3

我在一个名为 ulyDataLefs60_12 的数据框中有这个非常复杂的数据：

  Year Day Hour Min  Sec. E1.S1 E1.S2 E1.S3 E1.S4 E1.S5 E1.S6 E1.S7 E1.S8 E2.S1 E2.S2 E2.S3 E2.S4
1 2000 122    0   1 38.01  3.31 0.662 0.662  2.65  1.32 0.000 3.310  1.32 1.980 1.980 0.662 0.000
2 2000 122    0   1 50.10  1.98 3.310 1.980  1.98  1.98 1.320 4.630  1.32 1.320 0.662 0.000 3.310
3 2000 122    0   2  2.19  1.98 1.320 3.970  1.98  1.32 0.662 0.662  3.97 1.320 0.662 1.320 0.662
4 2000 122    0   2 14.28  2.65 1.320 2.650  3.31  2.65 1.320 3.970  2.65 2.650 0.000 0.662 2.650
5 2000 122    0   2 26.38  3.97 6.620 0.662  3.31  3.31 4.630 5.290  1.98 0.000 0.000 1.980 0.662
6 2000 122    0   2 38.47  2.65 0.662 3.310  1.98  1.32 1.980 1.980  2.65 0.662 1.320 1.980 1.320
  E2.S5 E2.S6 E2.S7 E2.S8 E3.S1 E3.S2 E3.S3 E3.S4 E3.S5 E3.S6 E3.S7 E3.S8 E4.S1 E4.S2 E4.S3 E4.S4
1 1.320  1.32  2.65 2.650 0.662 0.000 1.320 2.650 1.320 0.000 1.320 1.320 0.000 0.000 0.662 0.662
2 0.000  0.00  1.98 0.662 0.000 0.662 0.000 0.662 1.980 1.980 0.662 1.320 0.000 0.000 0.000 0.662
3 0.662  1.98  2.65 1.980 0.000 0.662 0.662 1.320 0.662 0.000 1.320 3.310 0.662 0.000 1.980 0.662
4 0.662  1.32  1.32 0.662 0.000 0.662 0.662 0.662 0.662 0.662 0.662 0.000 0.000 0.662 0.000 0.000
5 0.000  1.32  1.32 0.662 0.662 0.000 0.000 0.662 0.000 0.662 1.320 0.662 0.000 0.000 0.000 0.662
6 1.320  1.32  1.32 0.000 1.320 0.000 0.000 0.662 1.320 0.000 0.662 0.662 0.662 1.320 0.000 0.000
  E4.S5 E4.S6 E4.S7 E4.S8 FP5.S1 FP5.S2 FP5.S3 FP5.S4 FP5.S5 FP5.S6 FP5.S7 FP5.S8 FP6.S1 FP6.S2
1 0.000 0.662 0.662 0.000  0.331      0  0.662  0.000  0.662      0  0.000  0.331      0  0.331
2 0.000 0.000 0.662 0.662  0.331      0  0.662  0.000  0.662      0  0.000  0.331      0  0.331
3 0.662 0.000 0.662 1.320  0.000      0  0.662  0.000  0.331      0  0.000  0.000      0  0.000
4 0.662 0.662 0.000 0.662  0.000      0  0.662  0.000  0.331      0  0.000  0.000      0  0.000
5 0.000 0.000 0.662 0.000  0.331      0  0.000  0.331  0.331      0  0.331  0.000      0  0.000
6 0.000 0.000 0.662 0.662  0.331      0  0.000  0.331  0.331      0  0.331  0.000      0  0.000
  FP6.S3 FP6.S4 FP6.S5 FP6.S6 FP6.S7 FP6.S8 FP7.S1 FP7.S2 FP7.S3 FP7.S4 FP7.S5 FP7.S6 FP7.S7 FP7.S8
1  0.331  0.000  0.000  0.000      0  0.000      0  0.331  0.331  0.662      0  0.000  0.331  0.000
2  0.331  0.000  0.000  0.000      0  0.000      0  0.331  0.331  0.662      0  0.000  0.331  0.000
3  0.662  0.000  0.662  0.000      0  0.331      0  0.000  0.000  0.331      0  0.000  0.000  0.000
4  0.662  0.000  0.662  0.000      0  0.331      0  0.000  0.000  0.331      0  0.000  0.000  0.000
5  0.000  0.662  0.000  0.992      0  0.000      0  0.000  0.000  0.000      0  0.331  0.000  0.331
6  0.000  0.662  0.000  0.992      0  0.000      0  0.000  0.000  0.000      0  0.331  0.000  0.331
  PA.LEFS60S1 PA.LEFS60S2 PA.LEFS60S3 PA.LEFS60S4 PA.LEFS60S5 PA.LEFS60S6 PA.LEFS60S7 PA.LEFS60S8
1        64.2        52.0        70.9       105.0         144         170         134        96.2
2        62.6        49.5        68.8       104.0         142         168         134        95.4
3        62.7        47.7        66.2       101.0         140         167         135        96.5
4        62.4        46.3        64.4        99.3         138         166         135        96.7
5        59.9        43.7        63.2        98.8         138         164         133        94.8
6        62.3        45.7        63.7        98.7         137         166         136        96.9
    BX   BY   BZ Bmag....nT.  X            datetime
1 2.64 4.98 2.25        6.07 NA 2000-05-01 00:01:38
2 2.67 5.16 2.03        6.15 NA 2000-05-01 00:01:50
3 2.52 5.35 1.88        6.21 NA 2000-05-01 00:02:02
4 2.43 5.45 1.74        6.22 NA 2000-05-01 00:02:14
5 2.53 5.46 1.46        6.19 NA 2000-05-01 00:02:26
6 2.29 5.26 1.61        5.96 NA 2000-05-01 00:02:38

