Good evening,
I have a python variable like so
myList = ["['Ben'", " 'Dillon'", " 'Rawr'", " 'Mega'", " 'Tote'", " 'Case']"]
I would like it to look like this instead
myList = ['Ben', 'Dillon', 'Rawr', 'Mega', 'Tote', 'Case']
If I do something like this
','.join(myList)
It gives me what I want but the type is a String
I also would like it to keep the type of List. I have tried using the Join method and split method. And I have been debugging use the type() method. It tells me that the type in the original scenario is a list.
I appreciate any and all help on this.
Join the inner list elements, then call ast.literal_eval() to parse it as a list of strings.
import ast
myList = ast.literal_eval(",".join(myList))
Also can be done by truncating Strings, therefore avoiding the import of ast.
myList[5] = (myList[5])[:-1]
for n in range(0, len(myList)):
myList[n] = (myList[n])[2:-1]
Related
I have lists of strings, some are hashtags - like #rabbitsarecool others are short pieces of prose like "My rabbits name is fred."
I have written a program to seperate them:
def seperate_hashtags_from_prose(*strs):
props = []
hashtags = []
for x in strs:
if x[0]=="#" and x.find(' ')==-1:
hashtags += x
else:
prose += x
return hashtags, prose
seperate_hashtags_from_prose(["I like cats","#cats","Rabbits are the best","#Rabbits"])
This program does not work. in the above example when i debug it, it tells me that on the first loop:
x=["I like cats","#cats","Rabbits are the best",#Rabbits].
Thisis not what I would have expected - my intuition is that something about the way the loop over optional arguments is constructed is causing an error- but i can't see why.
There are several issues.
The most obvious is switching between props and prose. The code you posted does not run.
As others have commented, if you use the * in the function call, you should not make the call with a list. You could use seperate_hashtags_from_prose("I like cats","#cats","Rabbits are the best","#Rabbits") instead.
The line hashtags += x does not do what you think it does. When you use + as an operator on iterables (such as list and string) it will concatenate them. You probably meant hashtags.append(x) instead.
I need help using Python.
Supposing I have the list [22,23,45].
Is it possible to get an output like this: [22;23:45] ?
It's possible to change the delimiters if you display your list as a string. You can then use the join method. The following example will display your list with ; as a delimiter:
print(";".join(my_list))
This will only work if your list's items are string, by the way.
Even if you have more than one item
str(your_list[:1][0])+";" + ":".join(map(str,your_list[1:])) #'22;23:45'
Not sure why you want to wrap in the list but if you do just wrap around the above string in list()
my list which was [22,23,45] returned [;2;2;,; ;2;3;,; ;4;5;,] for both methods.
To bring more information, I have a variable:
ID= [elements['id'] for elements in country]
Using a print (ID), I get [22,23,45] so I suppose that the list is already to this form.
Problem is: I need another delimiters because [22,23,45] corresponds to ['EU, UK', 'EU, Italy', 'USA, California'].
The output I wish is [22,23,45] --> ['EU, UK'; 'EU, Italy'; 'USA, California']
I don't know if it's clearer but hope it could help
Try this I don't know exactly what do You want?
first solution:
list = [22,23,45]
str = ""
for i in list:
str += "{}{}".format(i, ";")
new_list=str[:-len(";")]
print(new_list)
and this is second solution
list = [22,23,45]
print(list)
list=str(list)
list=list.split(",")
list=";".join(list)
print(list)
If I want to define a function called match_numbers, which would match the area code from one list to the phone number of another list, how should I fix my code? For example:
match_phone(['666', '332'], ['(443)241-1254', '(666)313-2534', '(332)123-3332'])
would give me
(666)313-2534
(332)123-3332
My code is:
def phone (nlist, nlist1):
results = {}
for x in nlist1:
results.setdefault(x[0:3], [])
results[x[0:3]].append(x)
for x in nlist:
if x in results:
print(results[x])
The problem with this code is, however:
It gives me the outputs in brackets, whereas I want it to print
the output line by line like shown above, and
it won't work with the parantheses in the 2nd list (for example
(666)543-2322 must be converted as 666-543-2322 for the list to
work.
Now, there are better/faster approaches to do what you are trying to do, but let us focus on fixing your code.
The first issue you have is how you are slicing your string. Remember that you start at index 0. So if you do:
x[0:3]
What you are actually getting is something like this from your string:
(12
Instead of your intended:
123
So, knowing that indexes start at 0, what you actually want to do is slice your string as such:
x[1:4]
Finally, your line here:
results[x[0:3]].append(x)
There are two problems here.
First, as mentioned above, you are still trying to slice the wrong parts of your string, so fix that.
