我试图遍历几列数据，并根据另一列中的行来更新其中一列。
df <- mutate(df, A = ifelse(AA < 0, NA, A))
我想将A列设置为NA，因为AA列<0。B列和BB列等相同。
我有一条ifelse语句，一次可以处理一列，但是我需要将其放入purrr的循环或map（）函数中，以连续遍历所有列。但是，每当我尝试将其放入for循环或使用map_dbl（）时，都会遇到错误。这是我的ifelse声明。
cleaning <- function(df,x,y,z) {df <- mutate(df, df$x == ifelse(df$y < 2, NA, df$z))
for (i in seq_along(x)) {df[i] <-cleaning[[i]]}
}
cleaning(df,A:B,AA:BB,A:B)
这基本上是我尝试执行的循环：
这基本上就是我尝试这样做的方式。我是R的新手，所以我可能会完全离开。
structure(list(ID = c(74L, 11L, 66L, 125L, 89L, 25L, 57L, 43L,
114L, 47L), COUNTER = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L
), INTERPOLATED = c(4518.461428, 4573.333222, 4604.102452, 4655.384502,
4570.256298, 4473.846044, 4610.256298, 4585.128093, 4721.538346,
4653.84604), AF3 = c(3624.615296, 4025.640927, 4034.871696, 4004.615287,
3971.281954, 3868.717854, 3968.205031, 4005.128107, 3898.974264,
4058.461439), F7 = c(4345.128099, 4644.615271, 4665.128091, 4525.640915,
4571.281939, 4804.615267, 4479.48707, 4614.358862, 4482.563993,
4708.205013), F3 = c(3757.948626, 3978.974262, 4057.435798, 4118.974258,
4061.538362, 3591.794784, 4060.512721, 4019.999902, 4203.589641,
4068.717849), FC5 = c(4107.179387, 4126.153745, 4058.97426, 4085.640926,
4098.461438, 4245.128101, 4094.358874, 4133.333232, 3742.051191,
4152.307591), T7 = c(4316.410151, 4824.102446, 4765.128089, 4783.076806,
4685.640911, 4422.051174, 4742.051166, 4710.769116, 4850.256292,
4734.358859), P7 = c(4747.179371, 4458.97425, 4423.589635, 4497.948608,
4578.974247, 4752.307576, 4418.46143, 4599.999888, 4713.333218,
4579.487068), O1 = c(3947.179391, 3908.205033, 3966.66657, 4042.051183,
4008.20503, 4006.153748, 3972.820416, 3984.615287, 4167.692206,
3996.410159), O2 = c(4077.435798, 4171.281949, 4094.358874, 4147.179386,
4121.538361, 4138.461437, 4137.948617, 4151.79477, 4134.358873,
4118.974258), P8 = c(3606.666578, 3820.512727, 3874.35888, 4060.512721,
3775.897344, 3631.794783, 3959.999903, 3896.410161, 3858.974265,
3922.051186), T8 = c(4146.666565, 4330.256304, 4353.333227, 4415.384507,
4338.461432, 3941.538365, 4432.307584, 4382.051175, 4587.692196,
4419.999892), FC6 = c(4418.974251, 4632.8204, 4692.307578, 4634.871682,
4568.717837, 4684.61527, 4363.589637, 4615.897323, 4654.358861,
4631.794759), F4 = c(3808.205035, 4205.640923, 4373.846047, 4414.871687,
4293.846049, 3727.692217, 4151.79477, 4218.461435, 4284.61528,
4341.538355), F8 = c(4282.051177, 4243.076819, 4239.999896, 4437.435789,
4340.512714, 3809.743497, 4276.922972, 4269.743485, 4536.922966,
4360.512714), AF4 = c(487L, 484L, 513L, 444L, 444L, 0L, 0L, 0L,
482L, 0L), RAW_CQ = c(4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L),
CQ_AF3 = c(4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L), CQ_F7 = c(4L,
4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L), CQ_F3 = c(4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L, 4L, 4L), CQ_FC5 = c(4L, 4L, 4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L), CQ_T7 = c(4L, 4L, 4L, 4L, 4L, 4L, 4L,
