How to use cal command to add the calendar of next July to the end of the file, for example, myfile, and what day of the week the upcoming Canada Day fall on?
So far I just have this command:
cal July 2017 >> myfile
I feel like I am not doing it correct and I don't know which command to use, to find the day of the week for specific date.
Use this command:
cal 7 2017 >> file
The output is:
July 2017
Su Mo Tu We Th Fr Sa
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 28 29
30 31
You can find out day of week of a particular day with the GNU date command:
date -d"2017-07-01" # what day of week is Canada Day this year?
=> Sat Jul 1 00:00:00 UTC 2017
If you just want the week day, then:
date -d"2017-07-01" +%A
=> Saturday
You can check more about these commands with man cal or man date.
On a Mac, you could do this:
date -j -vJulm -v1d -v2017y +%A
See more on this post: date command on Mac OS
Related
As we know that each year have the following max day in each month as follows:
Jan - 31 days
Feb - 28 days / 29 days (leap year)
Mar - 31 days
Apr - 30 days
May - 31 days
Jun - 30 days
Jul - 31 days
Aug - 31 days
Sep - 30 days
Oct - 31 days
Nov - 30 days
Dec - 31 days
How to I get bash to return the value (last day of each month) for the current year without using if else or switch or while loop?
my take:
for m in {1..12}; do
date -d "$m/1 + 1 month - 1 day" "+%b - %d days";
done
To explain: for the first iteration when m=1 the -d argument is "1/1 + 1 month - 1 day" and "1/1" is interpreted as Jan 1st. So Jan 1 + 1 month - 1 day is Jan 31. Next iteration "2/1" is Feb 1st, add a month subtract a day to get Feb 28 or 29. And so on.
cat <<EOF
Jan - 31 days
Feb - `date -d "yesterday 3/1" +"%d"` days
Mar - 31 days
Apr - 30 days
May - 31 days
Jun - 30 days
Jul - 31 days
Aug - 31 days
Sep - 30 days
Oct - 31 days
Nov - 30 days
Dec - 31 days
EOF
cal $(date +"%m %Y") |
awk 'NF {DAYS = $NF}; END {print DAYS}'
This uses the standard cal utility to display the specified month, then runs a simple Awk script to pull out just the last day's number.
Assuming you allow "for", then the following in bash
for m in {1..12}; do
echo $(date -d $m/1/1 +%b) - $(date -d "$(($m%12+1))/1 - 1 days" +%d) days
done
produces this
Jan - 31 days
Feb - 29 days
Mar - 31 days
Apr - 30 days
May - 31 days
Jun - 30 days
Jul - 31 days
Aug - 31 days
Sep - 30 days
Oct - 31 days
Nov - 30 days
Dec - 31 days
Note: I removed the need for cal
For those that enjoy trivia:
Number months from 1 to 12 and look at the binary representation in four
bits {b3,b2,b1,b0}. A month has 31 days if and only if b3 differs from b0.
All other months have 30 days except for February.
So with the exception of February this works:
for m in {1..12}; do
echo $(date -d $m/1/1 +%b) - $((30+($m>>3^$m&1))) days
done
Result:
Jan - 31 days
Feb - 30 days (wrong)
Mar - 31 days
Apr - 30 days
May - 31 days
Jun - 30 days
Jul - 31 days
Aug - 31 days
Sep - 30 days
Oct - 31 days
Nov - 30 days
Dec - 31 days
Try using this code
date -d "-$(date +%d) days month" +%Y-%m-%d
Returns the number of days in the month compensating for February changes in leap years without looping or using an if statement
This code tests date to see if Feb 29th of the requested year is valid, if so then it updates the second character in the day offset string. The month argument selects the respective substring and adds the month difference to 28.
function daysin()
{
s="303232332323" # normal year
((!($2%4)&&($2%100||!($2%400)))) && s=313232332323 # leap year
echo $[ ${s:$[$1-1]:1} + 28 ]
}
daysin $1 $2 #daysin [1-12] [YYYY]
On a Mac which features BSD date you can just do:
for i in {2..12}; do date -v1d -v"$i"m -v-1d "+%d"; done
Quick Explanation
-v stands for adjust. We are adjusting the date to:
-v1d stands for first day of the month
-v"$i"m defined the month e.g. (-v2m for Feb)
-v-1d minus one day (so we're getting the last day of the previous month)
"+%d" print the day of the month
for i in {2..12}; do date -v1d -v"$i"m -v-1d "+%d"; done
31
28
31
30
31
30
31
31
30
31
30
You can add year of course. See examples in the manpage (link above).