Second, since you are trying to make a key value pair, what that above line is actually doing is making a key value pair where the value is a list. I don't think you want to do that. You want to do something like:
{'123': '(123)5556666'}
So, you don't want to use the append in this case. What you want to do is assign the string directly as the value for that key. You can do that as such:
results[x[1:4]] = x
Finally, another problem that was noticed, is in what you are doing here:
results.setdefault(x[1:4], [])
Based on the above explanation on how you want to store a string as your value in your dictionary instead of a list, so you don't need to be doing this. Therefore, you should simply be removing that line, it does not serve any purpose for what you are trying to do. You have already initialized your dictionary as results = {}
When you put it all together, your code will look like this:
def match_phone(nlist, nlist1):
results = {}
for x in nlist1:
results[x[1:4]] = x
for x in nlist:
if x in results:
print(results[x])
match_phone(['666', '332'], ['(443)241-1254', '(666)313-2534', '(332)123-3332'])
And will provide the following output:
(666)313-2534
(332)123-3332
If all the phone numbers will be in the format (ddd)ddd-dddd you can use
for number in (num for num in nlist1 if num[1:4] in nlist):
print(number)
You could use some better variable names than nlist and nlist1, in my view.
def match_phone(area_codes, numbers):
area_codes = set(area_codes)
for num in numbers:
if num in area_codes:
print num
You could do something like this:
phone_numbers = ['(443)241-1254', '(666)313-2534', '(332)123-3332']
area_codes = ['666', '332']
numbers = filter(lambda number: number[1:4] in area_codes, phone_numbers)
for number in numbers:
print(number)
Another similar way to do this without using a filter could be something like this:
for number in phone_numbers:
if number[1:4] in area_codes:
print(number)
Output in either case would be:
(666)313-2534
(332)123-3332
No one with regex solution! This may be an option too.
import re
def my_formatter(l1,l2):
mydic = {re.match(r'([(])([0-9]+)([)])([0-9]+[-][0-9]+)',i).group(2):re.match(r'([(])([0-9]+)([)])([0-9]+[-][0-9]+)',i).group(4) for i in l2}
for i in l1:
print "({0}){1}".format(str(i),str(mydic.get(i)))
my_formatter(['666', '332'], ['(443)241-1254', '(666)313-2534', '(332)123-3332'])
It prints-
(666)313-2534
(332)123-3332
I need to declare certain values in List.
Values looks like this:
["compute","controller"], ["compute"] ,["controller"]
I know the List syntax in python is
example = []
I am not sure how I will include square brackets and double quotes in the List.
Could anyone please help.
I tried the following:
cls.node = ["\["compute"\]","\["controller"\]"]
cls.node = ["[\"compute\"]","[\"controller\"]"]
Both did not work.
I think you mean list not dictionary because that is the syntax of a list:
You can simply do it using the following format '"Hello"':
cls.node = ['["compute"]','["controller"]']
cls.node = ['["compute"]','["controller"]']
Demo:
s = ['["hello"]', '["world"]']
for i in s:
print i
[OUTPUT]
["hello"]
["world"]
After a MySQL select statement, I am left with the following:
set([('1#a.com',), ('2#b.net',), ('3#c.com',), ('4#d.com',), ('5#e.com',), ('6#f.net',), ('7#h.net',), ('8#g.com',)])
What I would like to have is a
emaillist = "\n".join(queryresult)
to in the end, have a string:
1#a.com
2#b.net
3#c.com
etc
What would be the proper way to convert this nested tuple into string?
As long as you're sure you have just one element per tuple:
'\n'.join(elem[0] for elem in queryresult)
A list comprehension along the lines of
[item[0] for item in queryresult]
should help. eg,
emaillist = "\n".join([item[0] for item in queryresult])
.. this makes strong assumptions about the type and structure of queryresult, however.
edit
A generator expression is even better at doing this:
emaillist = "\n".join(item[0] for item in queryresult)
thanks to #delnan for this update
Try this and see
>>> something=set([('1#a.com',), ('2#b.net',), ('3#c.com',), ('4#d.com',), ('5#e.com',), ('6#f.net',), ('7#h.net',), ('8#g.com',)])
>>> print '\n'.join(''.join(s) for s in something)
6#f.net
7#h.net
2#b.net
1#a.com
8#g.com
5#e.com
4#d.com
3#c.com
Note, join can only work on strings, but the items of the set are tuples. To make the join work, you have to iterate over the set to convert each item as string. Finally once done you can join them in the way your heart desires
On a side note. A circumbendibus way of doing it :-)
>>> print "\n".join(re.findall("\'(.*?)\'",pprint.pformat(something)))
1#a.com
2#b.net
3#c.com
4#d.com
5#e.com
6#f.net
7#h.net
8#g.com
new = '\n'.join(x[0] for x in old)