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本文介绍了R使用dplyr来剪切包含2个或更多变量的修复时间间隔的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 我有一个数据框 df< - data.frame（time = c（2015-09-07 00 ：32：19，2015-09-07 01:02:30，2015-09-07 01:31:36，2015-09-07 01:47:45， 2015-09-07 02:00:172015-09-07 02:07:30，2015-09-07 03:39:41，2015-09-07 04:04:21 ，2015-09-07 04:04:21，2015-09-07 04:04:22）， inOut = c（IN，OUT，IN，IN ，IN，IN，IN，OUT，IN，OUT）） > df 时间inOut 1 2015-09-07 00:32:19 IN 2 2015-09-07 01:02:30 OUT 3 2015-09-07 01 ：31：36 IN 4 2015-09-07 01:47:45 IN 5 2015-09-07 02:00:17 IN 6 2015-09-07 02:07 ：30 IN 7 2015-09-07 03:39:41 IN 8 2015-09-07 04:04:21 OUT 9 2015-09-07 04:04:21 IN 10 2015-09-07 04:04:22 OUT > 我想计算每15分钟IN / OUT的计数，我可以这样做创建另一个in_df，out_df，每15分钟剪下这些数据帧，然后将其合并在一起，以获得我的结果。 outdf是我的预期结果。 in_df< - df [which（df $ inOut ==IN），] out_df< - df [which（df $ inOut ==OUT），] a< - data.frame（table（cut（as.POSIXct（in_df $ time） ，break =15 mins）））b< - data.frame（table（cut（as.POSIXct（out_df $ time），breaks =15 mins））） colnames ）< - c（Time，Out） colnames（a）< - c（Time，In） outdf< - merge ，b，all = TRUE） outdf [is.na（outdf）]< - 0 > outdf 时间出来 1 2015-09-07 00:32:00 1 0 2 2015-09-07 00:47:00 0 0 3 2015-09 -07 01:02:00 0 1 4 2015-09-07 01:17:00 1 0 5 2015-09-07 01:32:00 0 0 6 2015- 09-07 01:47:00 2 0 7 2015-09-07 02:02:00 1 0 8 2015-09-07 02:17:00 0 0 9 2015 -09-07 02:32:00 0 0 10 2015-09-07 02:47:00 0 0 11 2015-09-07 03:02:00 0 0 12 2015-09-07 03:17:00 0 0 13 2015-09-07 03:32:00 1 0 14 2015-09-07 03:47:00 0 0 15 2015-09-07 04:02:00 1 2 我已经在此链接上提出过类似的问题 R使用data.table可以剪切包含2个或更多变量的修复时间间隔，而Frank为data.table提供了很好的解决方案，而且我想知道有没有人可以解决问题？如果它具有类似的强大的命令，就像Frank data.table sol ==> df [J（levels（timeCut）），as.list（table（inOut）），by = .EACHI] 对于dplyr，我已经尝试下面了，但是看起来下面将缺少0值（即，2015-09-07 00:47:00 0 0），我也想改变一个IN和OUT计数与我的预期结果相同（outdf），请注释，谢谢。 as.data.frame（df％> ％group_by（inOut，timeCut = cut（as.POSIXct（time），breaks =15 min））％>％summarize（n（））） inOut timeCut n（） 1 IN 2015-09-07 00:32:00 1 2 IN 2015-09-07 01:17:00 1 3 IN 2015-09-07 01:47:00 2 4 IN 2015-09-07 02:02:00 1 5 IN 2015-09-07 03:32:00 1 6 IN 2015-09-07 04:02:00 1 7 OUT 2015-09-07 01:02:00 1 8 OUT 2015-09-07 04:02:00 2 解决方案使用 dplyr 和 reshape2 ： 天秤座ry（dplyr）库（reshape2） my_levels data_frame（timeCut = levels（cut（as.POSIXct（df $ time），breaks =15 min ））） my_df< - df％>％ mutate（timeCut = cut（as.POSIXct（time），breaks =15 min ％>％ mutate_each（funs（as.character））％>％ right_join（。，my_levels）％>％ select（-time）％>％ dcast（timeCut〜inOut，length） 结果 timeCut IN OUT NA 1 2015-09-07 00:32:00 1 0 0 2 2015-09-07 00:47： 00 0 0 1 3 2015-09-07 01:02:00 0 1 0 4 2015-09-07 01:17:00 1 0 0 5 2015-09-07 01:32:00 0 0 1 6 2015-09-07 01:47:00 2 0 0 7 2015-09-07 02:02:00 1 0 0 8 2015 -09-07 02:17:00 0 0 1 9 2015-09-07 02:32:00 0 0 1 10 2015-09-07 02:47:00 0 0 1 11 2015-09-07 03:02:00 0 0 1 12 2015-09-07 03:17:00 0 0 1 13 2015-09-07 03:32:00 10 0 14 2015-09-07 03:47:00 0 0 1 15 2015-09-07 04:02:00 1 2 0 / pre> I have a dataframedf <- data.frame(time = c("2015-09-07 00:32:19", "2015-09-07 01:02:30", "2015-09-07 01:31:36", "2015-09-07 01:47:45","2015-09-07 02:00:17", "2015-09-07 02:07:30", "2015-09-07 03:39:41", "2015-09-07 