time

问题 I'm trying to collect information from a system and I need to get the current time in hours and minutes. Currently I have: date | awk '{print $4}' which outputs something like: 16:18:54 How do I chop off the seconds? 回答1: Provide a format string: date +"%H:%M" Running man date will give all the format options %a locale's abbreviated weekday name (e.g., Sun) %A locale's full weekday name (e.g., Sunday) %b locale's abbreviated month name (e.g., Jan) %B locale's full month name (e.g., January)

问题 I'm trying to collect information from a system and I need to get the current time in hours and minutes. Currently I have: date | awk '{print $4}' which outputs something like: 16:18:54 How do I chop off the seconds? 回答1: Provide a format string: date +"%H:%M" Running man date will give all the format options %a locale's abbreviated weekday name (e.g., Sun) %A locale's full weekday name (e.g., Sunday) %b locale's abbreviated month name (e.g., Jan) %B locale's full month name (e.g., January)

问题 I'm currently working on a project for school in which I have a time complexity restriction. I'm still beginning to learn to program so apologies of this question is stupid. Essentially, let's say I have a constraint that my program must satisfy the time complexity O(n 2 + m log m) , where n and m are some sizes of input. Does that mean that's the maximum time complexity any specific component of my program can run in? That is, if I have a ton of, say, O(n 2 + m) functions, but the most

问题 I have survival data in this format, with a time-varying exposure to Intervention: ID start stop status Intervention 1 2 14 0 0 2 2 5 0 0 3 2 3 0 0 3 3 10 1 1 4 5 8 0 0 5 6 10 0 0 For example, for patient ID #3: from day 2 to day 3, the patient has not yet received the intervention (Intervention = 0), but starting on day 3 and lasting until day 10 (when the patient dies), the patient has received the intervention (Intervention = 1). I thought that I could then estimate the time-varying effect

问题 The bounty expires in 5 days . Answers to this question are eligible for a +50 reputation bounty. stats555 is looking for a canonical answer : I am using the R programming language. I am trying to use the "plotly" library to make interactive time series plots as shown in this tutorial over here : plotly.com/r/custom-buttons and plotly.com/r/cumulative-animations . I generated some artificial time series data that matches the structure of the data shown in the tutorial - now I am trying to

问题 I'm trying to run a script I wrote on my Raspberry Pi Zero, but I keep getting the error OverflowError: timestamp out of range for platform time_t . I'm relatively certain it's something with the 32-bit ARM architecture of the pi, but I can't seem to figure out a workaround. Here's the traceback: File "twitter.py", line 37, in <module> t.run.Search(c) File "/home/pi/.local/lib/python3.7/site-packages/twint/run.py", line 288, in Search run(config, callback) File "/home/pi/.local/lib/python3.7

问题 This question already has answers here : How to get time difference in minutes in PHP (17 answers) Closed 18 days ago . Well i have a time in UNIX format. i need to find the difference between a specific time and current time. if difference is 5 minutes. do something. $sp_time = 1400325952;//Specific time $currentTime = //current time in UNIX format $difference = $currentTime - $sp_time; if($difference >= 5)//if difference is less than or equal to 5minutes { //DO SOME STUFF } EDIT : How to

问题 | Month | day | hour | Temperature | |-----------|-----|------|-------------| | September | 01 | 0:00 | 19,11 | | September | 01 | 1:00 | 18,67 | | September | 01 | 2:00 | 18,22 | | September | 01 | 3:00 | 17,77 | convert to: | Month | day | hour | Temperature | |-----------|-----|------|-------------| | September | 01 | 0:00 | T = 19,11 | | September | 01 | 0:15 | T2 = T + (18,67 - 19,11)/ 4 | | September | 01 | 0:30 | T3 = T2 + (18,67 - 19,11)/4 | | September | 01 | 0:45 | T4 = T3 + (18,67

问题 | Month | day | hour | Temperature | |-----------|-----|------|-------------| | September | 01 | 0:00 | 19,11 | | September | 01 | 1:00 | 18,67 | | September | 01 | 2:00 | 18,22 | | September | 01 | 3:00 | 17,77 | convert to: | Month | day | hour | Temperature | |-----------|-----|------|-------------| | September | 01 | 0:00 | T = 19,11 | | September | 01 | 0:15 | T2 = T + (18,67 - 19,11)/ 4 | | September | 01 | 0:30 | T3 = T2 + (18,67 - 19,11)/4 | | September | 01 | 0:45 | T4 = T3 + (18,67

问题 I need to check if a time is between a start and end time. All times are on the same day, so date is not important. I'm able to compare hours, but I'm unsure how to add in minutes to the start and end times. var thedate = new Date(); var dayofweek = thedate.getUTCDay(); var hourofday = thedate.getUTCHours(); var minutesofday = date.getMinutes(); function inTime() { if (dayofweek != 0 && dayofweek != 7 && (hourofday > 13 && hourofday < 20)) { return true; } return false; } If I want to check