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IE9在开始从Amazon S3流式传输视频之前等待7-10秒(IE9 Waiting 7-10 seconds before starting to stream a video from Amazon S3)
IE9在开始从Amazon S3流式传输视频之前等待7-10秒(IE9 Waiting 7-10 seconds before starting to stream a video from Amazon S3)
http://screencast.com/t/NKb9erVyd0
我正在使用Projekktor HTML5视频播放器。 除了Internet Explorer之外,视频会立即在所有浏览器中播放。 在开发人员工具中,IE显示了巨大的等待时间 这个等待时间是什么?为什么IE感觉需要浪费10秒的用户生命?
http://screencast.com/t/NKb9erVyd0
I'm using the Projekktor HTML5 video player. The video starts playing immediately in all browsers except for internet explorer. In developer tools IE shows a huge Wait time. What is this wait time and why does IE feel the need to waste 10 seconds of my user's life?
原文:https://stackoverflow.com/questions/10239882
更新时间:2023-07-15 18:07
最满意答案
我个人会使用熔化/铸造整形策略
library(reshape2) melted <- rbind(melt(df1), melt(df2)) dcast(melted, Type~variable, sum) # Type CA AR OR Total # 1 alpha 5 7 1 13 # 2 beta 3 11 2 16 # 3 delta 12 2 1 15 # 4 gamma 9 1 0 10
I personally would use a melt/cast reshaping strategy
library(reshape2) melted <- rbind(melt(df1), melt(df2)) dcast(melted, Type~variable, sum) # Type CA AR OR Total # 1 alpha 5 7 1 13 # 2 beta 3 11 2 16 # 3 delta 12 2 1 15 # 4 gamma 9 1 0 10
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我个人会使用熔化/铸造整形策略 library(reshape2) melted <- rbind(melt(df1), melt(df2)) dcast(melted, Type~variable, sum) # Type CA AR OR Total # 1 alpha 5 7 1 13 # 2 beta 3 11 2 16 # 3 delta 12 2 1 15 # 4 gamma 9 1 0 10 I personally would use ...
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首先,你的机场经度是积极的,因为它们应该是负面的,这会导致结果。 让我们修复它们,这样结果更有意义: airports$long <- -airports$long 现在,您可以使用apply来评估每个机场的所有飞行员。 geosphere包有几个计算直线距离的函数,包括distGeo和distHaversine 。 library(geosphere) pilots$closest_airport <- apply(pilots[, 3:2], 1, function(x){ airports ...
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如果我们需要更快的选项,可以使用data.table连接以及将NA值( := )分配到0。 library(data.table) setDT(df2)[df1, on = "term"][is.na(freq), freq := 0][] 或者为了避免副本,正如@Arun所提到的,在'df1'中创建一个'freq'列,然后on 'term' on加入'freq'替换相应的'i.freq'值。 setDT(df1)[, freq := 0][df2, freq := i.freq, on = "term" ...
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可能你可以尝试重命名列名称,如下所示: names(XY)<- c("x","yz") names(XZ)<- c("x","yz") new <- rbind(XY,XZ) > new x yz 1 x1 y1 2 x2 y2 3 x3 y3 4 x4 y4 5 x5 y5 6 x1 z1 7 x2 z2 8 x3 z3 9 x4 z4 10 x5 z5 我希望它有帮助。 我相信为了使用rbind你需要有相同的列名。 May be you could try renaming ...
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这个怎么样? merge(df.1,df.2,by="V1",all=TRUE) V1 V2.x V2.y 1 1 0.9169835 0.9658816 2 2 1.0327111 1.0618083 3 3 0.8368222 NA 4 4 NA 1.2700018 5 5 1.0061137 1.0186826 6 6 1.0086698 1.1848159 7 7 1.0362074 NA 8 8 ...
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请尝试以下方法: D2[!is.na(D2)] <- D1[!is.na(D2)] D2 ID I1 I2 I3 1 001
2 2 002 3 5 3 003 9 Try the following: D2[!is.na(D2)] <- D1[!is.na(D2)] D2 ID I1 I2 I3 1 001 2 2 002 3 5 3 003 findInterval函数很好地完成了这项工作。 set.seed(1) df1 <- data.frame(val = runif(10)) df1$Category <- findInterval(df1$val, vec = c(0,0.2, 0.4, 0.6, 0.8), left.open = TRUE) df1 # val Category # 1 0.26550866 2 # 2 0.37212390 2 # 3 0.57285336 ...我们可以使用rbindlist的data.table 。 我们将data.frames保存在list并使用rbindlist和fill=TRUE选项为数据集中缺失的列生成NA。 library(data.table) rbindlist(list(a,b), fill=TRUE) # A B C D #1: 3 9 10 6 #2: 1 2 NA 5 另一个选项是来自dplyr ,它是do.call(rbind, ...)的有效实现,用于将许多数据帧绑定在一起。 library(dplyr) bin ...以下是我在na.locf文档的帮助下如何做到这一点。 有帮助吗? dat<- dget("yoursample") require(xts) datxts<- as.xts(dat[,-1],order.by = dat$date,frequency = 24) tzn<-tzone(datxts) g<- seq(start(datxts), end(datxts), "hour") gxts<- xts(rep(NA,length(g)),order.by = as.POSIXct(g), tzone ...您可以使用match来标识与d1每一行对应的d2行: d1$x1_rat <- d1$x1 / d2$rat[match(d1$date, d2$date)] d1 # x1 date x1_rat # 1 0 3652 0.000000 # 2 10 3652 8.333333 # 3 10 3653 7.692308 # 4 20 3654 20.000000 # 5 2 3655 2.000000 # 6 2 3656 1.666667 # 7 6 3657 4.6153 ...相关文章
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