Question

我正在尝试创建一个基本的for循环函数来打印某些数据的摘要统计信息。我有两个ID变量

k <- unique(df$ID_1)
m <- unique(df$ID_2)

func <- 
  for (i in 1:k) {
  for (j in 1:m) {
    tmp[i] <- summary([[i]])
    print(tmp[i])
    }
}

我在循环的j in 1:m部分遇到问题，以及如何处理它以打印摘要统计信息。

我有：

for i in 1:k(countries) for j in 1:m(industries) print summary statistics for each i across j基本上将其保存在新的数据框中。

稍微不确定如何处理此for循环中的j部分。

以下是随机抽样数据的前40个观察结果：

df = structure(list(Major.sectors.id = c(13L, 6L, 1L, 7L, 7L, 5L,  11L, 5L, 5L, 13L, 7L, 5L, 5L, 7L, 7L, 11L, 7L, 5L, 3L, 7L, 5L,  2L, 5L, 7L, 5L, 16L, 16L, 3L, 5L, 8L, 5L, 16L, 5L, 3L, 11L, 2L,  7L, 7L, 5L, 5L), Region.in.country.id = c(14L, 2L, 2L, 1L, 4L,  2L, 8L, 2L, 2L, 11L, 4L, 10L, 1L, 1L, 8L, 11L, 11L, 5L, 9L, 15L,  1L, 3L, 1L, 2L, 1L, 1L, 6L, 14L, 1L, 1L, 10L, 8L, 3L, 15L, 10L,  12L, 5L, 2L, 1L, 10L), CL.CA
= c(0.760887885213847, 1.65052687741763, 
0.0806197205130879, 0.81806086270666, 0.288078797135706, 4.40245492070389, 
1.04708699122107, 0.303834508544838, 0.660743728653869, 0.588710351051974, 
0.798830178759201, 0.927690515241424, 0.425398597752096, 3.94063680888526, 
0.23268050036403, 0.0476527647463019, 0.793725271068126, 7.57011718390699, 
0.313488176978694, 0.932388646476865, 0.528403681370993, 0.327018095170076, 
0.026490765171504, 0.0684209603311003, 0.199002799160252, 1.39359718123149, 
1.46636993076162, 1.4949863023741, 0.950167224080268, 0.976629282263177, 
0.0311026456992869, 0.850753860127157, 1.13918820019141, 0.801954640022603, 
71.4120603015075, 1.81240164673211, 0.813709616403117, 1.16837699936937, 
0.535646685572253, 0.54046260697011), WC.TA = c(0.912151080743853, 
-0.149811669390946, 0, 0.317269061162005, 0.404991527614234, 
0.0917180013689254, 0, 0.0363331984252548, 0.0581172322430804, 
0.0984968038873609, 0.404607993560882, 0.429639957001037, 0.0310040298241571, 
0.51689621232918, 0.20287556599724, 0.0768790516016589, 0.470663333767837, 
0.115332950488128, 0.959409600460894, 0.810159261144644, -0.000968329915698337, 
0.276768734669129, -0.00197889182058048, 0.841627179101607, 0.337732164517433, 
0.0312224689196786, 0.0905910619894282, 0.554192762156444, 0.0112172829248027, 
0.365589304257529, 0.155960321393463, 0.781180744777475, 0.129165904803528, 
0.815471911266304, -0.0833032382328157, 0.312503120641987, 0.637601499063086, 
0.482001582278481, 0.344956547341399, 0.142965448177925)), .Names = c("Major.sectors.id",  "Region.in.country.id", "CL.CA", "WC.TA"), row.names = c(102291L,  68029L, 67292L, 101860L, 94123L, 77664L, 18713L, 39361L, 61163L,  96185L, 92109L, 5411L, 52390L, 95008L, 12084L, 82522L, 102897L,  61661L, 33672L, 89727L, 76163L, 25942L, 10525L, 38400L, 24587L,  33894L, 20545L, 20563L, 10163L, 55586L, 77839L, 59394L, 86677L,  43471L, 77523L, 2190L, 61390L, 77768L, 95584L, 70096L), class = "data.frame")

