Python replacement for datetime.strftime that is able to handle dates before 1900. Written by Andrew Dalke on comp.lang.python (https://groups.google.com/forum/#!msg/comp.lang.python/v-9L1D6oc4k/3faS6YjTRg4J)

strftime.py

import time, datetime

def _findall(text, substr):
    # Also finds overlaps
    sites = []
    i = 0
    while 1:
        j = text.find(substr, i)
        if j == -1:
           break
        sites.append(j)
        i=j+1
    return sites

# I hope I did this math right.  Every 28 years the
# calendar repeats, except through century leap years
# excepting the 400 year leap years.  But only if
# you're using the Gregorian calendar.

def strftime(dt, fmt):
    # WARNING: known bug with "%s", which is the number
    # of seconds since the epoch.  This is too harsh
    # of a check.  It should allow "%%s".
    fmt = fmt.replace("%s", "s")
    if dt.year > 1900:
        return time.strftime(fmt, dt.timetuple())

    year = dt.year
    # For every non-leap year century, advance by
    # 6 years to get into the 28-year repeat cycle
    delta = 2000 - year
    off = 6*(delta // 100 + delta // 400)
    year = year + off

    # Move to around the year 2000
    year = year + ((2000 - year)//28)*28
    timetuple = dt.timetuple()
    s1 = time.strftime(fmt, (year,) + timetuple[1:])
    sites1 = _findall(s1, str(year))

    s2 = time.strftime(fmt, (year+28,) + timetuple[1:])
    sites2 = _findall(s2, str(year+28))

    sites = []
    for site in sites1:
        if site in sites2:
           sites.append(site)

    s = s1
    syear = "%4d" % (dt.year,)
    for site in sites:
        s = s[:site] + syear + s[site+4:]
    return s