R-如何删除单个列中某个特定短语之前和之后的行 [英] R - How to delete rows before a certain phrase and after a certain phrase in a single column

问题描述

我想删除某个短语之前的行，然后删除稍后出现的相同（几乎）短语之后的行。我猜想看待它的另一种方式是只保留某个部分开头和结尾的数据。

Hi I would like to delete rows before a certain phrase and then after the same (almost) phrase which appears later on. I guess another way to look at it would be keep only the data from the start and end of a certain section.

我的数据如下：

df <- data.frame(time = as.factor(c(1,2,3,4,5,6,7,8,9,10,11,12,13)), 
                 type = c("","","GMT:yyyy-mm-dd_HH:MM:SS_LT:2016-10-18_06:09:53","(K)","","","","(K)","(K)","","(K)","GMT:yyyy-mm-dd_HH:MM:SS_CAM:2016-10-18_06:20:03",""),
                 names = c("J","J","J","J","J","J","J","J","J","J","J","J","J"))

，我想删除第一个 GMT之前的所有内容：yyyy ... 短语以及第二个 GMT：yyyy ... 短语之后。因此最终产品将是

and I would like to delete everything before the first GMT:yyyy... phrase and after the second GMT:yyyy... phrase. So the end product would be

time   type                                                    names
3      GMT:yyyy-mm-dd_HH:MM:SS_LT:2016-10-18_06:09:53           J
4      (K)                                                      J
5                                                               J
6                                                               J
7                                                               J 
8      (K)                                                      J
9      (K)                                                      J
10                                                              J
11     (K)                                                      J
12     GMT:yyyy-mm-dd_HH:MM:SS_LT:2016-10-18_06:20:03           J

我认为子集可能有用，但这给我带来了问题。

I thought subset might work but it is giving me problems.

推荐答案

使用grep，您可以找到找到您的模式的行的索引：

Using grep, you can locate the indexes of the rows where your pattern is found:

ind=grep("^GMT",df$type)

然后，您只能保留两个索引之间的行：

Then you can keep only the rows between the two indexes:

df=df[ind[1]:ind[2],]

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