I'm very lucky to be able to see the Empire State Building from my apartment windows, and at night the lights are fantastic! But i'm also curious to know what's going to be today's lights, and tomorrow, etc.
I thought I'd easily find a calendar to add to gCal to show that, but i wasn't able to find any, so I made it myself: https://sandrotosi.github.io/esb-lights-calendar/
to send emails with images you need to use MIMEMultipart, but the basic approach:
import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.image import MIMEImage
msg = MIMEMultipart('alternative')
msg['Subject'] = "subject"
msg['From'] = from_addr
msg['To'] = to_addr
part = MIMEImage(open('/path/to/image', 'rb').read())
s = smtplib.SMTP('localhost')
s.sendmail(from_addr, to_addr, msg.as_string())
s.quit()
will produce an email with empty body and the image as an attachment.
The better way, ie to have the image as part of the body of the email, requires to write an HTML body that refers to that image:
import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.mime.image import MIMEImage
msg = MIMEMultipart('alternative')
msg['Subject'] = "subject
msg['From'] = from_addr
msg['To'] = to_addr
text = MIMEText('<img src="cid:image1">', 'html')
msg.attach(text)
image = MIMEImage(open('/path/to/image', 'rb').read())
# Define the image's ID as referenced in the HTML body above
image.add_header('Content-ID', '<image1>')
msg.attach(image)
s = smtplib.SMTP('localhost')
s.sendmail(from_addr, to_addr, msg.as_string())
s.quit()
# handy function to check if a number is palindrome
def is_palindrome(i):
return str(i) == str(i)[::-1]
# what's the max?
max_p = 0
# multiply all numbers `i` between 100 and 998, with all between i+1 and 999
for i in xrange(100, 999):
for j in xrange(i+1, 1000):
p = i * j
if is_palindrome(p) and p > max_p:
max_p = p
print max_p
def group_iter(iterator, n=2):
""" Given an iterator, it returns sub-lists made of n items
(except the last that can have len < n)
inspired by http://countergram.com/python-group-iterator-list-function"""
accumulator = []
for item in iterator:
accumulator.append(item)
if len(accumulator) == n: # tested as fast as separate counter
yield accumulator
accumulator = [] # tested faster than accumulator[:] = []
# and tested as fast as re-using one list object
if len(accumulator) != 0:
yield accumulator
>>> import time
>>> str = '2009-03-04'
>>> format = '%Y-%m-%d'
>>> time.mktime(time.strptime(str, format))
1236121200.0
>>> int(_)
1236121200
$ curl -H "Host: domain" http://ip_address/path/to/the/page.html
import httplib
conn = httplib.HTTPConnection("ip_address")
conn.putrequest("GET", "/", skip_host=True)
conn.putheader("Host", "domain.ext")
conn.endheaders()
res = conn.getresponse()
print res.read()
def collatz(n):very rapidly converges to... an error:
if n == 1:
return 1
if n % 2 == 0:
return 1 + collatz(n/2)
else:
return 1 + collatz(3*n + 1)
RuntimeError: maximum recursion depth exceededSo I've recoded it a bit using a cache dictionary to store all the intermediate values in it:
cache = {1: 1}
def collatz(n, res):
nn = n
while True:
if n in cache:
if nn != n:
cache[nn] = cache[n]
return cache[nn]
if n % 2 == 0:
cache[n] = 1 + collatz(n/2, cache)
else:
cache[n] = 1 + collatz(3*n + 1, cache)
Now loping for all the numbers lower than 1 million we find the solution to the problem: the number with the longest path is 837799 with a path length of 524. The code runs in about 4secs.
There are few short paths, several long paths but rare to occur, and most of the paths have length between 100 and 200 (more or less).