The following simple loop seems to skip elements in a data frame. I would appreciate any tips in figuring out where the problem with the data/code may lay.
下面的简单循环似乎跳过了数据框架中的元素。如果您能告诉我数据/代码的问题在哪里，我将不胜感激。
foo <- apply(data, 1, function(x) {
vec <- x
mylist <- list()
for (i in vec){
#print(i)
mylist[[i]]<-i
}
print(length(vec))
print(length(mylist))
})
My data frame has 25 columns. For some rows, length(vec) returns 25, while length(mylist) returns 24.
我的数据框有25列。对于某些行，length(vec)返回25,length(mylist)返回24。
[1] 25
[1] 24
If I use the hashed-out print(i) I can see 25 elements in all rows.
如果我使用hasheout print(I)，我可以在所有行中看到25个元素。
The above is a simplification of the actual code I want to use, but the problem already occurs in this simple format.
以上是我想要使用的实际代码的简化，但问题已经出现在这个简单的格式中。
Thanks in advance!
提前谢谢!
PS. I have tried having data as character or as factor. Neither seems to impact the problem.
我试过把数据作为性格或因素。两者似乎都不会对问题产生影响。
PPS. two lines of the data frame that give different results (although they contain the same number of elements):
pp。数据框的两行给出不同的结果(尽管它们包含相同数量的元素):
structure(list(data1.LOC = c("LL_A1_00000003068_686", "LL_A1_00000003538_274"), REF = c("G", "T"), ALT = c("C", "C"), L47.variant = c("0/1:28,34:62:99:1154,0,926", "0/0:9,0:9:21:0,21,276"), L51.variant = c("0/0:61,0:61:99:0,184,2417", "0/0:6,0:6:15:0,15,192"), LCro11.variant = c("0/0:24,0:24:72:0,72,951", "0/0:2,0:2:6:0,6,80"), LCro5.variant = c("0/0:48,0:48:99:0,141,1869", "0/0:5,0:5:15:0,15,173"), N01.variant = c("0/1:22,16:38:99:526,0,758", "1/1:0,2:2:6:63,6,0"), N09.variant = c("1/1:1,50:51:99:1885,110,0", "0/0:12,0:12:36:0,36,460"), Nor28.variant = c("1/1:0,23:23:66:874,66,0", "0/0:5,0:5:12:0,12,159"), P161.variant = c("1/1:0,54:55:99:2118,163,0", "0/0:2,0:2:6:0,6,80"), Rom155.variant = c("0/0:69,0:69:99:0,208,2749", "0/1:5,3:8:99:102,0,102"), Rom161.variant = c("0/0:75,0:75:99:0,226,2957", "0/0:5,0:5:15:0,15,196"), Rom303.variant = c("0/0:44,0:44:99:0,132,1739", "0/0:5,0:5:15:0,15,195"), Rus291.variant = c("0/1:43,30:73:99:972,0,1443", "0/1:1,3:4:28:108,0,28"), Rus292.variant = c("0/0:56,0:56:99:0,163,2139", "0/0:11,0:11:33:0,33,429"), Sl5t.variant = c("0/1:55,34:89:99:1003,0,1911", "0/0:10,0:10:30:0,30,379"), Sl6t.variant = c("0/0:89,0:89:99:0,268,3513", "0/0:10,0:10:30:0,30,383"), s037y.variant = c("0/0:63,0:63:99:0,190,2484", "0/0:8,0:8:18:0,18,236"), s087y.variant = c("0/0:72,0:72:99:0,211,2770", "0/0:6,0:6:15:0,15,179"), s2E03.variant = c("0/1:34,27:61:99:810,0,1175", "0/0:4,0:4:12:0,12,143"), s2L05.variant = c("0/0:56,0:56:99:0,169,2220", "0/1:4,4:8:95:139,0,95"), s2P01.variant = c("0/1:44,27:71:99:859,0,1519", "0/0:6,0:6:18:0,18,240"), s2R01.variant = c("1/1:0,68:68:99:2642,202,0", "0/1:5,6:11:99:202,0,130"), s2R05.variant = c("0/1:41,33:74:99:1012,0,1393", "0/0:8,0:8:24:0,24,312")), .Names = c("data1.LOC", "REF", "ALT", "L47.variant", "L51.variant", "LCro11.variant", "LCro5.variant", "N01.variant", "N09.variant", "Nor28.variant", "P161.variant", "Rom155.variant", "Rom161.variant", "Rom303.variant", "Rus291.variant", "Rus292.variant", "Sl5t.variant", "Sl6t.variant", "s037y.variant", "s087y.variant", "s2E03.variant", "s2L05.variant", "s2P01.variant", "s2R01.variant", "s2R05.variant"), row.names = 19:20, class = "data.frame")
1 个解决方案
#1
2
instead of using elements of vec to access elements of mylist, thus updating the same element in case of duplicates in vec, you should iterate through vec one by one via a standard integer index i running the length of it, like so:
不要使用vec的元素访问mylist的元素，从而在vec中出现重复时更新相同的元素，您应该通过运行其长度的标准整型索引i逐一遍历vec，如下所示:
foo <- apply(data, 1, function(x) {
vec <- x
mylist <- list()
for (i in seq(vec)){
#print(i)
mylist[[i]] <- vec[i]
}
print(length(vec))
print(length(mylist))
})
To summarise, your code did not work because: You may have duplicates in vec. If for instance, vec<-c(1,1,2), length(vec)==3 but it will result that length(mylist)==2. # Comment from nicola 1 hour ago
总之，您的代码不能工作，因为:您可能在vec中有副本。例如，如果vec<-c(1,1,2)， length(vec)= 3，则会导致长度(mylist)= 2。1小时前nicola的评论。
#1
2
instead of using elements of vec to access elements of mylist, thus updating the same element in case of duplicates in vec, you should iterate through vec one by one via a standard integer index i running the length of it, like so:
不要使用vec的元素访问mylist的元素，从而在vec中出现重复时更新相同的元素，您应该通过运行其长度的标准整型索引i逐一遍历vec，如下所示:
foo <- apply(data, 1, function(x) {
vec <- x
mylist <- list()
for (i in seq(vec)){
#print(i)
mylist[[i]] <- vec[i]
}
print(length(vec))
print(length(mylist))
})
To summarise, your code did not work because: You may have duplicates in vec. If for instance, vec<-c(1,1,2), length(vec)==3 but it will result that length(mylist)==2. # Comment from nicola 1 hour ago
总之，您的代码不能工作，因为:您可能在vec中有副本。例如，如果vec<-c(1,1,2)， length(vec)= 3，则会导致长度(mylist)= 2。1小时前nicola的评论。