计算目标月

本文介绍了计算目标月 - 12个月目标年的算法*日历*的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

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

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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 11

The 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)))