编辑@ nate.edwinton：以下评论;

使用原始帖子中的dput输出，运行以下代码：

k <- unique(df$Major.sectors.id)
 m <- unique(df$Region.in.country.id)

    colnames(df)
     for(i in 1:length(m)){
       country.ID <- m[i]    # provided ID_2 corresponds to countries
       for(j in 1:length(k)){
         sector.ID <- k[j]   # provided ID_1 corresponds to sectors
         S1 <- which(df$Major.sectors.id == sector.ID)
         S2 <- which(df$Region.in.country.id == country.ID)
         rows.2.consider <- intersect(S1, S2)
         tmp.summary <- summary(df[rows.2.consider,])
         print(tmp.summary)
         rm(country.ID, sector.ID, S1, S2, rows.2.consider, tmp.summary, i, j)
       }
     }
     `

Answer 1

所以这可能对你有帮助

for(i in 1:length(m)){
  country.ID <- m[i]    # provided ID_2 corresponds to countries
  for(j in 1:length(k)){
    sector.ID <- k[j]   # provided ID_1 corresponds to sectors
    S1 <- which(df$ID_1 == sector.ID)
    S2 <- which(df$ID_2 == country.ID)
    rows.2.consider <- intersect(S1, S2)
    tmp.summary <- summary(df[rows.2.consider,])
    print(tmp.summary)
    rm(sector.ID, S1, S2, rows.2.consider, tmp.summary, j)
    }
   rm(country.ID, i)
 }

<强>更新

这是输出的一部分：

 Major.sectors.id Region.in.country.id     CL.CA            WC.TA       
 Min.   :13       Min.   :14           Min.   :0.7609   Min.   :0.9122  
 1st Qu.:13       1st Qu.:14           1st Qu.:0.7609   1st Qu.:0.9122  
 Median :13       Median :14           Median :0.7609   Median :0.9122  
 Mean   :13       Mean   :14           Mean   :0.7609   Mean   :0.9122  
 3rd Qu.:13       3rd Qu.:14           3rd Qu.:0.7609   3rd Qu.:0.9122  
 Max.   :13       Max.   :14           Max.   :0.7609   Max.   :0.9122  
 Major.sectors.id Region.in.country.id     CL.CA         WC.TA    
 Min.   : NA      Min.   : NA          Min.   : NA   Min.   : NA  
 1st Qu.: NA      1st Qu.: NA          1st Qu.: NA   1st Qu.: NA  
 Median : NA      Median : NA          Median : NA   Median : NA  
 Mean   :NaN      Mean   :NaN          Mean   :NaN   Mean   :NaN  
 3rd Qu.: NA      3rd Qu.: NA          3rd Qu.: NA   3rd Qu.: NA  
 Max.   : NA      Max.   : NA          Max.   : NA   Max.   : NA

我看到它的方式，所有这些NA和NaN的解释都非常简单：循环遍历所有可能的（country.id，region.id）组合。但是，对于某些组合，没有数据，例如df中有一行与sector.id == 13和region.id == 14对应，但不存在sector.id == 6和region.id == 14的行。这解释了上面第二个摘要中的NA。

我仍然对NaN中的Mean感到有点疑惑（我问自己为什么不NA中的Median，但我猜它有一些东西与定义函数的方式有关。

最后，sum(is.na(df))返回0的原因是NA中没有df！ ~~确实，which(is.na(df)==TRUE)返回integer(0)。换句话说，你总结了什么。~~ is.na(df)是一个由TRUE s或FALSE组成的逻辑向量。在此，因为NA中没有df，所以它只包含FALSE个。现在，这些逻辑运算符可用于执行数学运算：TRUE被视为1而FALSE被视为0。这解释了它。如果有人试图总结NA s，则可获得以下内容：sum(c(NA,NA))返回NA。

for两个ID列的循环

1 个答案: