目标:此线程的目标是创建一个数学公式来替换@lawlist中的长方法,函数 lawlist-target-year-function (以下)。
注意:这个线程的解决方案有些类似,但是会有所不同, @AShelly在相关线程中编写的算法: stackoverflow/a/21709710/2112489
故障问题
现在,Emacs中存在一个12个月的日历,向前滚动一次(或更多)一次。一个名为 lawlist-target-year-function 的帮助函数由sevaral假日函数用于在每个假日上放置重叠。
可以在这里找到12个月滚动日历(包括长期解决方案)的完整工作草案 - [Github源代码已被修改包括@legoscia的简洁算法解决方案]:
github/lawlist/calendar-yearly-scroll-by-month/blob/master/lawlist-cal.el
LEGEND :
/ code>(数字1到12)是出现在缓冲区左上角的月份,并且随着12个月的日历向前或向后滚动而改变。
目标月份(数字1到12)是月份包含将标有叠加层的假期。有三(3)个可能的 x 轴坐标(即列1,列2或列3)。有四(4)个可能的 y 轴坐标(即行1,行2,行3或行4)。 [引用x / y坐标: www.mathsisfun/data/ cartesian-coordinates.html ]
显示年份是出现在上层的年份缓冲区的左上角,随着12个月的日历向前或向后滚动,这一变化。
目标年份是目标月份的年份。
示例:
-
当显示月份是1月(即1)那么所有目标月份的年份是一样的。
-
当显示月份是二月,2):
(if(memq target-month`(2 3 4 5 6 7 8 9 10 11 12)) 显示年(+显示年1)) -
当显示月份是3月(即3):
(if(memq ta rget-month`(3 4 5 6 7 8 9 10 11 12))显示年(+显示年1)) -
当显示月份是四月(即4)时:
(if(memq target-month`(4 5 6 7 8 9 10 11 12))显示年(+显示年份1)) -
当显示月份是May(即5)
(if(memq target-month`(5 6 7 8 9 10 11 12))显示年(+显示年1)) -
当显示月份是6月(即6):
code>(if(memq target-month`(6 7 8 9 10 11 12))显示年(+显示年1)) -
当显示月份是七月(即7) / p>
(if(memq target-month`(7 8 9 10 11 12))显示年(+显示年份1)) -
当莫第n个是八月(即8):
(if(memq target-month` 9 10 11 12))显示年(+显示年1)) -
当显示月份为9月(即9日)时:
(if(memq target-month`(9 10 11 12))显示年(+显示年份1)) -
当显示月份是十月(即10)时:
(if(memq target-month`(10 11 12))显示年(+年份1)) -
当显示月份是11月(即11):
(if(memq target-month`(11 12))显示年(+显示年1)) -
显示月份是十二月(即12):
(if (memq target-month`(12))显示年(+显示年1))
12个月的日历如下所示:布局一次滚动一个月:
;; 1 2 3 ;; 4 5 6 ;; 7 8 9 ;; 10 11 12 ;; 2 3 4 ;; 5 6 7 ;; 8 9 10 ;; 11 12 1 ;; 3 4 5 ;; 6 7 8 ;; 9 10 11 ;; 12 1 2 ;; 4 5 6 ;; 7 8 9 ;; 10 11 12 ;; 1 2 3 ;; 5 6 7 ;; 8 9 10 ;; 11 12 1 ;; 2 3 4 ;; 6 7 8 ;; 9 10 11 ;; 12 1 2 ;; 3 4 5 ;; 7 8 9 ;; 10 11 12 ;; 1 2 3 ;; 4 5 6 ;; 8 9 10 ;; 11 12 1 ;; 2 3 4 ;; 5 6 7 ;; 9 10 11 ;; 12 1 2 ;; 3 4 5 ;; 6 7 8 ;; 10 11 12 ;; 1 2 3 ;; 4 5 6 ;; 7 8 9 ;; 11 12 1 ;; 2 3 4 ;; 5 6 7 ;; 8 9 10 ;; 12 1 2 ;; 3 4 5 ;; 6 7 8 ;; 9 10 11@lawlist的长手解决方案如下:
(defun lawlist-target-year-function(target-month)(cond ;; 1 2 3 ;; 4 5 6 ;; 7 8 9 ;; 10 11 12 ((eq显示月1)显示年) ;; 2 3 4 ;; 5 6 7 ;; 8 9 10 ;; 11 12 1 ((eq显示月份2)(if(memq目标月份(2 3 4 5 6 7 8 9 10 11 12))显示年(+显示年1)) ;; 3 4 5 (6)()()(7)()()() 4 5 6 7 8 9 10 11 12))显示年(+显示年1)) ;; 4 5 6 ;; 7 8 9 ;; 10 11 12 ;; 1 2 3 ((eq显示月份4)(if(memq target-month`(4 5 6 7 8 9 10 11 12))显示年(+显示年1))) ;; 5 6 7 ;; 8 9 10 ;; 11 12 1 ;; 2 3 4 ((eq显示月份5)(if(memq target-month`(5 6 7 8 9 10 11 12))显示年 +显示年份1))) ;; 6 7 8 ;; 9 10 11 ;; 12 1 2 ;; 3 4 5 ((eq显示月份6)(if(memq target-month`(6 7 8 9 10 11 12))显示年(+显示年份1))) ;; 7 8 9 ;; 10 11 12 ;; 1 2 3 ;; 4 5 6 ((eq显示月份7)(if(memq