I have the follow python code which is a list. For some reason I get unwanted characters in the list how can we remove these characters from the beginning and end of list.
stuff = []
a = ('{"type": "push"}')
stuff.append(a)
print(stuff)
outputs
['{"type": "push"}']
How can we remove the [''] and output like
{"type": "push"}
By doing the following :
a = ('{"type": "push"}')
you are appending a string (type(a) returns str). What you might want instead is to append a dictionary
a = {"type": "push"} # ===> output = [{"type": "push"}]
which will give you the right output
I think you are assuming that a will be a tuple. However, that's not the case. In your current implementation it will be a string. That's why when you append(a) the whole string gets appended. Use any other data structures and there is no need of extra quotes that you don't require to be present.
Related
I've seen a lot of examples on how to remove brackets from a string in Python, but I've not seen any that allow me to remove the brackets and a number inside of the brackets from that string.
For example, suppose I've got a string such as "abc[1]". How can I remove the "[1]" from the string to return just "abc"?
I've tried the following:
stringTest = "abc[1]"
stringTestWithoutBrackets = str(stringTest).strip('[]')
but this only outputs the string without the final bracket
abc[1
I've also tried with a wildcard option:
stringTest = "abc[1]"
stringTestWithoutBrackets = str(stringTest).strip('[\w+\]')
but this also outputs the string without the final bracket
abc[1
You could use regular expressions for that, but I think the easiest way would be to use split:
>>> stringTest = "abc[1][2][3]"
>>> stringTest.split('[', maxsplit=1)[0]
'abc'
You can use regex but you need to use it with the re module:
re.sub(r'\[\d+\]', '', stringTest)
If the [<number>] part is always at the end of the string you can also strip via:
stringTest.rstrip('[0123456789]')
Though the latter version might strip beyond the [ if the previous character is in the strip list too. For example in "abc1[5]" the "1" would be stripped as well.
Assuming your string has the format "text[number]" and you only want to keep the "text", then you could do:
stringTest = "abc[1]"
bracketBegin = stringTest.find('[')
stringTestWithoutBrackets = stringTest[:bracketBegin]
When I'm splitting a string "abac" I'm getting undesired results.
Example
print("abac".split("a"))
Why does it print:
['', 'b', 'c']
instead of
['b', 'c']
Can anyone explain this behavior and guide me on how to get my desired output?
Thanks in advance.
As #DeepSpace pointed out (referring to the docs)
If sep is given, consecutive delimiters are not grouped together and are deemed to delimit empty strings (for example, '1,,2'.split(',') returns ['1', '', '2']).
Therefore I'd suggest using a better delimiter such as a comma , or if this is the formatting you're stuck with then you could just use the builtin filter() function as suggested in this answer, this will remove any "empty" strings if passed None as the function.
sample = 'abac'
filtered_sample = filter(None, sample.split('a'))
print(filtered_sample)
#['b', 'c']
When you split a string in python you keep everything between your delimiters (even when it's an empty string!)
For example, if you had a list of letters separated by commas:
>>> "a,b,c,d".split(',')
['a','b','c','d']
If your list had some missing values you might leave the space in between the commas blank:
>>> "a,b,,d".split(',')
['a','b','','d']
The start and end of the string act as delimiters themselves, so if you have a leading or trailing delimiter you will also get this "empty string" sliced out of your main string:
>>> "a,b,c,d,,".split(',')
['a','b','c','d','','']
>>> ",a,b,c,d".split(',')
['','a','b','c','d']
If you want to get rid of any empty strings in your output, you can use the filter function.
If instead you just want to get rid of this behavior near the edges of your main string, you can strip the delimiters off first:
>>> ",,a,b,c,d".strip(',')
"a,b,c,d"
>>> ",,a,b,c,d".strip(',').split(',')
['a','b','c','d']
In your example, "a" is what's called a delimiter. It acts as a boundary between the characters before it and after it. So, when you call split, it gets the characters before "a" and after "a" and inserts it into the list. Since there's nothing in front of the first "a" in the string "abac", it returns an empty string and inserts it into the list.
split will return the characters between the delimiters you specify (or between an end of the string and a delimiter), even if there aren't any, in which case it will return an empty string. (See the documentation for more information.)
In this case, if you don't want any empty strings in the output, you can use filter to remove them:
list(filter(lambda s: len(s) > 0, "abac".split("a"))
I have strings as tuples that I'm trying to remove quotation marks from. If there isn't a comma present in the string, then it works. But if there is a comma, then quotation marks still remain:
example = [('7-30-17','0x34','"Upload Complete"'),('7-31-17','0x35','"RCM","Interlock error"')]
example = [(x,y,(z.strip('"')))
for x,y,z in example]
The result is that quotation marks partially remain in the strings that had commas in them. The second tuple now reads RCM","Interlock error as opposed to RCM, Interlock error
('7-30-17','0x34','Upload Complete')
('7-31-17','0x35','RCM","Interlock error')
Any ideas what I'm doing wrong? Thanks!
You can use list comprehension to iterate the list items and similarly for the inner tuple items
>>> [tuple(s.replace('"','') for s in tup) for tup in example]
[('7-30-17', '0x34', 'Upload Complete'), ('7-31-17', '0x35', 'RCM,Interlock error')]
It seems like you're looking for the behaviour of replace(), rather than strip().
Try using replace('"', '') instead of strip('"'). strip only removes characters from the beginning and end of strings, while replace will take care of all occurrences.
Your example would be updated to look like this:
example = [('7-30-17','0x34','"Upload Complete"'),('7-31-17','0x35','"RCM","Interlock error"')]
example = [(x,y,(z.replace('"', '')))
for x,y,z in example]
example ends up with the following value:
[('7-30-17', '0x34', 'Upload Complete'), ('7-31-17', '0x35', 'RCM,Interlock error')]
The problem is because strip will remove only from ends of string.
Use a regex to replace ":
import re
example = [('7-30-17','0x34','"Upload Complete"'),('7-31-17','0x35','"RCM","Interlock error"')]
example = [(x,y,(re.sub('"','',z)))
for x,y,z in example]
print(example)
# [('7-30-17', '0x34', 'Upload Complete'), ('7-31-17', '0x35', 'RCM,Interlock error')]
I want to remove specific value from a Unicode list i.e field
u'abv,( field),apn,army,elev,fema'
But when i try something like result.remove ('(field)') it stops working and gives an error ?
Convert it into list and use remove
s = u'abv,( field),apn,army,elev,fema'
res = s.split(",")
res.remove("army") # lets assume we need to remove army
['abv', '( field)', 'apn', 'elev', 'fema']
You can make your output list back to string as well, if you wish
output = ",".join(res)
'abv,( field),apn,elev,fema'
Is there a way to pass in a list instead of a char to str.strip() in python? I have been doing it this way:
unwanted = [c for c in '!##$%^&*(FGHJKmn']
s = 'FFFFoFob*&%ar**^'
for u in unwanted:
s = s.strip(u)
print s
Desired output, this output is correct but there should be some sort of a more elegant way than how i'm coding it above:
oFob*&%ar
Strip and friends take a string representing a set of characters, so you can skip the loop:
>>> s = 'FFFFoFob*&%ar**^'
>>> s.strip('!##$%^&*(FGHJKmn')
'oFob*&%ar'
(the downside of this is that things like fn.rstrip(".png") seems to work for many filenames, but doesn't really work)
Since, you are looking to not delete elements from the middle, you can just use.
>>> 'FFFFoFob*&%ar**^'.strip('!##$%^&*(FGHJKmn')
'oFob*&%ar'
Otherwise, Use str.translate().
>>> 'FFFFoFob*&%ar**^'.translate(None, '!##$%^&*(FGHJKmn')
'oobar'