There is a string method called zfill:

>>> '12344'.zfill(10)
0000012344

It will pad the left side of the string with zeros to make the string length N (10 in this case).

It has been a few years since this was answered, but as of Python 3.6 (PEP498) you could use the new f-strings:

numbers = [23.23, 0.123334987, 1, 4.223, 9887.2]

for number in numbers:
    print(f'{number:9.4f}')

Prints:

  23.2300
   0.1233
   1.0000
   4.2230
9887.2000

The OP & accepted answer focus on formatting time, but the OP question itself discusses formatting numbers to strings in Python. In many cases, the output requires additional data fields be included along with timestamps, all of which include formatting numbers as strings.

Below are a variety of non time-based examples of formatting numbers as strings, and different ways to do so, starting with the existing string format operator (%) which has been around for as long as Python has been around (meaning this solution is compatible across Python 1.x, 2.x, and 3.x):

>>> "Name: %s, age: %d" % ('John', 35) 
'Name: John, age: 35' 
>>> i = 45 
>>> 'dec: %d/oct: %#o/hex: %#X' % (i, i, i) 
'dec: 45/oct: 055/hex: 0X2D' 
>>> "MM/DD/YY = %02d/%02d/%02d" % (12, 7, 41) 
'MM/DD/YY = 12/07/41' 
>>> 'Total with tax: $%.2f' % (13.00 * 1.0825) 
'Total with tax: $14.07' 
>>> d = {'web': 'user', 'page': 42} 
>>> 'http://xxx.yyy.zzz/%(web)s/%(page)d.html' % d 
'http://xxx.yyy.zzz/user/42.html' 

Starting in Python 2.6 (meaning it works for 2.x and 3.x), there is an alternative: the str.format() method. Here are the equivalent snippets to the above but using str.format():

>>> "Name: {0}, age: {1}".format('John', 35) 
'Name: John, age: 35' 
>>> i = 45 
>>> 'dec: {0}/oct: {0:#o}/hex: {0:#X}'.format(i) 
'dec: 45/oct: 0o55/hex: 0X2D' 
>>> "MM/DD/YY = {0:02d}/{1:02d}/{2:02d}".format(12, 7, 41) 
'MM/DD/YY = 12/07/41' 
>>> 'Total with tax: ${0:.2f}'.format(13.00 * 1.0825) 
'Total with tax: $14.07' 
>>> d = {'web': 'user', 'page': 42} 
>>> 'http://xxx.yyy.zzz/{web}/{page}.html'.format(**d) 
'http://xxx.yyy.zzz/user/42.html'

Like Python 2.6+, all Python 3 releases (so far) understand how to do both. I shamelessly ripped this stuff straight out of my hardcore Python intro book and the slides for the Intro+Intermediate Python courses I offer from time-to-time. :-)

Aug 2018 UPDATE: Of course, now that we have the f-string feature introduced in 3.6 (only works in 3.6 and newer), we need the equivalent examples of that; yes, another alternative:

>>> name, age = 'John', 35
>>> f'Name: {name}, age: {age}'
'Name: John, age: 35'

>>> i = 45
>>> f'dec: {i}/oct: {i:#o}/hex: {i:#X}'
'dec: 45/oct: 0o55/hex: 0X2D'

>>> m, d, y = 12, 7, 41
>>> f"MM/DD/YY = {m:02d}/{d:02d}/{y:02d}"
'MM/DD/YY = 12/07/41'

>>> f'Total with tax: ${13.00 * 1.0825:.2f}'
'Total with tax: $14.07'

>>> d = {'web': 'user', 'page': 42}
>>> f"http://xxx.yyy.zzz/{d['web']}/{d['page']}.html"
'http://xxx.yyy.zzz/user/42.html'

print("{0:02}".format(1))
>>0001

Just learnt from other answers and commentators that we don't need zfill but can use the expression :02 to give the padding.

Expand to more positions:

print("{0:02}_{1:02}".format(1, 2))
>>0001_0002

The simplest solution is probably to use the exponential format with one less than the number of digits.

"{0:.6e}".format(1234.457890123) = '1.234568e+03'

I ended up writing this solution that can print floats as well as exponentials but it's probably unnecessarily long for most needs.

 import numpy as np

 def sigprint(number,nsig):
     """
     Returns a string with the given number of significant digits.
     For numbers >= 1e5, and less than 0.001, it does exponential notation
     This is almost what ":.3g".format(x) does, but in the case
     of '{:.3g}'.format(2189), we want 2190 not 2.19e3. Also in the case of
     '{:.3g}'.format(1), we want 1.00, not 1
     """

     if ((abs(number) >= 1e-3) and (abs(number) < 1e5)) or number ==0:
         place = decplace(number) - nsig + 1
         decval = 10**place
         outnum = np.round(np.float(number) / decval) * decval
         ## Need to get the place again in case say 0.97 was rounded up to 1.0
         finalplace = decplace(outnum) - nsig + 1 
         if finalplace >= 0: finalplace=0
         fmt='.'+str(int(abs(finalplace)))+'f'
     else:
         stringnsig = str(int(nsig-1))
         fmt = '.'+stringnsig+'e'
         outnum=number
     wholefmt = "{0:"+fmt+"}"

     return wholefmt.format(outnum)

 def decplace(number):
     """
     Finds the decimal place of the leading digit of a number. For 0, it assumes
     a value of 0 (the one's digit)
     """
     if number == 0:
         place = 0
     else:
         place = np.floor(np.log10(np.abs(number)))
     return place