Develop Reference

Adding letters to each element of a list

I try to add "G:" in the beginning and a backslash before every point of each element in a list. Therefore I created this example list1:
list1 = ['AEX.EN', 'AXAL.OQ', 'AAPIOE.NW']
And I need something like list2:
list2 = ['G:AEX\.EN', 'G:AXAL\.OQ', 'G:AAPIOE\.NW']
Thank you very much for the help!

Use:
>>> ['G:' + i.replace('.', '\\.') for i in list1]
['G:AEX\\.EN', 'G:AXAL\\.OQ', 'G:AAPIOE\\.NW']
>>>
In this case I prefer re.escape:
>>> import re
>>> ['G:' + re.escape(i) for i in list1]
['G:AEX\\.EN', 'G:AXAL\\.OQ', 'G:AAPIOE\\.NW']
>>>

You can use + for join string then use replace() like below:
>>> list1 = ['AEX.EN', 'AXAL.OQ', 'AAPIOE.NW']
>>> [('G:'+l).replace('.','\.') for l in list1]
['G:AEX\\.EN', 'G:AXAL\\.OQ', 'G:AAPIOE\\.NW']

Replace in strings of list

I can use re.sub easily in single string like this:
>>> a = "ajhga':&+?%"
>>> a = re.sub('[.!,;+?&:%]', '', a)
>>> a
"ajhga'"
If I use this on list of strings then I am not getting the result. What I am doing is:
>>> a = ["abcd:+;", "(l&'kka)"]
>>> for x in a:
... x = re.sub('[\(\)&\':+]', '', x)
...
>>> a
['abcd:+;', "(l&'kka)"]
How can I strip expressions from strings in list?

>>> a = ["abcd:+;", "(l&'kka)"]
>>> a = [re.sub('[\(\)&\':+]', '', x) for x in a]
>>> a
['abcd;', 'lkka']
>>>

for index,x in enumerate(a):
a[index] = re.sub('[\(\)&\':+]', '', x)
Your changing the value but not updating your list. enumerate is function that return tuple (index,value) for each item of list

Replacing an item in a python list by index.. failing?

Any idea why when I call:
>>> hi = [1, 2]
>>> hi[1]=3
>>> print hi
[1, 3]
I can update a list item by its index, but when I call:
>>> phrase = "hello"
>>> for item in "123":
>>> list(phrase)[int(item)] = list(phrase)[int(item)].upper()
>>> print phrase
hello
It fails?
Should be hELLo

You haven't initialised phrase (The list you were intending to make) into a variable yet. So pretty much you have created a list in each loop, it being the exact same.
If you were intending to actually change the characters of phrase, well that's not possible, as in python, strings are immutable.
Perhaps make phraselist = list(phrase), then edit the list in the for-loop. Also, you can use range():
>>> phrase = "hello"
>>> phraselist = list(phrase)
>>> for i in range(1,4):
... phraselist[i] = phraselist[i].upper()
...
>>> print ''.join(phraselist)
hELLo

>>> phrase = "hello"
>>> list_phrase = list(phrase)
>>> for index in (1, 2, 3):
list_phrase[index] = phrase[index].upper()
>>> ''.join(list_phrase)
'hELLo'
If you prefer one-liner:
>>> ''.join(x.upper() if index in (1, 2, 3) else x for
index, x in enumerate(phrase))
'hELLo'

Another answer, just for fun :)
phrase = 'hello'
func = lambda x: x[1].upper() if str(x[0]) in '123' else x[1]
print ''.join(map(func, enumerate(phrase)))
# hELLo
To make this robust, I created a method: (because I am awesome, and bored)
def method(phrase, indexes):
func = lambda x: x[1].upper() if str(x[0]) in indexes else x[1]
return ''.join(map(func, enumerate(phrase)))
print method('hello', '123')
# hELLo

consider that strings are immutable in python You can't modify existing string can create new.
''.join([c if i not in (1, 2, 3) else c.upper() for i, c in enumerate(phrase)])

list() creates a new list. Your loop creates and instantly discards two new lists on each iteration. You could write it as:
phrase = "hello"
L = list(phrase)
L[1:4] = phrase[1:4].upper()
print("".join(L))
Or without a list:
print("".join([phrase[:1], phrase[1:4].upper(), phrase[4:]]))
Strings are immutable in Python therefore to change it, you need to create a new string.
Or if you are dealing with bytestrings, you could use bytearray which is mutable:
phrase = bytearray(b"hello")
phrase[1:4] = phrase[1:4].upper()
print(phrase.decode())
If indexes are not consecutive; you could use an explicit for-loop:
indexes = [1, 2, 4]
for i in indexes:
L[i] = L[i].upper()

Python Regular expression repeat

I have a string like this
--x123-09827--x456-9908872--x789-267504
I am trying to get all value like
123:09827
456:9908872
789:267504
I've tried (--x([0-9]+)-([0-9])+)+
but it only gives me last pair result, I am testing it through python
>>> import re
>>> x = "--x123-09827--x456-9908872--x789-267504"
>>> p = "(--x([0-9]+)-([0-9]+))+"
>>> re.match(p,x)
>>> re.match(p,x).groups()
('--x789-267504', '789', '267504')
How should I write with nested repeat pattern?
Thanks a lot!
David

Code it like this:
x = "--x123-09827--x456-9908872--x789-267504"
p = "--x(?:[0-9]+)-(?:[0-9]+)"
print re.findall(p,x)

Just use the .findall method instead, it makes the expression simpler.
>>> import re
>>> x = "--x123-09827--x456-9908872--x789-267504"
>>> r = re.compile(r"--x(\d+)-(\d+)")
>>> r.findall(x)
[('123', '09827'), ('456', '9908872'), ('789', '267504')]
You can also use .finditer which might be helpful for longer strings.
>>> [m.groups() for m in r.finditer(x)]
[('123', '09827'), ('456', '9908872'), ('789', '267504')]

Use re.finditer or re.findall. Then you don't need the extra pair of parentheses that wrap the entire expression. For example,
>>> import re
>>> x = "--x123-09827--x456-9908872--x789-267504"
>>> p = "--x([0-9]+)-([0-9]+)"
>>> for m in re.finditer(p,x):
>>> print '{0} {1}'.format(m.group(1),m.group(2))

try this
p='--x([0-9]+)-([0-9]+)'
re.findall(p,x)

No need to use regex :
>>> "--x123-09827--x456-9908872--x789-267504".replace('--x',' ').replace('-',':').strip()
'123:09827 456:9908872 789:267504'

You don't need regular expressions for this. Here is a simple one-liner, non-regex solution:
>>> input = "--x123-09827--x456-9908872--x789-267504"
>>> [ x.replace("-", ":") for x in input.split("--x")[1:] ]
['123:09827', '456:9908872', '789:267504']
If this is an exercise on regex, here is a solution that uses the repetition (technically), though the findall(...) solution may be preferred:
>>> import re
>>> input = "--x123-09827--x456-9908872--x789-267504"
>>> regex = '--x(.+)'
>>> [ x.replace("-", ":") for x in re.match(regex*3, input).groups() ]
['123:09827', '456:9908872', '789:267504']

How would you make a comma-separated string from a list of strings?

What would be your preferred way to concatenate strings from a sequence such that between every two consecutive pairs a comma is added. That is, how do you map, for instance, ['a', 'b', 'c'] to 'a,b,c'? (The cases ['s'] and [] should be mapped to 's' and '', respectively.)
I usually end up using something like ''.join(map(lambda x: x+',',l))[:-1], but also feeling somewhat unsatisfied.

my_list = ['a', 'b', 'c', 'd']
my_string = ','.join(my_list)
'a,b,c,d'
This won't work if the list contains integers
And if the list contains non-string types (such as integers, floats, bools, None) then do:
my_string = ','.join(map(str, my_list))

Why the map/lambda magic? Doesn't this work?
>>> foo = ['a', 'b', 'c']
>>> print(','.join(foo))
a,b,c
>>> print(','.join([]))
>>> print(','.join(['a']))
a
In case if there are numbers in the list, you could use list comprehension:
>>> ','.join([str(x) for x in foo])
or a generator expression:
>>> ','.join(str(x) for x in foo)

",".join(l) will not work for all cases. I'd suggest using the csv module with StringIO
import StringIO
import csv
l = ['list','of','["""crazy"quotes"and\'',123,'other things']
line = StringIO.StringIO()
writer = csv.writer(line)
writer.writerow(l)
csvcontent = line.getvalue()
# 'list,of,"[""""""crazy""quotes""and\'",123,other things\r\n'

Here is a alternative solution in Python 3.0 which allows non-string list items:
>>> alist = ['a', 1, (2, 'b')]
a standard way
>>> ", ".join(map(str, alist))
"a, 1, (2, 'b')"
the alternative solution
>>> import io
>>> s = io.StringIO()
>>> print(*alist, file=s, sep=', ', end='')
>>> s.getvalue()
"a, 1, (2, 'b')"
NOTE: The space after comma is intentional.

