 1.00551539597356,
1.0617929411748, 1.93397100841674, 1.00049922917461, 1.00391214501682,
1.04414180746079, 1.89014845620677, 1.1133844912819, 1.94890869596221,
1.99253538304467, 2.00313100995317, 2.00364556256633), fit.0.975quant = c(1.5548453965449,
3.17193689185671, 3.30217150519806, 3.30887034452248, 3.23188378430461,
2.32122877803568, 1.42445510409649, 1.33229333739629, 1.33797676580827,
1.33651739528035, 1.32146593655336, 1.32126741860939, 1.32126414685111,
1.32126408690283, 1.32126414685902, 1.32126741859377, 1.32146593653428,
1.33651739521255, 1.43806634680653, 2.23154393918938, 2.35542858727393,
2.23131692404623, 1.40982833868788, 1.33388203589275, 1.33103300867831,
1.41382643372889, 2.24644777498799, 1.33739126255841, 1.32804123237336,
1.38436190848754, 2.18404788457664, 1.4941065137638, 2.25306205937895,
2.30955823803133, 2.32596318832388, 2.34217586698821)), row.names = c(1L,
2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L,
16L, 17L, 18L, 91L, 92L, 93L, 94L, 95L, 96L, 97L, 98L, 99L, 100L,
101L, 102L, 103L, 104L, 105L, 106L, 107L, 108L), class = "data.frame")
问题如下：
对于日期为“1000”的意大利，“fit.0.025quant”、“fit.mean”和“fit.0.975quant”的值为
0.67363761.00000121.326401数据集中的每一行都应该如此。例如，对于日期“1200”的意大利，“fit.0.025quant”、“fit.mean”和“fit.0.975quant”的值为 2.5193345、2.8869927 和 3.168365。我希望所有三个值都提升 0.5，以便它们完全位于 3.01-3.99 之间。
因此，这就像将‘fit.0.025quant’、fit.mean’和‘fit.0.975quant’提升到当前位于其边界内的最高数字，以便所有三个都完全位于其边界内。
如果“fit.0.025quant”、“fit.mean”和“fit.0.975quant”的值已经完全位于某个数字的边界内（例如 5.01-5.99），那么这些值应该不会发生任何情况。
伙计们，我希望我的解释不会令人困惑。
您认为有合适的代码可以按预期自动转换整个数据集中的值吗？
使用
apply()add <- apply(dat[, 3:5], 1, \(x) floor(max(x))) - apply(dat[, 3:5], 1, min)
add <- ifelse(add < -0.01, 0, add + 0.01)
dat[, 3:5] <- dat[, 3:5] + add
结果：
#> head(dat, 10)
country date fit.0.025quant fit.mean fit.0.975quant
1 Italy 1000 1.01 1.315146 1.735987
2 Italy 1200 3.01 3.398131 3.691768
3 Italy 1300 3.01 3.340113 3.653883
4 Italy 1400 3.01 3.336898 3.653059
5 Italy 1500 3.01 3.367435 3.666398
6 Italy 1600 2.01 2.331160 2.652375
7 Italy 1700 1.01 1.309067 1.666517
8 Italy 1750 1.01 1.327176 1.653279
9 Italy 1800 1.01 1.345652 1.682840
10 Austria 1000 1.01 1.346345 1.682818