Question

我正在尝试查找R中匹配的实例的平均数量。

我想知道何时所有3列== 1，何时所有3列== 0，最后是两者。

这不起作用：

mean(test$direction == test$pred.lm == test$pred.svm)

这给了我direction列等于pred.lm列的实例总数。

mean(test$direction == test$pred.lm)

示例：

           direction pred.lm pred.svm
2018-07-20         0       0        0
2018-07-23         1       0        0
2018-07-24         0       0        1
2018-07-25         1       1        1
2018-07-26         1       1        1
2018-07-27         0       0        0

第1行，第4行，第5行和第6行都匹配。我想要它们在== 0且== 1时匹配的平均次数，最后是所有匹配，无论0或1。

数据：

library(xts)
df <- structure(c(0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 
    1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 
    1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 
    1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 
    0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 
    1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 
    0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 
    1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 
    1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1), index = structure(c(1532044800, 
    1532304000, 1532390400, 1532476800, 1532563200, 1532649600, 1532908800, 
    1532995200, 1533081600, 1533168000, 1533254400, 1533513600, 1533600000, 
    1533686400, 1533772800, 1533859200, 1534118400, 1534204800, 1534291200, 
    1534377600, 1534464000, 1534723200, 1534809600, 1534896000, 1534982400, 
    1535068800, 1535328000, 1535414400, 1535500800, 1535587200, 1535673600, 
    1536019200, 1536105600, 1536192000, 1536278400, 1536537600, 1536624000, 
    1536710400, 1536796800, 1536883200, 1537142400, 1537228800, 1537315200, 
    1537401600, 1537488000, 1537747200, 1537833600, 1537920000, 1538006400, 
    1538092800, 1538352000, 1538438400, 1538524800, 1538611200, 1538697600, 
    1538956800, 1539043200, 1539129600, 1539216000, 1539302400, 1539561600, 
    1539648000, 1539734400, 1539820800, 1539907200, 1540166400, 1540252800, 
    1540339200, 1540425600, 1540512000, 1540771200, 1540857600, 1540944000, 
    1541030400, 1541116800, 1541376000, 1541462400, 1541548800, 1541635200, 
    1541721600, 1541980800, 1542067200, 1542153600, 1542240000, 1542326400, 
    1542585600, 1542672000, 1542758400, 1542931200, 1543190400, 1543276800, 
    1543363200, 1543449600, 1543536000), tzone = "UTC", tclass = "Date"), class = c("xts", 
    "zoo"), .indexCLASS = "Date", .indexTZ = "UTC", tclass = "Date", tzone = "UTC", src = "yahoo", updated = structure(1544903554.77594, class = c("POSIXct", 
    "POSIXt")), .Dim = c(94L, 3L), .Dimnames = list(NULL, c("direction", 
    "pred.lm", "pred.svm")))

Answer 1

你很近；主要问题是所有比较都必须是二进制的，如：

# All 1's
with(df, mean(direction == pred.lm & pred.lm == pred.svm & pred.svm == 1))
# [1] 0.1276596

# All 0's
with(df, mean(direction == pred.lm & pred.lm == pred.svm & pred.svm == 0))
# [1] 0.393617

# All equal
with(df, mean(direction == pred.lm & pred.lm == pred.svm))
# [1] 0.5212766

但是，您可能会做得更好：

# All 1's
mean(rowSums(df) == ncol(df))
# [1] 0.1276596

# All 0's
mean(rowSums(df) == 0)
# [1] 0.393617

# All equal
mean(rowSums(df) %in% c(0, ncol(df)))
# [1] 0.5212766

这不仅更短，而且在df中允许多于三列。

Answer 2

我从您的问题中得到的是您想要实例为0,0,0或1,1,1的所有行的均值

> rowno.<-NULL
> for(i in 1:nrow(df))
+ if(all(df[i,]==0) || all(df[i,]==1))
+ rowno.<-c(rowno.,i)
> rowno.
[1]  1  4  5  6  7  8  9 11 12 17 19 20 21 23 25 31 32 33
[19] 34 36 38 39 45 48 51 53 54 55 56 57 58 65 68 69 71 73
[37] 74 75 77 78 80 82 83 86 89 91 92 93 94
> mean(rowno.)
[1] 47.91837

如果您想获得所有观察结果的均值，则可以通过

     > mean(df[rowno.,])
     [1] 0.244898

数据与各列匹配的平均时间

2 个答案: