Question

是否有一项职能,至少保留一个新国家,如果没有任何新国家......在<代码>na.omit(对面的,则将抛弃物。我尝试了<条码>!na.omit(),但那是 t。

Answer 1

Use the negation of complete.cases, i.e. !complete.cases(x)

改编自?complete.cases:

data(airquality)
head(airquality[!complete.cases(airquality), ])

   Ozone Solar.R Wind Temp Month Day
5     NA      NA 14.3   56     5   5
6     28      NA 14.9   66     5   6
10    NA     194  8.6   69     5  10
11     7      NA  6.9   74     5  11
25    NA      66 16.6   57     5  25
26    NA     266 14.9   58     5  26

Answer 2

这里是将 du数据用于矩阵的一种方法,但可以很容易地适应数据框架情况。

mat <- matrix(runif(100), ncol = 10)
set.seed(2)
mat[sample(100, 10)] <- NA

我们可以使用<条码>is.na(),根据<条码>将矩阵转换成一个逻辑格式。 NA出席:

> is.na(mat)
       [,1]  [,2]  [,3]  [,4]  [,5]  [,6]  [,7]  [,8]  [,9] [,10]
 [1,] FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE  TRUE
 [2,] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
 [3,] FALSE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
 [4,] FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE
 [5,] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
 [6,] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
 [7,] FALSE  TRUE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE
 [8,] FALSE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE
 [9,] FALSE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
[10,] FALSE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE  TRUE FALSE

From that we can apply() the any() function to the rows, to return a logical index for rows with one or more NAs:

> apply(is.na(mat), 1, any)
 [1]  TRUE FALSE  TRUE  TRUE FALSE FALSE  TRUE  TRUE  TRUE  TRUE

共同实现这一目标:

ind <- apply(is.na(mat), 1, any)
mat[ind, ]

提供:

> ind <- apply(is.na(mat), 1, any)
> mat[ind, ]
          [,1]       [,2]      [,3]        [,4]       [,5]       [,6]
[1,] 0.6618988 0.01041453 0.9817279 0.007109038 0.77002786         NA
[2,] 0.8368892         NA 0.1150841 0.683403423 0.62512173 0.04217553
[3,] 0.1505014 0.86886104 0.1632009 0.929720222         NA 0.18467346
[4,] 0.1492469         NA 0.9746879 0.785878913 0.38814476         NA
[5,] 0.3570626 0.28487057 0.3490884 0.988902156 0.46150111 0.86784466
[6,] 0.9626440         NA 0.5019699 0.613952910 0.21867519 0.40264274
[7,] 0.1323720 0.15046975 0.8103973 0.710185730 0.06593551 0.57268500
           [,7]      [,8]      [,9]     [,10]
[1,] 0.35064257 0.9767552 0.2009347        NA
[2,] 0.02505036 0.3799989 0.9806000 0.3733586
[3,] 0.40110104 0.5603876 0.8289221 0.5743769
[4,] 0.97151543 0.4269434 0.8989719 0.8726963
[5,] 0.32372244        NA 0.4533770 0.1105549
[6,] 0.73319143 0.1153091 0.1474178 0.9527002
[7,]         NA 0.4400317        NA 0.5690021

Answer 3

简单的在线选择

is.na() returns a matrix of booleans detecting NA values (same dimensions as input data frame)

<代码>rowSums(s number of TRUE Value perrow

最后,在发现至少1个国民时,使用本地的留级。

data[rowSums(is.na(data)) > 0, ]

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