我有dataframe幾行和多列，如果行值高于該行最后一列中的值，我想檢查每一行。我的資料如下所示：
date ABI.BR NOVN.S ROG.S NESN.S NOVOb.CO HRMS.PA TTEF.PA OREP.PA SASY.PA LVMH.PA MAX
1 31.12.2001 7342987743 21072681660 19737660401 57324018814 3199830778 1226900000 105895000000 13740400000 6.4880e 09 1.2229e 10 163033294916
2 31.12.2002 6280730128 17748921146 20316428627 61501808862 3350848339 1242300000 92108991318 14288000000 7.4480e 09 1.2693e 10 144604417185
3 31.12.2003 6324404232 19755193920 20002601887 56367998444 3510991763 1230000000 93961037997 14029100000 8.0480e 09 1.1962e 10 122995279989
4 31.12.2004 7850834622 20947725570 19134043533 54889985325 3905732023 1331400000 87346032957 13641300000 1.5733e 10 1.2481e 10 123515812436
5 31.12.2005 12233279545 26451714210 22833829836 58587603997 4527870421 1427400000 122854000000 14532500000 2.8513e 10 1.3910e 10 171566971202
6 31.12.2006 12660668341 26602920050 26148301223 61238063838 5198638861 1514900000 126235000000 15790100000 2.9489e 10 1.5306e 10 173975809100
我現在的目標是檢查列的每一行中的值2:10是否高于列中的相應值11（即MAX）。如果在一行中，列中的值高于列中的值，MAX我想用列中的值覆寫列中的實際值MAX。
例如，如果我有以下資料：
a b c MAX
2 3 4 3
4 5 6 6
6 7 9 8
我想在更正后的資料中有以下輸出
a b c MAX
2 3 **3** 3
4 5 6 6
6 7 **8** 8
非常感謝您的幫助！
uj5u.com熱心網友回復：
使用您的簡短示例和 library dplyr，您可以執行以下操作：
代表
- 資料
df <- structure(list(a = c(2L, 4L, 6L), b = c(3L, 5L, 7L), c = c(4L,
6L, 9L), MAX = c(3L, 6L, 8L)), class = "data.frame", row.names = c(NA,
-3L))
- 代碼
library(dplyr)
df %>% transmute(across(.cols = 1:3, ~ ifelse(.x > MAX, MAX, .x)))
- 輸出
#> a b c
#> 1 2 3 3
#> 2 4 5 6
#> 3 6 7 8
由reprex 包于 2022-03-31 創建(v2.0.1)
編輯： 使用您的真實資料并遵循@langtang 的評論以保留第一列和最后一列，您可以這樣做：
代表
- 資料
structure(list(date = c("31.12.2001", "31.12.2002", "31.12.2003",
"31.12.2004", "31.12.2005", "31.12.2006"), ABI.BR = c(7342987743,
6280730128, 6324404232, 7850834622, 12233279545, 12660668341),
NOVN.S = c(21072681660, 17748921146, 19755193920, 20947725570,
26451714210, 26602920050), ROG.S = c(19737660401, 20316428627,
20002601887, 19134043533, 22833829836, 26148301223), NESN.S = c(57324018814,
61501808862, 56367998444, 54889985325, 58587603997, 61238063838
), NOVOb.CO = c(3199830778, 3350848339, 3510991763, 3905732023,
4527870421, 5198638861), HRMS.PA = c(1226900000L, 1242300000L,
1230000000L, 1331400000L, 1427400000L, 1514900000L), TTEF.PA = c(1.05895e 11,
92108991318, 93961037997, 87346032957, 1.22854e 11, 1.26235e 11
), OREP.PA = c(13740400000, 1.4288e 10, 14029100000, 13641300000,
14532500000, 15790100000), SASY.PA = c(6.488e 09, 7.448e 09,
8.048e 09, 1.5733e 10, 2.8513e 10, 2.9489e 10), LVMH.PA = c(1.2229e 10,
1.2693e 10, 1.1962e 10, 1.2481e 10, 1.391e 10, 1.5306e 10
), MAX = c(163033294916, 144604417185, 122995279989, 123515812436,
171566971202, 173975809100)), class = "data.frame", row.names = c("1",
"2", "3", "4", "5", "6"))
- 代碼
library(dplyr)
df %>% mutate(across(.cols = 2:10, ~ ifelse(.x > MAX, MAX, .x)))
- 輸出
#> date ABI.BR NOVN.S ROG.S NESN.S NOVOb.CO
#> 1 31.12.2001 7342987743 21072681660 19737660401 57324018814 3199830778
#> 2 31.12.2002 6280730128 17748921146 20316428627 61501808862 3350848339
#> 3 31.12.2003 6324404232 19755193920 20002601887 56367998444 3510991763
#> 4 31.12.2004 7850834622 20947725570 19134043533 54889985325 3905732023
#> 5 31.12.2005 12233279545 26451714210 22833829836 58587603997 4527870421
#> 6 31.12.2006 12660668341 26602920050 26148301223 61238063838 5198638861
#> HRMS.PA TTEF.PA OREP.PA SASY.PA LVMH.PA MAX
#> 1 1226900000 105895000000 13740400000 6.4880e 09 1.2229e 10 163033294916
#> 2 1242300000 92108991318 14288000000 7.4480e 09 1.2693e 10 144604417185
#> 3 1230000000 93961037997 14029100000 8.0480e 09 1.1962e 10 122995279989
#> 4 1331400000 87346032957 13641300000 1.5733e 10 1.2481e 10 123515812436
#> 5 1427400000 122854000000 14532500000 2.8513e 10 1.3910e 10 171566971202
#> 6 1514900000 126235000000 15790100000 2.9489e 10 1.5306e 10 173975809100
由reprex 包于 2022-03-31 創建(v2.0.1)
uj5u.com熱心網友回復：
cols = c('a', 'b', 'c')
df[cols] = lapply(df[cols], \(x) ifelse(x > df$MAX, df$MAX, x))
# a b c MAX
# 1 2 3 3 3
# 2 4 5 6 6
# 3 6 7 8 8
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