我有一个月的日历项目字典（日期作为“键”，列表形式的项目作为“值”），我想以某种方式打印出来（该字典包含在代码中，分配给dct）。我只想显示当前日期或之后的项目（即今天）。显示格式为：
day: item1, item2
我还希望这些项目仅跨越 5 行标准输出，每行 49 个字符宽（包括空格）。这是必要的，因为输出将显示在 conky（适用于 linux 的应用程序）中。
由于一天可以有多个议程项目，因此必须将输出包装并打印在多行上。我希望代码通过仅选择那些项目可以放入 5 行或更少行的日期来解决这一问题，而不是打印 5 天且相关项目超过 5 行。例如
day1: item1, item2
item3
day2: item1
day3: item1,
item2
那是当天/之后的 3 天，打印 5 行，每行 49 个字符宽。超过 49 个字符的字符串用换行符换行。
这是我为此编写的代码：
#!/usr/bin/env python
from datetime import date, timedelta, datetime
import heapq
import re
import textwrap
pattern_string = '(1[012]|[1-9]):[0-5][0-9](\\s)?(?i)(am|pm)'
pattern = re.compile(pattern_string)
# Explanation of pattern_string:
# ------------------------------
#( #start of group #1
#1[012] # start with 10, 11, 12
#| # or
#[1-9] # start with 1,2,...9
#) #end of group #1
#: # follow by a semi colon (:)
#[0-5][0-9] # follw by 0..5 and 0..9, which means 00 to 59
#(\\s)? # follow by a white space (optional)
#(?i) # next checking is case insensitive
#(am|pm) # follow by am or pm
# The 12-hour clock format is start from 0-12, then a semi colon (:) and follow by 00-59 , and end with am or pm.
# Time format that match:
# 1. "1:00am", "1:00 am","1:00 AM" ,
# 2. "1:00pm", "1:00 pm", "1:00 PM",
# 3. "12:50 pm"
d = date.today() # datetime.date(2013, 8, 11)
e = datetime.today() # datetime.datetime(2013, 8, 11, 5, 56, 28, 702926)
today = d.strftime('%a %b %d') # 'Sun Aug 11'
dct = {
'Thu Aug 01' : [' Weigh In'],
'Thu Aug 08' : [' 8:00am', 'Serum uric acid test', '12:00pm', 'Make Cheesecake'],
'Sun Aug 11' : [" Awais chotu's birthday", ' Car wash'],
'Mon Aug 12' : ['10:00am', 'Start car for 10 minutes'],
'Thu Aug 15' : [" Hooray! You're Facebook Free!", '10:00am', 'Start car for 10 minutes'],
'Mon Aug 19' : ['10:00am', 'Start car for 10 minutes'],
'Thu Aug 22' : ['10:00am', 'Start car for 10 minutes'],
'Mon Aug 26' : ['10:00am', 'Start car for 10 minutes'],
'Thu Aug 29' : ['10:00am', 'Start car for 10 minutes']
}
def join_time(lst):
'''Searches for a time format string in supplied list and concatenates it + the event next to it as an single item
to a list and returns that list'''
mod_lst = []
for number, item in enumerate(lst):
if re.search(pattern, item):
mod_lst.append(item + ' ' + lst[number+1]) # append the item (i.e time e.g '1:00am') and the item next to it (i.e. event)
del lst[number+1]
else:
mod_lst.append(item)
return mod_lst
def parse_date(datestring):
return datetime.strptime(datestring + ' ' + str(date.today().year), "%a %b %d %Y") # returns a datetime obj for the time string; "Sun Aug 11" = datetime.datetime(1900, 8, 11, 0, 0)
deltas = [] # holds datetime.timedelta() objs; timedelta(days, seconds, microseconds)
val_len = []
key_len = {}
for key in dct:
num = len(''.join(item for item in dct[key]))
val_len.append(num) # calculate the combined len of all items in the
# list which are the val of a key and add them to val_len
if num > 37:
key_len[key] = 2
else:
key_len[key] = 1
# val_len = [31, 9, 61, 31, 31, 49, 31, 32, 31]
# key_len = {'Sun Aug 11': 1, 'Mon Aug 12': 1, 'Thu Aug 01': 1, 'Thu Aug 15': 2, 'Thu Aug 22': 1, 'Mon Aug 19': 1, 'Thu Aug 08': 2, 'Mon Aug 26': 1, 'Thu Aug 29': 1}
counter = 0
for eachLen in val_len:
if eachLen > 37:
counter = counter + 2
else:
counter = counter + 1
# counter = 11
if counter > 5: # because we want only those 5 events in our conky output which are closest to today
n = counter - 5 # n = 6, these no of event lines should be skipped
for key in dct:
deltas.append(e - parse_date(key)) # today - key date (e.g. 'Sun Aug 11') ---> datetime.datetime(2013, 8, 11, 5, 56, 28, 702926) - datetime.datetime(1900, 8, 11, 0, 0)
# TODO: 'n' no of event lines should be skipped, NOT n no of days!
for key in sorted(dct, key=parse_date): # sorted() returns ['Thu Aug 01', 'Thu Aug 08', 'Sun Aug 11', 'Mon Aug 12', 'Thu Aug 15', 'Mon Aug 19', 'Thu Aug 22', 'Mon Aug 26', 'Thu Aug 29']
tdelta = e - parse_date(key)
if tdelta in heapq.nlargest(n, deltas): # heapq.nlargest(x, iterable[, key]); returns list of 'x' no. of largest items in iterable
pass # In this case it should return a list of top 6 largest timedeltas; if the tdelta is in
# that list, it means its not amongst the 5 events we want to print
else:
if key == today:
value = dct[key]
val1 = '${color green}' + key + '$color: '
mod_val = join_time(value)
val2 = textwrap.wrap(', '.join(item for item in mod_val), 37)
print val1 + '${color 40E0D0}' + '$color\n ${color 40E0D0}'.join(item for item in val2) + '$color'
else:
value = dct[key]
mod_val = join_time(value)
output = key + ': ' + ', '.join(item for item in mod_val)
print '\n '.join(textwrap.wrap(output, 49))
else:
for key in sorted(dct, key=parse_date):
if key == today:
value = dct[key]
val1 = '${color green}' + key + '$color: '
mod_val = join_time(value)
val2 = textwrap.wrap(', '.join(item for item in mod_val), 37)
print val1 + '${color 40E0D0}' + '$color\n ${color 40E0D0}'.join(item for item in val2) + '$color'
else:
value = dct[key]
mod_val = join_time(value)
output = key + ': ' + ', '.join(item for item in mod_val)
print '\n '.join(textwrap.wrap(output, 49))
结果是：
Thu Aug 22: 10:00am Start car for 10 minutes
Mon Aug 26: 10:00am Start car for 10 minutes
Thu Aug 29: 10:00am Start car for 10 minutes
我已经对代码进行了大量注释，因此不难弄清楚它是如何工作的。我基本上是使用 datetime 计算距当天最远的日子，并跳过那些日子及其项目。该代码通常运行良好，但有时却不行。在这种情况下，输出应该是：
Mon Aug 19: 10:00am Start car for 10 minutes
Thu Aug 22: 10:00am Start car for 10 minutes
Mon Aug 26: 10:00am Start car for 10 minutes
Thu Aug 29: 10:00am Start car for 10 minutes
因为这些是当天（8 月 16 日星期五）之后的日子，其项目适合 5 行。如何修复它以跳过n离今天最远的天数而不是行数？
我正在考虑使用key_lendict 以某种方式进一步过滤输出，方法是仅打印项目长度总和为 < 或 = 5的那些日子的项目...
我被困住了。