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
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
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
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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