How do you print range a to z in python?

1. Print a-n: a b c d e f g h i j k l m n

2. Every second in a-n: a c e g i k m

3. Append a-n to index of urls{hello.com/, hej.com/, ..., hallo.com/}: hello.com/a hej.com/b ... hallo.com/n

dreftymac

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asked Jul 6, 2010 at 20:51

2

>>> import string >>> string.ascii_lowercase[:14] 'abcdefghijklmn' >>> string.ascii_lowercase[:14:2] 'acegikm'

To do the urls, you could use something like this

[i + j for i, j in zip(list_of_urls, string.ascii_lowercase[:14])]

answered Jul 6, 2010 at 21:01

John La RooyJohn La Rooy

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3

Assuming this is a homework ;-) - no need to summon libraries etc - it probably expect you to use range() with chr/ord, like so:

for i in range(ord('a'), ord('n')+1): print chr(i),

For the rest, just play a bit more with the range()

answered Jul 6, 2010 at 23:55

Nas BanovNas Banov

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Hints:

import string print string.ascii_lowercase

and

for i in xrange(0, 10, 2): print i

and

"hello{0}, world!".format('z')

answered Jul 6, 2010 at 21:01

Wayne WernerWayne Werner

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for one in range(97,110): print chr(one)

answered Jul 6, 2010 at 21:02

yedpodtrzitkoyedpodtrzitko

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Get a list with the desired values

small_letters = map(chr, range(ord('a'), ord('z')+1)) big_letters = map(chr, range(ord('A'), ord('Z')+1)) digits = map(chr, range(ord('0'), ord('9')+1))

or

import string string.letters string.uppercase string.digits

This solution uses the ASCII table. ord gets the ascii value from a character and chr vice versa.

Apply what you know about lists

>>> small_letters = map(chr, range(ord('a'), ord('z')+1)) >>> an = small_letters[0:(ord('n')-ord('a')+1)] >>> print(" ".join(an)) a b c d e f g h i j k l m n >>> print(" ".join(small_letters[0::2])) a c e g i k m o q s u w y >>> s = small_letters[0:(ord('n')-ord('a')+1):2] >>> print(" ".join(s)) a c e g i k m >>> urls = ["hello.com/", "hej.com/", "hallo.com/"] >>> print([x + y for x, y in zip(urls, an)]) ['hello.com/a', 'hej.com/b', 'hallo.com/c']

answered Jun 22, 2014 at 15:03

Martin ThomaMartin Thoma

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1

import string print list(string.ascii_lowercase) # ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']

answered Feb 26, 2016 at 8:24

1

import string print list(string.ascii_lowercase) # ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']

and

for c in list(string.ascii_lowercase)[:5]: ...operation with the first 5 characters

answered Jul 21, 2018 at 4:42

myList = [chr(chNum) for chNum in list(range(ord('a'),ord('z')+1))] print(myList)

Output

['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']

Jeroen Heier

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answered Sep 6, 2019 at 11:03

2

Try:

strng = "" for i in range(97,123): strng = strng + chr(i) print(strng)

S. Salman

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answered Aug 13, 2015 at 20:51

0

import string string.printable[10:36] # abcdefghijklmnopqrstuvwxyz string.printable[10:62] # abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ

answered Sep 25, 2020 at 8:49

WeiloryWeilory

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#1) print " ".join(map(chr, range(ord('a'),ord('n')+1))) #2) print " ".join(map(chr, range(ord('a'),ord('n')+1,2))) #3) urls = ["hello.com/", "hej.com/", "hallo.com/"] an = map(chr, range(ord('a'),ord('n')+1)) print [ x + y for x,y in zip(urls, an)]

answered Nov 29, 2013 at 16:48

carlos_lmcarlos_lm

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list(string.ascii_lowercase) ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']

Dhia

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answered Jun 22, 2016 at 5:11

townietownie

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The answer to this question is simple, just make a list called ABC like so:

ABC = ['abcdefghijklmnopqrstuvwxyz']

And whenever you need to refer to it, just do:

print ABC[0:9] #prints abcdefghij print ABC #prints abcdefghijklmnopqrstuvwxyz for x in range(0,25): if x % 2 == 0: print ABC[x] #prints acegikmoqsuwy (all odd numbered letters)

Also try this to break ur device :D

##Try this and call it AlphabetSoup.py: ABC = ['abcdefghijklmnopqrstuvwxyz'] try: while True: for a in ABC: for b in ABC: for c in ABC: for d in ABC: for e in ABC: for f in ABC: print a, b, c, d, e, f, ' ', except KeyboardInterrupt: pass

answered Dec 18, 2016 at 18:17

0

This is your 2nd question: string.lowercase[ord('a')-97:ord('n')-97:2] because 97==ord('a') -- if you want to learn a bit you should figure out the rest yourself ;-)

answered Jul 6, 2010 at 21:04

Jochen RitzelJochen Ritzel

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I hope this helps:

import string alphas = list(string.ascii_letters[:26]) for chr in alphas: print(chr)

answered Aug 9, 2019 at 18:51

About gnibbler's answer.

Zip -function, full explanation, returns a list of tuples, where the i-th tuple contains the i-th element from each of the argument sequences or iterables. [...] construct is called list comprehension, very cool feature!

answered Dec 23, 2010 at 2:22

hhhhhh

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# Assign the range of characters first_char_start = 'a' last_char = 'n' # Generate a list of assigned characters (here from 'a' to 'n') alpha_list = [chr(i) for i in range(ord(first_char), ord(last_char) + 1)] # Print a-n with spaces: a b c d e f g h i j k l m n print(" ".join(alpha_list)) # Every second in a-n: a c e g i k m print(" ".join(alpha_list[::2])) # Append a-n to index of urls{hello.com/, hej.com/, ..., hallo.com/} # Ex.hello.com/a hej.com/b ... hallo.com/n #urls: list of urls results = [i+j for i, j in zip(urls, alpha_list)] #print new url list 'results' (concatenated two lists element-wise) print(results)

answered Jun 12, 2021 at 16:02

docjagdocjag

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Another way to do it

import string aalist = list(string.ascii_lowercase) aaurls = ['alpha.com','bravo.com','chrly.com','delta.com',] iilen = aaurls.__len__() ans01 = "".join( (aalist[0:14]) ) ans02 = "".join( (aalist[0:14:2]) ) ans03 = "".join( "{vurl}/{vl}\n".format(vl=vlet,vurl=aaurls[vind % iilen]) for vind,vlet in enumerate(aalist[0:14]) ) print(ans01) print(ans02) print(ans03)

Result

abcdefghijklmn acegikm alpha.com/a bravo.com/b chrly.com/c delta.com/d alpha.com/e bravo.com/f chrly.com/g delta.com/h alpha.com/i bravo.com/j chrly.com/k delta.com/l alpha.com/m bravo.com/n

How this differs from the other replies

• iterate over an arbitrary number of base urls
• cycle through the urls using modular arithmetic, and do not stop until we run out of letters
• use enumerate in conjunction with list comprehension and str.format

answered Jun 12, 2019 at 0:08

dreftymacdreftymac

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