Question

在R中寻找时差的最快方法是什么：第1行到第2行，第2行到第3行，第3行到第4行等等。或者从1到2的时间跨度，2到3，......

最后，我前往一个显示所有差异/时间范围的功能，例如＆GT; 7000ms

HH：MM：SS：MIS * *毫秒


18:41:24.244
18:41:29.290
18:41:34.259
18:41:55.040
18:42:15.556
18:42:21.587
18:42:25.509
18:42:31.009
18:42:39.072
18:42:59.025
18:43:03.134
18:43:06.712
18:43:47.244
18:43:53.353
18:43:59.181
18:44:14.744
18:44:22.572
18:44:40.040
18:44:44.900
18:44:48.884
18:44:53.744
18:45:01.134
18:45:56.884
18:46:01.384
18:46:05.915
18:46:10.025
18:46:13.837
18:46:18.275
18:46:28.931
18:46:41.259
18:46:44.619
18:46:50.619

Answer 1

目前还不清楚数据的格式是什么。我已将其作为字符向量导入：

head(times)
# [1] "18:41:24.244" "18:41:29.290" "18:41:34.259" "18:41:55.040" ...

然后，由于您需要差异，我们可以预先添加任意日期并转换为POSIXct

times <- as.POSIXct(paste("2014-01-01",times),format="%Y-%m-%d %H:%M:%OS")
diff(times)
# Time differences in secs
# [1]  5.046  4.969 20.781 20.516  6.031  3.922  5.500  8.063 19.953  4.109 ...

Answer 2

从字符向量x

开始

head(x)
# [1] "18:41:24.244" "18:41:29.290" "18:41:34.259" "18:41:55.040"
# [5] "18:42:15.556" "18:42:21.587"

您可以将strptime与diff

一起使用

st <- strptime(x, "%H:%M:%OS")
st[diff(st) > 7]
#  [1] "2014-11-02 18:41:34.259 PST" "2014-11-02 18:41:55.040 PST"
#  [3] "2014-11-02 18:42:31.009 PST" "2014-11-02 18:42:39.072 PST"
#  [5] "2014-11-02 18:43:06.712 PST" "2014-11-02 18:43:59.181 PST"
#  [7] "2014-11-02 18:44:14.744 PST" "2014-11-02 18:44:22.572 PST"
#  [9] "2014-11-02 18:44:53.744 PST" "2014-11-02 18:45:01.134 PST"
# [11] "2014-11-02 18:46:18.275 PST" "2014-11-02 18:46:28.931 PST"

什么是找到时间差异的最快方法

2 个答案: