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如何统计一列中数字值的数量？

rstatistics
9
我有一个数据框，想要生成一张包含每个三列的组别中有效数字值数量、平均数和标准差的汇总统计表。在R中，似乎找不到任何函数来计算数字值的数量。可以使用length()函数来告诉我们有多少个值，也可以使用colSums(is.na(x))函数来计算NA值的数量，但是colSums(is.numeric(x))函数不能像这些函数一样工作。
可以使用tapply函数 {长度 - NA值的数量 - 空白值的数量 - 文本值的数量}，但肯定有更简单的方法。
我的数据（我想按Nominal分组，并在Actual、LinPred和QualPred上生成汇总统计信息）。
structure(list(Nominal = c(1, 3, 6, 10, 30, 50, 150, 250, 1, 
3, 6, 10, 30, 50, 150, 250, 1, 3, 6, 10, 30, 50, 150, 250, 1, 
3, 6, 10, 30, 50, 150, 250, 1, 3, 6, 10, 30, 50, 150, 250, 1, 
3, 6, 10, 30, 50, 150, 250, 1, 3, 6, 10, 30, 50, 150, 250, 1, 
3, 6, 10, 30, 50, 150, 250, 1, 3, 6, 10, 30, 50, 150, 250), Actual = c(NA, 
0.422, 0.782, 1.25, 3.85, 6.94, 18.8, 31.2, 0.118, 0.361, 0.747, 
1.18, 3.58, 5.82, 16.7, 29, 0.113, 0.382, 0.692, 1.12, 3.51, 
5.43, 17.1, 28.7, 0.134, 0.402, 0.718, 1.25, 3.65, 6.52, NA, 
28.8, 0.123, 0.396, 0.664, 1.12, 3.83, 5.6, NA, 28.1, 0.112, 
0.341, 0.7, 1.08, 3.25, 5.97, NA, 27.1, 0.106, 0.35, 0.674, 1.14, 
3.28, 5.5, 17.3, 30, 0.122, 0.321, 0.673, 1.22, 3.41, 5.85, 17.6, 
28.1, 0.129, 0.351, 0.737, 1.06, 3.39, 5.53, 15.9, 28.5), LinPred = c(NA, 
3.49519490135683, 6.4706724568458, 10.3387932789814, 31.8283534019573, 
57.3678690865708, 155.393324109068, 257.881995464799, 0.982569410055046, 
2.99101676001009, 6.18138991672881, 9.76022819874748, 29.5967452353405, 
48.1108278028274, 138.036371702049, 239.698521514589, 0.941243332895477, 
3.16458628408028, 5.72680306797355, 9.26431527283265, 29.0181801551066, 
44.887393784381, 141.342457874815, 237.218956885015, 1.07941778099747, 
3.36900393602722, 6.0686652233011, 10.6136646056736, 31.1174212178803, 
55.6364968333108, NA, 245.979704049963, 0.98544222985819, 3.3177445444967, 
5.60733069952645, 9.50304445584572, 32.6552029637958, 47.7767234652982, 
NA, 239.999441704736, 0.89146667871891, 2.8478667888003, 5.91488704870955, 
9.1613151789756, 27.7001284491792, 50.9377192763467, NA, 231.456209782983, 
0.887738051402174, 3.04188235451485, 5.9023034783202, 10.0163659588551, 
28.9092709123842, 48.5084526866061, 152.684283738776, 264.805729023739, 
1.02899341554071, 2.78585700701375, 5.89347501806154, 10.7226427795477, 
30.0569707460098, 51.5984137771366, 155.332821816374, 248.031654532288, 
1.09079263735132, 3.05071081477351, 6.45849647461568, 9.31008913816238, 
29.8804015408367, 48.7733064943658, 140.324439376654, 251.563038635751
), QuadPred = c(NA, 3.46077095737974, 6.38659713413108, 10.1956079501556, 
31.4700369979564, 57.0089799611706, 157.775316006369, 268.303966059862, 
0.99289436409299, 2.96536517477853, 6.10198249392715, 9.62549220297933, 
29.2517496204359, 47.7196128593832, 139.600469198163, 248.272682787657, 
0.95232583127381, 3.13590297331348, 5.65480031033985, 9.13693141349813, 
28.6769820181676, 44.4936547741659, 143.050878627236, 245.555818447238, 
1.08417831830729, 3.33895371044810, 6.00044125019758, 10.4882228621509, 
30.8451526869812, 55.4331759085967, NA, 256.446833964951, 0.991679220421247, 
3.28844923081897, 5.54540949253351, 9.3907657095483, 32.3793538902883, 
47.5218142460371, NA, 249.828516445647, 0.899183876120787, 2.82554368740693, 
5.84875388286628, 9.05319326862309, 27.4395572248486, 50.7001828907023, 
NA, 240.411024762687, 0.884412915928806, 3.05257006009469, 5.93046554432476, 
10.0673979669, 29.0311859234644, 48.645035648271, 151.914544909710, 
261.273991566153, 1.02660962824666, 2.79491765184684, 5.92158513760114, 
10.7773327827008, 30.1813919027873, 51.7318741314584, 154.518856412401, 
245.027488125567, 1.08881969774848, 3.06145444119556, 6.48990638077339, 
9.35738460692028, 30.0044505131336, 48.9096796323938, 139.747394069421, 
248.451100154569)), .Names = c("Nominal", "Actual", "LinPred", 
"QuadPred"), row.names = c(NA, -72L), class = "data.frame")
5个回答
10