#Peter Hoffmann
Using generator expressions has the benefit of also producing an iterator but saves importing itertools. Furthermore, list comprehensions are generally preferred to map, thus, I'd expect generator expressions to be preferred to imap.
>>> l = [1, "foo", 4 ,"bar"]
>>> ",".join(str(bit) for bit in l)
'1,foo,4,bar'

Don't you just want:
",".join(l)
Obviously it gets more complicated if you need to quote/escape commas etc in the values. In that case I would suggest looking at the csv module in the standard library:
https://docs.python.org/library/csv.html

>>> my_list = ['A', '', '', 'D', 'E',]
>>> ",".join([str(i) for i in my_list if i])
'A,D,E'
my_list may contain any type of variables. This avoid the result 'A,,,D,E'.

l=['a', 1, 'b', 2]
print str(l)[1:-1]
Output: "'a', 1, 'b', 2"

#jmanning2k using a list comprehension has the downside of creating a new temporary list. The better solution would be using itertools.imap which returns an iterator
from itertools import imap
l = [1, "foo", 4 ,"bar"]
",".join(imap(str, l))

Here is an example with list
>>> myList = [['Apple'],['Orange']]
>>> myList = ','.join(map(str, [i[0] for i in myList]))
>>> print "Output:", myList
Output: Apple,Orange
More Accurate:-
>>> myList = [['Apple'],['Orange']]
>>> myList = ','.join(map(str, [type(i) == list and i[0] for i in myList]))
>>> print "Output:", myList
Output: Apple,Orange
Example 2:-
myList = ['Apple','Orange']
myList = ','.join(map(str, myList))
print "Output:", myList
Output: Apple,Orange

If you want to do the shortcut way :) :
','.join([str(word) for word in wordList])
But if you want to show off with logic :) :
wordList = ['USD', 'EUR', 'JPY', 'NZD', 'CHF', 'CAD']
stringText = ''
for word in wordList:
stringText += word + ','
stringText = stringText[:-2] # get rid of last comma
print(stringText)

Unless I'm missing something, ','.join(foo) should do what you're asking for.
>>> ','.join([''])
''
>>> ','.join(['s'])
's'
>>> ','.join(['a','b','c'])
'a,b,c'
(edit: and as jmanning2k points out,
','.join([str(x) for x in foo])
is safer and quite Pythonic, though the resulting string will be difficult to parse if the elements can contain commas -- at that point, you need the full power of the csv module, as Douglas points out in his answer.)

I would say the csv library is the only sensible option here, as it was built to cope with all csv use cases such as commas in a string, etc.
To output a list l to a .csv file:
import csv
with open('some.csv', 'w', newline='') as f:
writer = csv.writer(f)
writer.writerow(l) # this will output l as a single row.
It is also possible to use writer.writerows(iterable) to output multiple rows to csv.
This example is compatible with Python 3, as the other answer here used StringIO which is Python 2.

mmm also need for SQL is :
l = ["foo" , "baar" , 6]
where_clause = "..... IN ("+(','.join([ f"'{x}'" for x in l]))+")"
>> "..... IN ('foo','baar','6')"
enjoit

My two cents. I like simpler an one-line code in python:
>>> from itertools import imap, ifilter
>>> l = ['a', '', 'b', 1, None]
>>> ','.join(imap(str, ifilter(lambda x: x, l)))
a,b,1
>>> m = ['a', '', None]
>>> ','.join(imap(str, ifilter(lambda x: x, m)))
'a'
It's pythonic, works for strings, numbers, None and empty string. It's short and satisfies the requirements. If the list is not going to contain numbers, we can use this simpler variation:
>>> ','.join(ifilter(lambda x: x, l))
Also this solution doesn't create a new list, but uses an iterator, like #Peter Hoffmann pointed (thanks).