dput(head(ulyDataLefs60_12))
structure(list(Year = c(2000L, 2000L, 2000L, 2000L, 2000L, 2000L
), Day = c(122L, 122L, 122L, 122L, 122L, 122L), Hour = c(0L, 
0L, 0L, 0L, 0L, 0L), Min = c(1L, 1L, 2L, 2L, 2L, 2L), Sec. = c(38.01, 
50.1, 2.19, 14.28, 26.38, 38.47), E1.S1 = c(3.31, 1.98, 1.98, 
2.65, 3.97, 2.65), E1.S2 = c(0.662, 3.31, 1.32, 1.32, 6.62, 0.662
), E1.S3 = c(0.662, 1.98, 3.97, 2.65, 0.662, 3.31), E1.S4 = c(2.65, 
1.98, 1.98, 3.31, 3.31, 1.98), E1.S5 = c(1.32, 1.98, 1.32, 2.65, 
3.31, 1.32), E1.S6 = c(0, 1.32, 0.662, 1.32, 4.63, 1.98), E1.S7 = c(3.31, 
4.63, 0.662, 3.97, 5.29, 1.98), E1.S8 = c(1.32, 1.32, 3.97, 2.65, 
1.98, 2.65), E2.S1 = c(1.98, 1.32, 1.32, 2.65, 0, 0.662), E2.S2 = c(1.98, 
0.662, 0.662, 0, 0, 1.32), E2.S3 = c(0.662, 0, 1.32, 0.662, 1.98, 
1.98), E2.S4 = c(0, 3.31, 0.662, 2.65, 0.662, 1.32), E2.S5 = c(1.32, 
0, 0.662, 0.662, 0, 1.32), E2.S6 = c(1.32, 0, 1.98, 1.32, 1.32, 
1.32), E2.S7 = c(2.65, 1.98, 2.65, 1.32, 1.32, 1.32), E2.S8 = c(2.65, 
0.662, 1.98, 0.662, 0.662, 0), E3.S1 = c(0.662, 0, 0, 0, 0.662, 
1.32), E3.S2 = c(0, 0.662, 0.662, 0.662, 0, 0), E3.S3 = c(1.32, 
0, 0.662, 0.662, 0, 0), E3.S4 = c(2.65, 0.662, 1.32, 0.662, 0.662, 
0.662), E3.S5 = c(1.32, 1.98, 0.662, 0.662, 0, 1.32), E3.S6 = c(0, 
1.98, 0, 0.662, 0.662, 0), E3.S7 = c(1.32, 0.662, 1.32, 0.662, 
1.32, 0.662), E3.S8 = c(1.32, 1.32, 3.31, 0, 0.662, 0.662), E4.S1 = c(0, 
0, 0.662, 0, 0, 0.662), E4.S2 = c(0, 0, 0, 0.662, 0, 1.32), E4.S3 = c(0.662, 
0, 1.98, 0, 0, 0), E4.S4 = c(0.662, 0.662, 0.662, 0, 0.662, 0
), E4.S5 = c(0, 0, 0.662, 0.662, 0, 0), E4.S6 = c(0.662, 0, 0, 
0.662, 0, 0), E4.S7 = c(0.662, 0.662, 0.662, 0, 0.662, 0.662), 
    E4.S8 = c(0, 0.662, 1.32, 0.662, 0, 0.662), FP5.S1 = c(0.331, 
    0.331, 0, 0, 0.331, 0.331), FP5.S2 = c(0, 0, 0, 0, 0, 0), 
    FP5.S3 = c(0.662, 0.662, 0.662, 0.662, 0, 0), FP5.S4 = c(0, 
    0, 0, 0, 0.331, 0.331), FP5.S5 = c(0.662, 0.662, 0.331, 0.331, 
    0.331, 0.331), FP5.S6 = c(0, 0, 0, 0, 0, 0), FP5.S7 = c(0, 
    0, 0, 0, 0.331, 0.331), FP5.S8 = c(0.331, 0.331, 0, 0, 0, 
    0), FP6.S1 = c(0, 0, 0, 0, 0, 0), FP6.S2 = c(0.331, 0.331, 
    0, 0, 0, 0), FP6.S3 = c(0.331, 0.331, 0.662, 0.662, 0, 0), 
    FP6.S4 = c(0, 0, 0, 0, 0.662, 0.662), FP6.S5 = c(0, 0, 0.662, 
    0.662, 0, 0), FP6.S6 = c(0, 0, 0, 0, 0.992, 0.992), FP6.S7 = c(0, 
    0, 0, 0, 0, 0), FP6.S8 = c(0, 0, 0.331, 0.331, 0, 0), FP7.S1 = c(0, 
    0, 0, 0, 0, 0), FP7.S2 = c(0.331, 0.331, 0, 0, 0, 0), FP7.S3 = c(0.331, 
    0.331, 0, 0, 0, 0), FP7.S4 = c(0.662, 0.662, 0.331, 0.331, 
    0, 0), FP7.S5 = c(0, 0, 0, 0, 0, 0), FP7.S6 = c(0, 0, 0, 
    0, 0.331, 0.331), FP7.S7 = c(0.331, 0.331, 0, 0, 0, 0), FP7.S8 = c(0, 
    0, 0, 0, 0.331, 0.331), PA.LEFS60S1 = c(64.2, 62.6, 62.7, 
    62.4, 59.9, 62.3), PA.LEFS60S2 = c(52, 49.5, 47.7, 46.3, 
    43.7, 45.7), PA.LEFS60S3 = c(70.9, 68.8, 66.2, 64.4, 63.2, 
    63.7), PA.LEFS60S4 = c(105, 104, 101, 99.3, 98.8, 98.7), 
    PA.LEFS60S5 = c(144, 142, 140, 138, 138, 137), PA.LEFS60S6 = c(170, 
    168, 167, 166, 164, 166), PA.LEFS60S7 = c(134, 134, 135, 
    135, 133, 136), PA.LEFS60S8 = c(96.2, 95.4, 96.5, 96.7, 94.8, 
    96.9), BX = c(2.64, 2.67, 2.52, 2.43, 2.53, 2.29), BY = c(4.98, 
    5.16, 5.35, 5.45, 5.46, 5.26), BZ = c(2.25, 2.03, 1.88, 1.74, 
    1.46, 1.61), Bmag....nT. = c(6.07, 6.15, 6.21, 6.22, 6.19, 
    5.96), X = c(NA, NA, NA, NA, NA, NA), datetime = structure(list(
        sec = c(38, 50, 2, 14, 26, 38), min = c(1L, 1L, 2L, 2L, 
        2L, 2L), hour = c(0L, 0L, 0L, 0L, 0L, 0L), mday = c(1L, 
        1L, 1L, 1L, 1L, 1L), mon = c(4L, 4L, 4L, 4L, 4L, 4L), 
        year = c(100L, 100L, 100L, 100L, 100L, 100L), wday = c(1L, 
        1L, 1L, 1L, 1L, 1L), yday = c(121L, 121L, 121L, 121L, 
        121L, 121L), isdst = c(1L, 1L, 1L, 1L, 1L, 1L)), .Names = c("sec", 
    "min", "hour", "mday", "mon", "year", "wday", "yday", "isdst"
    ), class = c("POSIXlt", "POSIXt"))), .Names = c("Year", "Day", 
"Hour", "Min", "Sec.", "E1.S1", "E1.S2", "E1.S3", "E1.S4", "E1.S5", 
"E1.S6", "E1.S7", "E1.S8", "E2.S1", "E2.S2", "E2.S3", "E2.S4", 
"E2.S5", "E2.S6", "E2.S7", "E2.S8", "E3.S1", "E3.S2", "E3.S3", 
"E3.S4", "E3.S5", "E3.S6", "E3.S7", "E3.S8", "E4.S1", "E4.S2", 
"E4.S3", "E4.S4", "E4.S5", "E4.S6", "E4.S7", "E4.S8", "FP5.S1", 
"FP5.S2", "FP5.S3", "FP5.S4", "FP5.S5", "FP5.S6", "FP5.S7", "FP5.S8", 
"FP6.S1", "FP6.S2", "FP6.S3", "FP6.S4", "FP6.S5", "FP6.S6", "FP6.S7", 
"FP6.S8", "FP7.S1", "FP7.S2", "FP7.S3", "FP7.S4", "FP7.S5", "FP7.S6", 
"FP7.S7", "FP7.S8", "PA.LEFS60S1", "PA.LEFS60S2", "PA.LEFS60S3", 
"PA.LEFS60S4", "PA.LEFS60S5", "PA.LEFS60S6", "PA.LEFS60S7", "PA.LEFS60S8", 
"BX", "BY", "BZ", "Bmag....nT.", "X", "datetime"), row.names = c(NA, 
6L), class = "data.frame")

而我想要的是获得一定数量行的平均值和中位数。可以说，我想要一个新的数据框，而不是所有这些值，它具有所有列（或至少从 E1.S1 列开始的所有列）中每 5 行的平均值或中位数。

我从查看示例开始：计算行的平均值，它确实让我能够获得数据框单列的 N 行的平均值。

ulyDataLefs60_12_avg = colSums(matrix(ulyDataLefs60_12$E1.S1, nrow=5))

问题是我想使用的 R 函数colSums不适用于某些字段，即 datetime 字段（原因很明显），所以我不能将它应用于所有列并获得一个很好的平均数据帧。

ulyDataLefs60_12_avg = colSums(matrix(ulyDataLefs60_12, nrow=5))
Error in colSums(matrix(ulyDataLefs60_12, nrow = 5)) : 
  'x' must be numeric

我很高兴在我用来获取平均值和中位数的每 5 行的开头有一个 datetime 字段（如果我在 5 个值区间的中心获得 datetime 会更好），但到目前为止我没有t 来一个同时做两件事的答案。

也许这是一件很容易做到的事情，但它让我头疼。

4

2 回答 2

3

对于此数据：

> dput(df)
df <- structure(list(Year = c(2000L, 2000L, 2000L, 2000L, 2000L, 2000L
), Day = c(122L, 122L, 122L, 122L, 122L, 122L), Hour = c(0L, 
0L, 0L, 0L, 0L, 0L), Min = c(1L, 1L, 2L, 2L, 2L, 2L), Sec. = c(38.01, 
50.1, 2.19, 14.28, 26.38, 38.47), E1.S1 = c(3.31, 1.98, 1.98, 
2.65, 3.97, 2.65), E1.S2 = c(0.662, 3.31, 1.32, 1.32, 6.62, 0.662
), E1.S3 = c(0.662, 1.98, 3.97, 2.65, 0.662, 3.31), E1.S4 = c(2.65, 
1.98, 1.98, 3.31, 3.31, 1.98), E1.S5 = c(1.32, 1.98, 1.32, 2.65, 
3.31, 1.32), E1.S6 = c(0, 1.32, 0.662, 1.32, 4.63, 1.98), E1.S7 = c(3.31, 
4.63, 0.662, 3.97, 5.29, 1.98), E1.S8 = c(1.32, 1.32, 3.97, 2.65, 
1.98, 2.65), E2.S1 = c(1.98, 1.32, 1.32, 2.65, 0, 0.662), E2.S2 = c(1.98, 
0.662, 0.662, 0, 0, 1.32), E2.S3 = c(0.662, 0, 1.32, 0.662, 1.98, 
1.98), E2.S4 = c(0, 3.31, 0.662, 2.65, 0.662, 1.32)), .Names = c("Year", 
"Day", "Hour", "Min", "Sec.", "E1.S1", "E1.S2", "E1.S3", "E1.S4", 
"E1.S5", "E1.S6", "E1.S7", "E1.S8", "E2.S1", "E2.S2", "E2.S3", 
"E2.S4"), class = "data.frame", row.names = c("1", "2", "3", 
"4", "5", "6"))

这有效：

lapply(split(df, ceiling(seq_len(nrow(df)) / 5)), colMeans)
# $`1`
#      Year       Day      Hour       Min      Sec.     E1.S1     E1.S2     E1.S3     E1.S4     E1.S5     E1.S6     E1.S7 
# 2000.0000  122.0000    0.0000    1.6000   26.1920    2.7780    2.6464    1.9848    2.6460    2.1160    1.5864    3.5724 
#     E1.S8     E2.S1     E2.S2     E2.S3     E2.S4 
#    2.2480    1.4540    0.6608    0.9248    1.4568 
# 
# $`2`
#     Year      Day     Hour      Min     Sec.    E1.S1    E1.S2    E1.S3    E1.S4    E1.S5    E1.S6    E1.S7    E1.S8 
# 2000.000  122.000    0.000    2.000   38.470    2.650    0.662    3.310    1.980    1.320    1.980    1.980    2.650 
#    E2.S1    E2.S2    E2.S3    E2.S4 
#    0.662    1.320    1.980    1.320 
#

然后你可以只使用bind它们：

do.call(rbind, lapply(split(df, ceiling(seq_len(nrow(df)) / 5)), colMeans))
#   Year Day Hour Min   Sec. E1.S1  E1.S2  E1.S3 E1.S4 E1.S5  E1.S6  E1.S7 E1.S8 E2.S1  E2.S2  E2.S3  E2.S4
# 1 2000 122    0 1.6 26.192 2.778 2.6464 1.9848 2.646 2.116 1.5864 3.5724 2.248 1.454 0.6608 0.9248 1.4568
# 2 2000 122    0 2.0 38.470 2.650 0.6620 3.3100 1.980 1.320 1.9800 1.9800 2.650 0.662 1.3200 1.9800 1.3200

注意：它还有助于检查您要使用mean的所有列是否是integers或numeric通过执行以下操作：

> sapply(df, class)
#      Year       Day      Hour       Min      Sec.     E1.S1     E1.S2     E1.S3     E1.S4     E1.S5     E1.S6     E1.S7 
# "integer" "integer" "integer" "integer" "numeric" "numeric" "numeric" "numeric" "numeric" "numeric" "numeric" "numeric" 
#     E1.S8     E2.S1     E2.S2     E2.S3     E2.S4 
# "numeric" "numeric" "numeric" "numeric" "numeric"

编辑：遵循OP的评论：

idx <- ceiling(seq_len(nrow(dd)) / 5)
# do colMeans on all columns except last one.
res <- lapply(split(dd[-(ncol(dd))], idx), colMeans, na.rm = TRUE)
# assign first value of "datetime" in each 5-er group as names to list
names(res) <- dd$datetime[seq(1, nrow(df), by=5)]
# bind them to give a matrix
res <- do.call(rbind, res)

或者，如果你想要一个data.frame and datetime作为一个列：

idx <- ceiling(seq_len(nrow(dd)) / 5)
res <- as.data.frame(do.call(rbind, lapply(split(dd[-(ncol(dd))], idx), 
                 colMeans, na.rm = TRUE)))
res$datetime <- dd$datetime[seq(1, nrow(dd), by=5)]
于 2013-03-13T15:31:25.423 回答
1

您可以将数据视为时间序列。然后使用xts包你可以使用period.apply功能

dat.xts <- xts(dat[,-ncol(dat)],dat$datetime)
## here I take every minutes because I don't have enouhgt data
## I think in your case 5 rows is equal to 5*12 mintues = 1 hour
pts <- endpoints(dat.xts,on='mins')
period.apply(dat.xts,pts,mean)

                   Year Day Hour Min   Sec.  E1.S1  E1.S2 E1.S3 E1.S4 E1.S5 E1.S6  E1.S7  E1.S8 E2.S1  E2.S2  E2.S3  E2.S4 E2.S5 E2.S6
2000-05-01 00:01:50 2000 122    0   1 44.055 2.6450 1.9860 1.321 2.315  1.65 0.660 3.9700 1.3200 1.650 1.3210 0.3310 1.6550 0.660 0.660
2000-05-01 00:02:38 2000 122    0   2 20.330 2.8125 2.4805 2.648 2.645  2.15 2.148 2.9755 2.8125 1.158 0.4955 1.4855 1.3235 0.661 1.485

编辑显示如何将xts对象转换为data.frame：

要绘制数据，ggplot2您需要将 xts 对象强制为 data.frame。例如，您可以这样做：

  dat <- data.frame(date=index(dat.xts),coredata(dat.xts))

然后绘制 E1.S1 与日期：

library(ggplot2)
ggplot(data=dat)+ 
 geom_line(aes(x=date,y=E1.S1))
于 2013-03-13T15:59:19.600 回答