Contents of script.sh:
#!/bin/bash
begin="-$(date +'%-m') + 2"
end="10+$begin"
for ((i=$begin; i<=$end; i++)); do
echo $(date -d "$i month -$(date +%d) days" | awk '{ printf "%s - %s days", $2, $3 }')
done
Results:
Jan - 31 days
Feb - 29 days
Mar - 31 days
Apr - 30 days
May - 31 days
Jun - 30 days
Jul - 31 days
Aug - 31 days
Sep - 30 days
Oct - 31 days
Nov - 30 days
for m in $(seq 1 12); do cal $(date +"$m %Y") | grep -v "^$" |tail -1|grep -o "..$"; done
iterate from 1 to 12 (for...)
print calendar table for each month (cal...)
remove empty lines from output (grep -v...)
print last number in the table (tail...)
There is no sense, to avoid using cal, because it is required by POSIX, so should be there
A variation for the accepted answer to show the use of "yesterday"
$ for m in {1..12}; do date -d "yesterday $m/1 + 1 month" "+%b - %d days"; done
Jan - 31 days
Feb - 28 days
Mar - 31 days
Apr - 30 days
May - 31 days
Jun - 30 days
Jul - 31 days
Aug - 31 days
Sep - 30 days
Oct - 31 days
Nov - 30 days
Dec - 31 days
How it works?
Show the date of yesterday for the date "month/1" after adding 1 month
I needed this few times, so when in PHP comes with easy in bash is not,
so I used this till throw me error "invalid arithemtic operator"
and even with warrings in spellcheck ( "mt" stands for month, "yr" for year )
last=$(echo $(cal ${mt} ${yr}) | awk '{print $NF}')
so this works fine...
### get last day of month
#
# implement from PHP
# src: https://www.php.net/manual/en/function.cal-days-in-month.php
#
if [ $mt -eq 2 ];then
if [[ $(bc <<< "${yr} % 4") -gt 0 ]];then
last=28
else
if [[ $(bc <<< "${yr} % 100") -gt 0 ]];then
last=29
else
[[ $(bc <<< "${yr} % 400") -gt 0 ]] && last=28 || last=29
fi
fi
else
[[ $(bc <<< "(${mt}-1) % 7 % 2") -gt 0 ]] && last=30 || last=31
fi
Building on patm's answer using BSD date for macOS (patm's answer left out December):
for i in {1..12}; do date -v1m -v1d -v+"$i"m -v-1d "+%b - %d days"; done
Explanation:
-v, when using BSD date, means adjust date to:
-v1m means go to first month (January of current year).
-v1d means go to first day (so now we are in January 1).
-v+"$i"m means go to next month.
-v-1d means subtract one day. This gets the last day of the previous month.
"+%b - %d days" is whatever format you want the output to be in.
This will output all the months of the current year and the number of days in each month. The output below is for the as-of-now current year 2022:
Jan - 31 days
Feb - 28 days
Mar - 31 days
Apr - 30 days
May - 31 days
Jun - 30 days
Jul - 31 days
Aug - 31 days
Sep - 30 days
Oct - 31 days
Nov - 30 days
Dec - 31 days
System: WIN10
IDE: MS Visual Studio COde
Language: Python version 3.7.3
Library: pandas version 1.0.1
Data source: supplied in the example below
Dataset: supplied in the example below
Ask:
I need to split the date and time string out of a column from a data frame that has rows of uneven delimiters i.e. some with three and some with four commas.
I am trying to figure out how to strip the date and time values: 'Nov 11 2013 12:00AM', and 'Apr 11 2013 12:00AM' respectively off the back of these two records in one column into a new column given the second row in the example below has fewer commas.
Code:
df['sample field'].head(2)
4457-I need, this, date, Nov 11 2013 12:00AM ,
2359-I need this, date, Apr 11 2013 12:00AM ,
While the below method expands the data into different columns and staggers which column houses the date, this does not work. I need the date and time (or even just date) information in one column so that I can use the date values in further analysis (for example time-series).
Code:
df['sample field'].str.split(",", expand=True)
Data
df=pd.DataFrame({'Text':['4457-I need, this, date, Nov 11 2013 12:00AM ,','2359-I need this, date, Apr 11 2013 12:00AM ,']})
df
Use df.extract with a regex epression
df['Date']= df.Text.str.extract('([A-Za-z]+\s+\d+\s+\d+\s+\d+:[0-9A-Z]+(?=\s+\,+))')
df
#df.Date=pd.to_datetime(df.Date).dt.strftime('%b %d %Y %H:%M%p')
#df['date'] = pd.to_datetime(df['date'] ,format='%b %d %Y %H:%M%p')
df['Date']=pd.to_datetime(df['Date'])#This or even df['Date']=pd.to_datetime(df['Date'], format=('%b %d %Y %I:%M%p')) could work. Just remmeber because your time is 12AM use 12 clock hour system %I not %H and also hour 00.00 likely to be trncated, If have say11.00AM, the time will appear
IIUC you need str.extract with a regular expression.
Regex Demo Here
print(df)
0
0 4457-I need, this, date, Nov 11 2013 12:00AM
1 2359-I need this, date, Apr 11 2013 12:00AM
df['date'] = df[0].str.extract('(\w{3}\s\d.*\d{4}\s\d{2}:\d{2}\w{2})')
df['date'] = pd.to_datetime(df['date'] ,format='%b %d %Y %H:%M%p')
print(df)
0 date
0 4457-I need, this, date, Nov 11 2013 12:00AM 2013-11-11 12:00:00
1 2359-I need this, date, Apr 11 2013 12:00AM 2013-04-11 12:00:00
I'll use #wwnde's data :
df=pd.DataFrame({'Text':['4457-I need, this, date, Nov 11 2013 12:00AM ,','2359-I need this, date, Apr 11 2013 12:00AM ,']})
df['Date'] = df.Text.str.strip(',').str.split(',').str[-1].str.strip()
df['Date_formatted'] = pd.to_datetime(df.Date, format = '%b %d %Y %H:%M%p')
Text Date Date_formatted
0 4457-I need, this, date, Nov 11 2013 12:00AM , Nov 11 2013 12:00AM 2013-11-11 12:00:00
1 2359-I need this, date, Apr 11 2013 12:00AM , Apr 11 2013 12:00AM 2013-04-11 12:00:00
I have a simple inventory table in excel that looks like this:
Number of Items | Date Incoming | Date Out
-------------------------------------------------------
10 | 1 Jan 2018 | 30 Jan 2018
30 | 15 Jan 2018 | 1 May 2018
20 ! 1 Feb 2018 | 15 Mar 2018
I would like something that can give me the the total number of items that are present in the inventory at each date, that is:
1 Jan 2018 | 10
15 Jan 2018 | 40
30 Jan 2018 | 30
1 Feb 2018 | 50
15 Mar 2018 | 30
1 May 2018 | 0
What I was thing is some sort of cumulative sum where the number of items are added at "Date Incoming" and substracted at "Date Out".
Can you help me? I would prefer to avoid macros but even a vba solution if fine.
For a given date, you can do:
=sumif(#DateIn, "<="&#CellWithGivenDate, #NumberOfItems) - sumif(#DateOut, "<="&#CellWithGivenDate, #NumberOfItems)
With #NumberOfItems, #DateIn, and #DateOut being columns 1 to 3 of your sample, and #CellWithGivenDate being the relevant cell in column 1 of your expected result sample.
I'm trying to understand length of an outage that occurs within time range.
In once cell I have the support period: Level 1 Monday-Saturday 0800-1800
In two other cells I have the start time and end time of the outage
Example:
Outage Start: Wednesday, 24 August 2016 16:47
Outage End: Monday, 29 August 2016 10:15
Result should return: 33:28 hours
Workings:
Wednesday, 24 August 2016 16:47 - 18:00 = + 01:13
Thursday , 25 August 2016 08:00 - 18:00 = + 10:00
Friday , 26 August 2016 08:00 - 18:00 = + 10:00
Saturday , 27 August 2016 08:00 - 18:00 = + 10:00
Monday , 29 August 2016 08:00 - 10:15 = + 02:15
Any advice appreciated!
Updated
Please check the above image, I hope you want something similar to this.
Enter the following formula in the column C2
=INT(IF(DAYS(B2,A2)=0,IF(TEXT(A2,"ddd")="Sun",0,ROUND((B2-A2)/(1/24/60),0)
),IF(DAYS(B2,A2)=1,IF(TEXT(A2,"ddd")="Sun",0,(DATE(YEAR(A2),MONTH(A2),DAY(A2))+1/24*18)-A2
)/(1/24/60) +
IF(TEXT(B2,"ddd")="Sun",0,B2-(DATE(YEAR(B2),MONTH(B2),DAY(B2))+1/24*8)
)/(1/24/60),IF(TEXT(A2,"ddd")="Sun",0,(DATE(YEAR(A2),MONTH(A2),DAY(A2))+1/24*18)-A2
)/(1/24/60) +
IF(TEXT(B2,"ddd")="Sun",0,B2-(DATE(YEAR(B2),MONTH(B2),DAY(B2))+1/24*8)
)/(1/24/60) +
(IF(DAYS((DATE(YEAR(B2-1),MONTH(B2-1),DAY(B2-1))),A2)<0,0,DAYS((DATE(YEAR(B2-1),MONTH(B2-1),DAY(B2-1))),A2))-
INT(((DATE(YEAR(B2-1),MONTH(B2-1),DAY(B2-1)))-(DATE(YEAR(A2),MONTH(A2),DAY(A2)) + MOD(DATE(YEAR(A2),MONTH(A2),DAY(A2)),7)))/7+1)) * 10 * 60
)
))
Enter the following formula in the cell D2
=INT(C2/60) &":" & INT(MOD(C2,60)+0.5)
Then copy the C2 and D2 to all the cells that you want, it will give the outage value that you are looking for.
For example if cell A2 contains the Outage Start date Wednesday, 24 August 2016 16:47 and cell B2 contains the Outage End date Monday, 29 August 2016 10:15, then you can use the following:
=DAY(B2-A2)
This will return 4 days.
=HOUR(B2-A2)
Will return 17 hours
=MINUTE(B2-A2)
Will return 28 minutes.
Then you could use the following:
="The outage lasted for "& DAY(B2-A2) & " days, " & HOUR(B2-A2) & " hours, and " & MINUTE(B2-A2) & " minutes."
To produce this:
The outage lasted for 4 days, 17 hours, and 28 minutes.
How to get next year period based on current month and year, for example:
Jan 2014 - Dec 2014
Feb 2014 - Jan 2015
Mar 2014 - Feb 2015
Apr 2014 - Mar 2015
May 2014 - Apr 2015
Jun 2014 - May 2015
Jul 2014 - Jun 2015
Aug 2014 - Jul 2015
Sep 2014 - Aug 2015
Oct 2014 - Sep 2015
Nov 2014 - Oct 2015
Dec 2014 - Nov 2015
Next period
Jan 2015 - Dec 2015
Feb 2015 - Jan 2016
etc.
I have tried with the following formula:
=UPPER(TEXT(NOW();"MMM")) &" "& TEXT(NOW();"YY")-1
It works fine for Jan 2014 but can't figure out how to get Dec 2014; Feb 2014 - Jan 2015 and so on?
I think you need the EOMonth formula.
=EOMONTH(NOW(),-13) +1 and =EOMONTH(NOW(),-2) +1 should give give you JAN 2014 to DEC 2014
from the MS Excel documentation
Microsoft Excel stores dates as sequential serial numbers so they can
be used in calculations. By default, January 1, 1900 is serial number
1, and January 1, 2008 is serial number 39448 because it is 39,448
days after January 1, 1900.
To get the text formatting you are after, I would suggest that you stick with formatting the cell/column as #Makyen has suggested. Having said that this is the formula that you can use to format the text.
=UPPER(TEXT(EOMONTH(NOW(),-13) +1, "MMM YY"))
Assuming that the date (as a date serial number) for which you desire to find the year period is in cell A1, the following should provide the next year period starting from that day:
=EOMONTH(A1,11) +DAY(A1) -1
Examples:
Input Output
1/18/2014 1/17/2015
2/18/2014 2/17/2015
3/18/2014 3/17/2015
4/18/2014 4/17/2015
5/18/2014 5/17/2015
6/18/2014 6/17/2015
7/18/2014 7/17/2015
8/18/2014 8/17/2015
9/18/2014 9/17/2015
10/18/2014 10/17/2015
11/18/2014 11/17/2015
12/18/2014 12/17/2015
1/18/2015 1/17/2016
2/18/2015 2/17/2016
3/18/2015 3/17/2016
4/18/2015 4/17/2016
5/18/2015 5/17/2016
6/18/2015 6/17/2016
7/18/2015 7/17/2016
8/18/2015 8/17/2016
9/18/2015 9/17/2016
10/18/2015 10/17/2016
11/18/2015 11/17/2016
12/18/2015 12/17/2016
1/18/2016 1/17/2017
If you want the year period to start from the current day:
=EOMONTH(NOW(),11) + DAY(NOW()) -1
If you want the year period to start from the first day of the current month:
=EOMONTH(EOMONTH(NOW(),-1) + 1,11)
or
=EOMONTH(NOW() - DAY(NOW()) + 1,11)
The EOMONTH() function:
EOMONTH(start_date,months)
Returns the serial number for the last day of the month that is the
indicated number of months before or after start_date. Use EOMONTH to
calculate maturity dates or due dates that fall on the last day of the
month.
If this function is not available, and returns the #NAME? error,
install and load the Analysis ToolPak add-in.