04:04:21", "2015-09-07 04:04:21", "2015-09-07 04:04:22"), inOut = c("IN", "OUT", "IN", "IN", "IN", "IN", "IN", "OUT", "IN", "OUT")) > df time inOut1 2015-09-07 00:32:19 IN2 2015-09-07 01:02:30 OUT3 2015-09-07 01:31:36 IN4 2015-09-07 01:47:45 IN5 2015-09-07 02:00:17 IN6 2015-09-07 02:07:30 IN7 2015-09-07 03:39:41 IN8 2015-09-07 04:04:21 OUT9 2015-09-07 04:04:21 IN10 2015-09-07 04:04:22 OUT> I want to calculate the number of counts for IN/OUT per 15 mins, I can do this by creating another in_df, out_df, cut these dataframe per 15 mins, and then merge this together to obtain my result. The outdf is my expected result.in_df <- df[which(df$inOut== "IN"),]out_df <- df[which(df$inOut== "OUT"),]a <- data.frame(table(cut(as.POSIXct(in_df$time), breaks="15 mins")))b <- data.frame(table(cut(as.POSIXct(out_df$time), breaks="15 mins")))colnames(b) <- c("Time", "Out")colnames(a) <- c("Time", "In")outdf <- merge(a,b, all=TRUE)outdf[is.na(outdf)] <- 0> outdf Time In Out1 2015-09-07 00:32:00 1 02 2015-09-07 00:47:00 0 03 2015-09-07 01:02:00 0 14 2015-09-07 01:17:00 1 05 2015-09-07 01:32:00 0 06 2015-09-07 01:47:00 2 07 2015-09-07 02:02:00 1 08 2015-09-07 02:17:00 0 09 2015-09-07 02:32:00 0 010 2015-09-07 02:47:00 0 011 2015-09-07 03:02:00 0 012 2015-09-07 03:17:00 0 013 2015-09-07 03:32:00 1 014 2015-09-07 03:47:00 0 015 2015-09-07 04:02:00 1 2I have asked similar questions on this link R using data.table to cut fix time interval that contain 2 or more variables and Frank have provide good sol for data.table, and I wonder if someone have sol for dplyr? And if it have similar powerful command just like Frank data.table sol ==> df[J(levels(timeCut)), as.list(table(inOut)), by=.EACHI]And for dplyr, I have try below, but it seem below will missing 0 value(ie, 2015-09-07 00:47:00 0 0), also I want to mutate a IN and OUT count that same as my expect result(outdf), please comment, Thanks.as.data.frame(df %>% group_by(inOut, timeCut= cut(as.POSIXct(time), breaks="15 min")) %>% summarise(n())) inOut timeCut n()1 IN 2015-09-07 00:32:00 12 IN 2015-09-07 01:17:00 13 IN 2015-09-07 01:47:00 24 IN 2015-09-07 02:02:00 15 IN 2015-09-07 03:32:00 16 IN 2015-09-07 04:02:00 17 OUT 2015-09-07 01:02:00 18 OUT 2015-09-07 04:02:00 2 解决方案 Another solution using dplyr and reshape2:library(dplyr)library(reshape2)my_levels <- data_frame(timeCut = levels(cut(as.POSIXct(df$time), breaks="15 min")))my_df <- df %>% mutate(timeCut = cut(as.POSIXct(time), breaks = "15 min")) %>% mutate_each(funs(as.character)) %>% right_join(., my_levels) %>% select(-time) %>% dcast(timeCut ~ inOut, length)Result timeCut IN OUT NA1 2015-09-07 00:32:00 1 0 02 2015-09-07 00:47:00 0 0 13 2015-09-07 01:02:00 0 1 04 2015-09-07 01:17:00 1 0 05 2015-09-07 01:32:00 0 0 16 2015-09-07 01:47:00 2 0 07 2015-09-07 02:02:00 1 0 08 2015-09-07 02:17:00 0 0 19 2015-09-07 02:32:00 0 0 110 2015-09-07 02:47:00 0 0 111 2015-09-07 03:02:00 0 0 112 2015-09-07 03:17:00 0 0 113 2015-09-07 03:32:00 1 0 014 2015-09-07 03:47:00 0 0 115 2015-09-07 04:02:00 1 2 0 这篇关于R使用dplyr来剪切包含2个或更多变量的修复时间间隔的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 11-02 20:13