intervals

Summing up values within an interval from another data.frame in R

扶醉桌前 提交于 2019-12-06 15:17:52
I have thousands of entries in hkdata.2 and I want to create a loop that can help me to sum up the total exposed mxtemp from another data frame data.1 for each member in each houseID in data.2 Could any expert give me a hand on this? weather.data date mpressure mxtemp 1 2008-01-01 1025.3 15.7 2 2008-01-02 1025.6 16.0 3 2008-01-03 1023.6 18.1 4 2008-01-04 1021.8 18.4 5 2008-01-05 1020.1 20.9 6 2008-01-06 1019.7 20.7 7 2008-01-07 1018.4 24.0 8 2008-01-08 1016.7 23.7 hkdata.2 row.names houseID member male date.end date.begin 1 1 1 1 2008-01-07 2008-01-02 2 1 2 0 2008-01-06 2008-01-04 I want to

Append div with interval in Jquery

梦想的初衷 提交于 2019-12-06 14:05:33
问题 Hi I need help regarding appending div while using 2 types of speed interval via 2 loops Here is my sample code <script type="text/javascript"> $(document).ready(function() { for (var i = 0; i <= 300; i++) { $(".wrapper").append("<div class='item' id='" + i + "'>" + i + "</div>"); if (i==300) { //I need this for loop to slow down my //interval so div will display slower compared to the first 300 for (var i = 300; i <= 500; i++) { $(".wrapper").append("<div class='item' id='" + i + "'>" + i +

Qt seconds to DD HH SS

孤人 提交于 2019-12-06 07:27:51
using Qt 4.8 how can I print the time in the format DD HH SS? I have the seconds and I want to get back a string in that format. cor3ntin QDateTime::fromTime_t(seconds).toString("ss hh DD"); see http://qt-project.org/doc/qt-5.0/qdatetime.html#toString If you want a duration ( your question was really unclear) try something like : QString seconds_to_DHMS(quint32 duration) { QString res; int seconds = (int) (duration % 60); duration /= 60; int minutes = (int) (duration % 60); duration /= 60; int hours = (int) (duration % 24); int days = (int) (duration / 24); if((hours == 0)&&(days == 0)) return

How do run a Unix command at a given time interval?

大兔子大兔子 提交于 2019-12-06 06:21:52
问题 I want to run a Unix command (e.g. ls ) at 5 minute intervals through a script. Explanation: I have a Unix script. In that script I have a command called "ls". I want that "ls" command to run every 5 minutes from that script. 回答1: Use watch . The -n flag specifies interval in seconds, so watch -n 300 ls 回答2: while true; do ls sleep 300 done ? 回答3: Put your script into the Crontab via crontab -e More information about Cron you can find at wikipedia 回答4: You could use crontab for example. See

How to find if a date fits an interval in either PHP or MySQL?

冷暖自知 提交于 2019-12-06 04:06:41
Let's say I have a datetime, June 16 2011 at 7:00. I want to be able to check at, say, August 5 2011 at 7:00 and be able to tell that it is exactly a multiple of 1 day since the first date, whereas 7:01 would not count, since it is not an exact multiple. Another test set: Let's say we have June 16 2011 at 7:00, and I want to check if a particular minute is within an interval of exactly 2 hours since then. So 9:00, 11:00, 13:00, etc. would count, but 9:30 and 10:00 would not. And this could continue for days and months - September 1 at 7:00 would still count as within every 2 hours. (And no, at

using intervals to assign categorical values

橙三吉。 提交于 2019-12-06 04:05:33
问题 Take the following generic data A <- c(5,7,11,10,23,30,24,6) B <- c(1,2,3,1,2,3,1,2) C <- data.frame(A,B) and the following intervals library(intervals) interval1 <- Intervals( matrix( c( 5, 15, 15, 25, 25, 35, 35, 100 ), ncol = 2, byrow = TRUE ), closed = c( TRUE, FALSE ), type = "Z" ) rownames(interval1) <- c("A","B","C", "D") interval2 <- Intervals( matrix( c( 0, 10, 12, 20, 22, 30, 30, 100 ), ncol = 2, byrow = TRUE ), closed = c( TRUE, FALSE ), type = "Z" ) rownames(interval2) <- c("P","Q

Java check if number in interval [duplicate]

◇◆丶佛笑我妖孽 提交于 2019-12-06 03:58:58
问题 This question already has answers here : Closed 7 years ago . Possible Duplicate: Does an open-ended interval implementation exist for Java? i have an int variable and i'd like to check if it's value is in an interval [a,b]. I know it's a simple matter of using x>=a and x<=b or implementing a simple method which can do this, but i'd like to know if there is something already done. Searched in Math class, but i wasn't able to find one. It's not that important and not that big of an issue, but

python time interval algorithm sum

自作多情 提交于 2019-12-06 03:49:54
问题 Assume I have 2 time intervals,such as 16:30 - 20:00 AND 15:00 - 19:00, I need to find the total time between these two intervals so the result is 5 hours (I add both intervals and subtract the intersecting interval), how can I write a generic function which also deals with all cases such as one interval inside other(so the result is the interval of the bigger one), no intersection (so the result is the sum of both intervals). My incoming data structure is primitive, simply string like "15:30

Create monthly mean by time intervals

烂漫一生 提交于 2019-12-05 21:54:19
Sorry if this has already been posted but I looked really hard and could not find anything. I am working with monthly temperature observations for 30 years, comprising January 1960 to December 1989. It looks like this: > head(df) date temp 1 1960-01-01 22.92235 2 1960-02-01 23.07059 3 1960-03-01 23.10941 4 1960-04-01 20.78353 5 1960-05-01 17.45176 6 1960-06-01 17.31765 First, what I need to do is to average all januaries, februaries, marches and etc for the whole period. Then, I would like to do the same for specific periods of time (3 years, 5 years, 10 years etc). For example, The average of

Finding overlapping intervals when overlaps are rare

£可爱£侵袭症+ 提交于 2019-12-05 21:12:06
I have a huge database table with n integer intervals (for instance {1-5}, {4-16}, {6434-114343}) and need to find out which intervals overlap each other . There's a wealth of similar questions on SO , but the difference is I need returned, for each interval respectively, the set of overlapping intervals. ------------------ A ------------------- ------ B ------- ----- D ----- --------- C --------- For this example, output would be A:{B,C,D} B:{A,C} C:{A,B} D:{A} Worst case, all intervals could overlap each other, producing an output of size O( n 2 ). This is no better than the naïve solution