这是一些可能有帮助的附加包（请参见Quick-R）

使用Hmisc包

library(Hmisc)

describe(mydata) 
# n, nmiss, unique, mean, 5,10,25,50,75,90,95th percentiles 
# 5 lowest and 5 highest scores

使用pastecs包

library(pastecs)

stat.desc(mydata) 
# nbr.val, nbr.null, nbr.na, min max, range, sum, 
# median, mean, SE.mean, CI.mean, var, std.dev, coef.var

使用psych包

library(psych)
describe(mydata)
# item name ,item number, nvalid, mean, sd, 
# median, mad, min, max, skew, kurtosis, se

我会使用 psych 包中的 describe.by 方法；

> describe.by(biastable, as.factor(Nominal))
group: 1
         var n mean   sd median trimmed  mad  min  max range  skew kurtosis   se
Nominal    1 9 1.00 0.00   1.00    1.00 0.00 1.00 1.00  0.00   NaN      NaN 0.00
Actual     2 8 0.12 0.01   0.12    0.12 0.01 0.11 0.13  0.03  0.09    -1.47 0.00
LinPred    3 8 0.99 0.08   0.98    0.99 0.10 0.89 1.09  0.20  0.04    -1.70 0.03
QuadPred   4 8 0.99 0.08   0.99    0.99 0.10 0.88 1.09  0.20 -0.04    -1.64 0.03
------------------------------------------------------------------------ 
group: 3
         var n mean   sd median trimmed  mad  min  max range skew kurtosis   se
Nominal    1 9 3.00 0.00   3.00    3.00 0.00 3.00 3.00  0.00  NaN      NaN 0.00
Actual     2 9 0.37 0.03   0.36    0.37 0.03 0.32 0.42  0.10 0.15    -1.50 0.01
LinPred    3 9 3.12 0.24   3.05    3.12 0.30 2.79 3.50  0.71 0.15    -1.52 0.08
QuadPred   4 9 3.10 0.23   3.06    3.10 0.34 2.79 3.46  0.67 0.12    -1.51 0.08
------------------------------------------------------------------------ 
group: 6
         var n mean   sd median trimmed  mad  min  max range skew kurtosis   se
Nominal    1 9 6.00 0.00   6.00    6.00 0.00 6.00 6.00  0.00  NaN      NaN 0.00
Actual     2 9 0.71 0.04   0.70    0.71 0.04 0.66 0.78  0.12 0.46    -1.30 0.01
LinPred    3 9 6.02 0.30   5.91    6.02 0.28 5.61 6.47  0.86 0.28    -1.43 0.10
QuadPred   4 9 5.99 0.31   5.93    5.99 0.25 5.55 6.49  0.94 0.26    -1.26 0.10
------------------------------------------------------------------------ 
group: 10
         var n  mean   sd median trimmed  mad   min   max range skew kurtosis   se
Nominal    1 9 10.00 0.00  10.00   10.00 0.00 10.00 10.00  0.00  NaN      NaN 0.00
Actual     2 9  1.16 0.07   1.14    1.16 0.09  1.06  1.25  0.19 0.09    -1.71 0.02
LinPred    3 9  9.85 0.60   9.76    9.85 0.74  9.16 10.72  1.56 0.24    -1.76 0.20
QuadPred   4 9  9.79 0.62   9.63    9.79 0.72  9.05 10.78  1.72 0.27    -1.65 0.21
------------------------------------------------------------------------ 
group: 30
         var n  mean   sd median trimmed  mad   min   max range skew kurtosis   se
Nominal    1 9 30.00 0.00  30.00   30.00 0.00 30.00 30.00  0.00  NaN      NaN 0.00
Actual     2 9  3.53 0.22   3.51    3.53 0.21  3.25  3.85  0.60 0.23    -1.58 0.07
LinPred    3 9 30.08 1.55  29.88   30.08 1.44 27.70 32.66  4.96 0.21    -1.27 0.52
QuadPred   4 9 29.92 1.51  30.00   29.92 1.44 27.44 32.38  4.94 0.04    -1.22 0.50
------------------------------------------------------------------------ 
group: 50
         var n  mean   sd median trimmed  mad   min   max range skew kurtosis   se
Nominal    1 9 50.00 0.00  50.00   50.00 0.00 50.00 50.00  0.00  NaN      NaN 0.00
Actual     2 9  5.91 0.51   5.82    5.91 0.43  5.43  6.94  1.51 0.90    -0.73 0.17
LinPred    3 9 50.40 3.98  48.77   50.40 3.21 44.89 57.37 12.48 0.49    -1.16 1.33
QuadPred   4 9 50.24 3.97  48.91   50.24 2.65 44.49 57.01 12.52 0.39    -1.21 1.32
------------------------------------------------------------------------ 
group: 150
         var n   mean   sd median trimmed   mad    min    max range  skew kurtosis   se
Nominal    1 9 150.00 0.00 150.00  150.00  0.00 150.00 150.00  0.00   NaN      NaN 0.00
Actual     2 6  17.23 0.97  17.20   17.23  0.67  15.90  18.80  2.90  0.25    -1.23 0.39
LinPred    3 6 147.19 8.11 147.01  147.19 11.13 138.04 155.39 17.36 -0.01    -2.22 3.31
QuadPred   4 6 147.77 7.95 147.48  147.77 10.95 139.60 157.78 18.17  0.07    -2.10 3.25
------------------------------------------------------------------------ 
group: 250
         var n   mean    sd median trimmed  mad    min    max range skew kurtosis   se
Nominal    1 9 250.00  0.00 250.00  250.00 0.00 250.00 250.00  0.00  NaN      NaN 0.00
Actual     2 9  28.83  1.18  28.70   28.83 0.89  27.10  31.20  4.10 0.59    -0.57 0.39
LinPred    3 9 246.29 10.57 245.98  246.29 9.31 231.46 264.81 33.35 0.33    -1.26 3.52
QuadPred   4 9 251.51  8.84 248.45  251.51 5.08 240.41 268.30 27.89 0.62    -1.04 2.95
>
7

colSums(!is.na(x))应该可以正常工作。

5
你能使用类似这样的东西吗？
length(unique(x))
5
如果您有数字向量，则可以有NA（is.na()）、Inf（is.infinite()）、NaN（is.nan()）和“有效”数字值。什么是“空白值”和“文本值”？对于“有效”的数字值（以上意义），您可以使用is.finite()：
is.finite(c(1,NA,Inf,NaN))
# [1]  TRUE FALSE FALSE FALSE
sum( is.finite(c(1,NA,Inf,NaN)) )
# [1] 1

所以colSums(is.numeric(x))可以像colSums(is.finite(x))一样完成。
+1 注意到了OP所要求的内容，正是我所寻找的 :-) - Shalom Craimer
0

complete.cases（或sum(complete.cases)）是否符合您的需求？

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