问题描述
头(腹腔)
chr Pos Val
X 129271111 10
X 129271112 10
X 129271113 10
X 129271114 10
X 129271115 10
X 129271116 11
X 129271117 11
X 129271118 11
X 129271119 11
X 129271120 11
X 129271121 11
X 129271122 11
X 129271123 11
X 129271124 11
X 129271125 11
X 129271126 11
X 129271127 11
X 129271128 11
X 129271129 11
X 129271130 11
X 129271131 11
X 129271132 11
X 129271133 11
head(注释)
chr Region start end Gene status
X Exon 129271053 129271110 AIFM1 NO
X Exon 129270618 129270706 AIFM1 NO
X Exon 129270020 129270160 AIFM1 NO
X Exon 129267288 129267430 AIFM1 NO
X Exon 129265650 129265774 AIFM1 NO
X Exon 129263945 129264141 AIFM1 NO
X Exon 129263532 129263603 AIFM1 NO
3 Exon 15643358 15643401 BTD NO
3 Exon 15676931 15677195 BTD NO
3 Exon 15683415 15683564 BTD NO
尝试在第一个文件中用第二个位置的开始和结尾之间的位置创建具有相应基因名称的基因名称的新列.
Trying to create a new column with the Gene name in the first file for the positions between start and end of the second position with respective gene names.
covreage$Gene <- ifelse(covreage$chr == annotation$chr & covreage$pos >= annotation$start & covreage$pos <= annotation$end,annotation$Gene,"NA")
问题是第二个文件在范围内具有file1 pos的值，并且chr和position在两个文件中均应匹配.chr可以具有23个不同的值，并且Pos在所有不同的chr值中都具有相似的值.一起将chr和原始位置定位成为唯一元素
The problem is the second file have the value for file1 pos in range and chr and position should match in both files. The chr can have 23 different values and Pos will have similar values in all of the different chr values. Together chr and position the raw become unique element
上面的代码给出了此错误
The above code gives this error
Warning messages:
1: In is.na(e1) | is.na(e2) :
longer object length is not a multiple of shorter object length
2: In `==.default`(covreage$chr, annotation$chr) :
longer object length is not a multiple of shorter object length
3: In covreage$pos >= annotation$start :
longer object length is not a multiple of shorter object length
4: In covreage$pos <= annotation$end :
longer object length is not a multiple of shorter object length
推荐答案
通过评估诸如 covreage $ pos> =注记$ start 之类的内容，您可以逐行比较两个data.frames，这不是你想要的.您想使用一些R所不知道的分组规则，将第一行的第二行与第二行的第二行进行比较.
By evaluating something like covreage$pos >= annotation$start, you're comparing both data.frames row by row, which is not what you want. You want to compare several rows from the first against one row from the second, using some grouping rule R does not know about.
您仍然会得到一些输出，因为R通常会根据需要尝试回收元素:
You still get some output because R in general tries to recycle elements as needed:
>1:6< c(2,6,6)[1]是是是否是否
>1:5< c(2,6,6)[1]是是是否是警告信息:在1:5中c(2，6，6):较长的对象长度不是较短的对象长度的倍数
在第一种情况下，由于元素被均匀地重用，因此不会打印警告.在第二种情况下，这是不可能的(因为正如R所说，较长的对象长度不是较短的对象长度的倍数)，因此出现警告.
In the first case, no warning is printed because elements are evenly reused; in the second case, that is not possible (because as R says, longer object length is not a multiple of shorter object length), so a warning shows up.
即使在您所介绍的情况下回收被认为是错误，R还是允许它，因为在某些情况下它可能有用.
Even though recycling is to be considered an error in the context you presented, R allows it because it may be useful in some situations.
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