我从名为 dateCount.df 的数据框中获得以下数据（见下文）。
您可以看到记录号19是星期日，而前一记录是星期五。我正在尝试编写一个嵌套的if else语句，该语句将采用星期五，星期六和星期日nrec字段值并对它们取平均值并用该平均值替换星期日的nrec。它不适用于记录19.任何人都可以在我的嵌套if中找到逻辑中的错误吗？
suns = which(format(dateCount.df$Date, "%w") == 0)
for (i in suns) {
# this first "if" should check if the prior date Date[i - 1] in dateCount is equal to the actual day before Date[i]-1 and same for Friday
if (i <= 2 && dateCount.df$Date[i - 1] == (dateCount.df$Date[i] - 1) && dateCount.df$Date[i - 2] == (dateCount.df$Date[i] - 2)) {
dateCount.df$nrec[i] <- (dateCount.df$nrec[i] + dateCount.df$nrec[i - 1] + dateCount.df$nrec[i - 2]) / 3
}
else {
# this "if" should check if the prior date Date[i - 1] in dateCount is equal to the actual day before Date[i]-1 but not checking about Friday
#because to get into this block of the embedded if one of the 3 conditions failed and I am now dealing with missing Friday data
if (i>=2 && dateCount.df$Date[i-1] == (dateCount.df$Date[i] - 1)) {
# this should be the case where Saturday is available but not Friday, so add and divide by 2
dateCount.df$nrec[i] <- (dateCount.df$nrec[i] + dateCount.df$nrec[i - 1]) / 2
}
else {
# this "if" should check if the prior date Date[i-2] in dateCount is equal to the actual day before Date[i]-2 but not checking about Saturday
#because to get into this block of the embedded if one of the 3 conditions failed and also the case where Saturday is available but not Friday
# and I am now dealing with missing Saturday data but Friday is available.
if (i <= 2 && (dateCount.df$Date[i - 2] == dateCount.df$Date[i] - 2)) {
# this should be the case where Friday is available but not Saturday, so add and divide by 2
dateCount.df$nrec[i] <- (dateCount.df$nrec[i] + dateCount.df$nrec[i - 2]) / 2
}
else {
if (i >= 2) {
# this should be the case where neither Friday or Saturday is available so do nothing
dateCount.df$nrec[i]<-dateCount.df$nrec[i]
}
else {
dateCount.df$nrec[i]<-dateCount.df$nrec[i]
}
}
}
}
}
数据：
Date nrec DayOfWeek
--------------------------------
1 7/17/2011 220 Sun
2 7/18/2011 267 Mon
3 10/29/2009 30 Thu
4 10/30/2009 212 Fri
5 10/31/2009 238 Sat
6 11/1/2009 424 Sun
7 11/2/2009 423 Mon
8 11/3/2009 268 Tue
9 11/4/2009 445 Wed
10 11/5/2009 331 Thu
11 11/6/2009 241 Fri
12 11/7/2009 236 Sat
13 11/8/2009 332 Sun
14 11/9/2009 421 Mon
15 11/10/2009 399 Tue
16 11/11/2009 323 Wed
17 11/12/2009 358 Thu
18 11/13/2009 238 Fri
19 11/15/2009 301 Sun
20 11/16/2009 439 Mon
21 11/17/2009 374 Tue
22 11/23/2009 145 Mon
23 11/24/2009 472 Tue
24 11/25/2009 331 Wed
25 11/26/2009 327 Thu
26 11/27/2009 261 Fri
27 11/28/2009 296 Sat
28 11/29/2009 461 Sun
29 11/30/2009 514 Mon
30 12/1/2009 656 Tue
31 12/2/2009 505 Wed
32 12/3/2009 535 Thu
33 12/4/2009 331 Fri
34 12/5/2009 213 Sat
35 12/6/2009 444 Sun
36 12/7/2009 483 Mon
37 12/8/2009 225 Tue
38 12/9/2009 386 Wed
39 12/10/2009 102 Thu
40 12/11/2009 301 Fri
41 12/12/2009 375 Sat
42 12/13/2009 458 Sun
43 12/14/2009 332 Mon
44 12/15/2009 526 Tue
45 12/16/2009 515 Wed
46 12/17/2009 459 Thu
47 12/18/2009 312 Fri
48 12/19/2009 330 Sat
49 12/20/2009 34 Sun
50 1/16/2010 63 Sat
51 1/17/2010 238 Sun
52 1/19/2010 12 Tue
53 1/20/2010 481 Wed
54 1/21/2010 671 Thu
55 1/22/2010 439 Fri
56 1/23/2010 448 Sat
57 1/24/2010 648 Sun
58 1/25/2010 708 Mon
59 1/26/2010 695 Tue
60 1/27/2010 617 Wed
61 1/28/2010 499 Thu
62 2/25/2010 189 Thu
63 2/26/2010 551 Fri
64 2/27/2010 441 Sat
65 2/28/2010 716 Sun
66 3/1/2010 877 Mon
67 3/2/2010 758 Tue
68 3/3/2010 767 Wed
69 3/4/2010 721 Thu
70 3/5/2010 504 Fri
71 3/6/2010 36 Sat
72 4/5/2010 105 Mon
73 4/6/2010 885 Tue
答案 0 :(得分：0)
我认为问题是基于您的which声明以及您如何使用for声明。 which()应该为星期日的行提供行索引（数字），因此它将是：1,6,13,19，...
您的for语句会为i提供相同的值，因此第一次循环会给您i=1，第二次会给您i=6等。
因此，当您测试if (i <= 2 && ...时，只有i的第一个值符合<=2要求，其余的将不会执行此部分测试。因此行19会遇到第一个else语句，即使它有星期五的数据。
也许第一个if应该具有i>2的条件，以便您忽略前两行？
*阅读评论后编辑*
在前两个if语句失败后，第19天点击：
if (i <= 2 && (dateCount.df$Date[i - 2] == dateCount.df$Date[i] - 2))
19大于2，因此不符合此要求。我认为您需要将其切换为i>=2。从
if( i<-2) {do something - This is when a Sunday if the first or
second day of the data}
else{
if (dateCount.df$Date[i-1] == (dateCount.df$Date[i] - 1)).... etc
只需保留相同的代码，但删除所有部分，将i与2进行比较，然后再处理。