target-month`(7 8 9 10 11 12))显示年(+显示-year 1))) ;; 8 9 10 ;; 11 12 1 ;; 2 3 4 ;; 5 6 7 ((eq显示月份8)(if(memq target-month`(8 9 10 11 12))显示年(+年1))) ;; 9 10 11 ;; 12 1 2 ;; 3 4 5 ;; 6 7 8 ((eq显示月份9)(if(memq target-month`(9 10 11 12))显示年(+显示年份1))) ;; 10 11 12 ;; 1 2 3 ;; 4 5 6 ;; 7 8 9 ((eq显示月份10)(if(memq target-month`(10 11 12))显示年(+显示年1 ))) ;; 11 12 1 ;; 2 3 4 ;; 5 6 7 ;;新评新新新新新新新旗新新新新旗新新新新旗新新新新旗新新旗200新新新新旗新新旗200新新新新旗新新旗200新新新新新新旗新新旗200新新新新旗新新旗200新新新新旗新新旗200新新新新200新新新新旗新新旗新新旗旗)) ;; 12 1 2 ;; 3 4 5 ;; 6 7 8 ;;新评新新新新新新旗新新新新旗新新新新旗新新旗200新新新新新新旗新新旗200新新新新旗新新旗200新新新新旗新新旗200新新新新新旗200新新新新新旗200 Chan X- ))解决方案
(defun lawlist-target-year-function(target-month)(if(> = target-month shows-month新评新新新旗新新新新旗新新新新旗新新旗新新旗新新旗旗新p新新新新新旗新新旗新新旗新新旗旗醒新年200新新新新旗新新旗旗醒新年200旗新新旗旗时旗新新新新旗新StrEC新新新旗200 200 200新新新新旗新StrEC新新新旗200新新新新旗200 200新新Version 200新新新新新旗新StrippECol
GOAL: The goal of this thread is to create a mathematical formula to replace the long-hand solution by @lawlist in the function lawlist-target-year-function (below).
NOTE: The solution to this thread is somewhat similar, but will nevertheless be different, than the algorithm written by @AShelly in a related thread: stackoverflow/a/21709710/2112489
STORY PROBLEM
There now exists a 12-month calendar in Emacs that scrolls forwards and backwards one month (or more) at a time. A helper function called lawlist-target-year-function is used by sevaral holiday functions to place an overlay on each holiday.
A full working draft of the 12-month scrolling calendar (including the long-hand solution) may be found here -- [the Github source code has been revised to include the concise algorithm solution by @legoscia]:
github/lawlist/calendar-yearly-scroll-by-month/blob/master/lawlist-cal.el
LEGEND:
displayed-month (numbers 1 through 12) is the month that appears in the upper left-hand corner of the buffer, and this changes as the 12-month calendar is scrolled forwards or backwards.
The target-month (numbers 1 through 12) is the month that contains the holiday that will be marked with an overlay. There are three (3) possible x axis coordinates (i.e., column 1, column 2, or column 3). There are four (4) possible y axis coordinates (i.e., row 1, row 2, row 3, or row 4). [Citation to x / y coordinates: www.mathsisfun/data/cartesian-coordinates.html ]
The displayed-year is the year that appears in the upper left-hand corner of the buffer, and this changes as the 12-month calendar is scrolled forwards or backwards.
The target year is the year of the target-month.
EXAMPLE:
When displayed-month is January (i.e., 1), then the year is the same for all target months.
When displayed-month is February (i.e., 2):
(if (memq target-month `(2 3 4 5 6 7 8 9 10 11 12)) displayed-year (+ displayed-year 1))When displayed-month is March (i.e., 3):
(if (memq target-month `(3 4 5 6 7 8 9 10 11 12)) displayed-year (+ displayed-year 1))When displayed-month is April (i.e., 4):
(if (memq target-month `(4 5 6 7 8 9 10 11 12)) displayed-year (+ displayed-year 1))When displayed-month is May (i.e., 5)
(if (memq target-month `(5 6 7 8 9 10 11 12)) displayed-year (+ displayed-year 1))When displayed-month is June (i.e., 6):
(if (memq target-month `(6 7 8 9 10 11 12)) displayed-year (+ displayed-year 1))When displayed-month is July (i.e., 7):
(if (memq target-month `(7 8 9 10 11 12)) displayed-year (+ displayed-year 1))When displayed-month is August (i.e, 8):
(if (memq target-month `(8 9 10 11 12)) displayed-year (+ displayed-year 1))When displayed-month is September (i.e., 9):
(if (memq target-month `(9 10 11 12)) displayed-year (+ displayed-year 1))When displayed-month is October (i.e., 10):
(if (memq target-month `(10 11 12)) displayed-year (+ displayed-year 1))When displayed-month is November (i.e., 11):
(if (memq target-month `(11 12)) displayed-year (+ displayed-year 1))When displayed-month is December (i.e., 12):
(if (memq target-month `(12)) displayed-year (+ displayed-year 1))
The 12-month calendar looks like the following as the layout scrolls forward one month at a time:
;; 1 2 3 ;; 4 5 6 ;; 7 8 9 ;; 10 11 12 ;; 2 3 4 ;; 5 6 7 ;; 8 9 10 ;; 11 12 1 ;; 3 4 5 ;; 6 7 8 ;; 9 10 11 ;; 12 1 2 ;; 4 5 6 ;; 7 8 9 ;; 10 11 12 ;; 1 2 3 ;; 5 6 7 ;; 8 9 10 ;; 11 12 1 ;; 2 3 4 ;; 6 7 8 ;; 9 10 11 ;; 12 1 2 ;; 3 4 5 ;; 7 8 9 ;; 10 11 12 ;; 1 2 3 ;; 4 5 6 ;; 8 9 10 ;; 11 12 1 ;; 2 3 4 ;; 5 6 7 ;; 9 10 11 ;; 12 1 2 ;; 3 4 5 ;; 6 7 8 ;; 10 11 12 ;; 1 2 3 ;; 4 5 6 ;; 7 8 9 ;; 11 12 1 ;; 2 3 4 ;; 5 6 7 ;; 8 9 10 ;; 12 1 2 ;; 3 4 5 ;; 6 7 8 ;; 9 10 11The long-hand solution by @lawlist is as follows:
(defun lawlist-target-year-function (target-month) (cond ;; 1 2 3 ;; 4 5 6 ;; 7 8 9 ;; 10 11 12 ((eq displayed-month 1) displayed-year) ;; 2 3 4 ;; 5 6 7 ;; 8 9 10 ;; 11 12 1 ((eq displayed-month 2) (if (memq target-month `(2 3 4 5 6 7 8 9 10 11 12)) displayed-year (+ displayed-year 1))) ;; 3 4 5 ;; 6 7 8 ;; 9 10 11 ;; 12 1 2 ((eq displayed-month 3) (if (memq target-month `(3 4 5 6 7 8 9 10 11 12)) displayed-year (+ displayed-year 1))) ;; 4 5 6 ;; 7 8 9 ;; 10 11 12 ;; 1 2 3 ((eq displayed-month 4) (if (memq target-month `(4 5 6 7 8 9 10 11 12)) displayed-year (+ displayed-year 1))) ;; 5 6 7 ;; 8 9 10 ;; 11 12 1 ;; 2 3 4 ((eq displayed-month 5) (if (memq target-month `(5 6 7 8 9 10 11 12)) displayed-year (+ displayed-year 1))) ;; 6 7 8 ;; 9 10 11 ;; 12 1 2 ;; 3 4 5 ((eq displayed-month 6) (if (memq target-month `(6 7 8 9 10 11 12)) displayed-year (+ displayed-year 1))) ;; 7 8 9 ;; 10 11 12 ;; 1 2 3 ;; 4 5 6 ((eq displayed-month 7) (if (memq target-month `(7 8 9 10 11 12)) displayed-year (+ displayed-year 1))) ;; 8 9 10 ;; 11 12 1 ;; 2 3 4 ;; 5 6 7 ((eq displayed-month 8) (if (memq target-month `(8 9 10 11 12)) displayed-year (+ displayed-year 1))) ;; 9 10 11 ;; 12 1 2 ;; 3 4 5 ;; 6 7 8 ((eq displayed-month 9) (if (memq target-month `(9 10 11 12)) displayed-year (+ displayed-year 1))) ;; 10 11 12 ;; 1 2 3 ;; 4 5 6 ;; 7 8 9 ((eq displayed-month 10) (if (memq target-month `(10 11 12)) displayed-year (+ displayed-year 1))) ;; 11 12 1 ;; 2 3 4 ;; 5 6 7 ;; 8 9 10 ((eq displayed-month 11) (if (memq target-month `(11 12)) displayed-year (+ displayed-year 1))) ;; 12 1 2 ;; 3 4 5 ;; 6 7 8 ;; 9 10 11 ((eq displayed-month 12) (if (memq target-month `(12)) displayed-year (+ displayed-year 1))) ))解决方案
Would this work?
(defun lawlist-target-year-function (target-month) (if (>= target-month displayed-month) displayed-year (1+ displayed-year)))
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