4L, 4L, 4L), CQ_P7 = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L), CQ_O1 = c(4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L), CQ_O2 = c(0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), CQ_P8 = c(4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L, 4L, 4L), CQ_T8 = c(4L, 4L, 4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L), CQ_FC6 = c(4L, 4L, 4L, 4L, 4L, 4L, 4L,
4L, 4L, 4L), CQ_F4 = c(4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L,
4L), CQ_F8 = c(4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L), CQ_AF4 = c(4L,
4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L), CQ_CMS = c(4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L, 4L, 4L), CQ_DRL = c(1768L, 1770L, 1767L,
1768L, 1768L, 1768L, 1768L, 1771L, 1767L, 1770L), GYROX = c(1515L,
1517L, 1511L, 1512L, 1516L, 1514L, 1514L, 1515L, 1515L, 1515L
), `GYROY MARKER` = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L)), row.names = c(NA, -10L), class = c("tbl_df", "tbl",
"data.frame"))
这是我的实际数据的前10行的输出。需要根据值<2的CQ_AF3：CQ_AF4更新AF3：AF4列。
[(Int, Int, Int)]答案 0 :(得分：0)
这个怎么样？这个df是用set.seed(123)创建的：
> df
# A tibble: 10 x 4
A B AA BB
<dbl> <dbl> <dbl> <dbl>
1 1.26 1.66 -0.675 0.183
2 1.52 -1.63 -1.16 -0.0934
3 0.346 0.890 2.02 0.368
4 0.723 -0.532 -2.37 -0.580
5 -0.434 0.158 2.23 0.553
6 1.57 0.385 0.511 1.84
7 -0.516 1.25 -0.765 -2.05
8 0.561 0.313 0.0828 -0.214
9 -0.717 -1.46 -0.408 -0.166
10 -1.84 -0.267 -0.414 -1.05
然后修改后的df：
> df %>% mutate_at(vars(A), funs(ifelse(AA < 0, NA, .))) %>%
mutate_at(vars(B), funs(ifelse(BB < 0, NA, .)))
# A tibble: 10 x 4
A B AA BB
<dbl> <dbl> <dbl> <dbl>
1 NA 1.22 -1.07 0.426
2 NA NA -0.218 -0.295
3 NA 0.401 -1.03 0.895
4 NA 0.111 -0.729 0.878
5 NA -0.556 -0.625 0.822
6 NA 1.79 -1.69 0.689
7 0.461 0.498 0.838 0.554
8 -1.27 NA 0.153 -0.0619
9 NA NA -1.14 -0.306
10 -0.446 NA 1.25 -0.380
如果您需要保存修改后的df，只需将newname <-放在此命令的前面即可。
答案 1 :(得分：0)
简单的for循环解决方案，无需加载外部库：
> df
A B AA BB
1 -1.13879933 -0.05219939 1.40599762 0.277708500
2 -1.60341369 -1.23928624 0.25177871 0.009469529
3 NA 0.44474461 -0.68301589 1.914665838
4 NA NA -0.49642122 -0.555400536
5 -2.83194181 NA 0.09429098 -0.324189223
6 1.06254068 NA 0.84104522 -1.121906866
7 0.04871273 NA 1.13978470 -1.527878009
8 NA NA -1.29621577 -0.818883608
9 0.43346025 NA 0.45678599 -0.673519674
10 -0.08255017 NA 0.83648472 -1.851880253
for(i in 1:ncol(df)){
for(j in 1:ncol(df))(
if(paste0(colnames(df[i]), colnames(df[i])) == colnames(df[j]))(
for(k in 1:nrow(df))(
if(df[k, j] < 0)(
df[k, i] <- NA
)
)
)
)
}
set.seed(1701)
df <- data.frame("A" = c(rnorm(10)), "B" = c (rnorm(10)),
"AA" = c(rnorm(10)), "BB" = c(rnorm(10)))
================================================ =======
# Iterate through all columns for base i and match j
for(i in 1:ncol(df)){
for(j in 1:ncol(df))(
# Check if name of base + prefix is the same as match
if(paste0("CQ_", colnames(df[i])) == colnames(df[j]))(
# Iterate through rows
for(k in 1:nrow(df))(
# Check value in match row
if(df[k, j] < 2)(
# Update value in same row of base
df[k, i] <- NA
)
)
)
)
}
这同时适用于tibbles和data.frames。如果您有NA数据，则还应该使用!is.na()或其他方法排除这些情况，否则值检查将失败。