I am new to python and I have a string that looks like this
Temp = "', '/1412311.2121\n"
my desired output is just getting the numbers and decimal itself.. so im looking for
1412311.2121
as the output.. trying to get rid of the ', '/\n in the string.. I have tried Temp.strip("\n") and Temp.rstrip("\n") for trying to remove \n but i still seems to remain in my string. :/... Does anyone have any ideas? Thanks for your help.
Strings are immutable. string.strip() doesn't change string, it's a function that returns a value. You need to do:
Temp = Temp.strip()
Note also that calling strip() without any parameters causes it to remove all whitespace characters, including \n
As stalk said, you can achieve your desired result by calling strip("',/\n") on Temp.
If the data are like you show, numbers that are wrapped from right and left with non-number data, you can use a very simple regular expression:
g = re.search('[0-9.]+', s) # capture the inner number only
print g.group(0)
I would use a regular expression to do this:
In [8]: s = "', '/1412311.2121\n"
In [9]: re.findall(r'([+-]?\d+(?:\.\d+)?(?:[eE][+-]\d+)?)', s)
Out[9]: ['1412311.2121']
This returns a list of all floating-point numbers found in the string.
Related
string = "hi())("
string = string.rstrip("abcdefghijklmnoprstuwxyz")
print(string)
I want to remove every letter from given string using rstrip method, however it does not change the string in the slightest.
Output:
'hi())('
What i Want:
'())('
I know that I can use regex, but I really don't understand why it doesn't work.
Note : It is a part of the Valid Parentheses challenge on code-wars
You have to use lstrip instead of rstrip:
>>> string = "hi())("
>>> string = string.lstrip("abcdefghijklmnoprstuwxyz")
>>> string
'())('
I have the following file names that exhibit this pattern:
000014_L_20111007T084734-20111008T023142.txt
000014_U_20111007T084734-20111008T023142.txt
...
I want to extract the middle two time stamp parts after the second underscore '_' and before '.txt'. So I used the following Python regex string split:
time_info = re.split('^[0-9]+_[LU]_|-|\.txt$', f)
But this gives me two extra empty strings in the returned list:
time_info=['', '20111007T084734', '20111008T023142', '']
How do I get only the two time stamp information? i.e. I want:
time_info=['20111007T084734', '20111008T023142']
I'm no Python expert but maybe you could just remove the empty strings from your list?
str_list = re.split('^[0-9]+_[LU]_|-|\.txt$', f)
time_info = filter(None, str_list)
Don't use re.split(), use the groups() method of regex Match/SRE_Match objects.
>>> f = '000014_L_20111007T084734-20111008T023142.txt'
>>> time_info = re.search(r'[LU]_(\w+)-(\w+)\.', f).groups()
>>> time_info
('20111007T084734', '20111008T023142')
You can even name the capturing groups and retrieve them in a dict, though you use groupdict() rather than groups() for that. (The regex pattern for such a case would be something like r'[LU]_(?P<groupA>\w+)-(?P<groupB>\w+)\.')
If the timestamps are always after the second _ then you can use str.split and str.strip:
>>> strs = "000014_L_20111007T084734-20111008T023142.txt"
>>> strs.strip(".txt").split("_",2)[-1].split("-")
['20111007T084734', '20111008T023142']
Since this came up on google and for completeness, try using re.findall as an alternative!
This does require a little re-thinking, but it still returns a list of matches like split does. This makes it a nice drop-in replacement for some existing code and gets rid of the unwanted text. Pair it with lookaheads and/or lookbehinds and you get very similar behavior.
Yes, this is a bit of a "you're asking the wrong question" answer and doesn't use re.split(). It does solve the underlying issue- your list of matches suddenly have zero-length strings in it and you don't want that.
>>> f='000014_L_20111007T084734-20111008T023142.txt'
>>> f[10:-4].split('-')
['0111007T084734', '20111008T023142']
or, somewhat more general:
>>> f[f.rfind('_')+1:-4].split('-')
['20111007T084734', '20111008T023142']
I have a number of strings from which I am aiming to remove charactars using replace. However, this dosent seem to wake. To give a simplified example, this code:
row = "b'James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38'"
row = row.replace("b'", "").replace("'", "").replace('b"', '').replace('"', '')
print(row.encode('ascii', errors='ignore'))
still ouputs this b'James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38' wheras I would like it to output James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38. How can I do this?
Edit: Updataed the code with a better example.
You seem to be mistaking single quotes for double quotes. Simple replace 'b:
>>> row = "xyz'b"
>>> row.replace("'b", "")
'xyz'
As an alternative to str.replace, you can simple slice the string to remove the unwanted leading and trailing characters:
>>> row[2:-1]
'James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38'
In your first .replace, change b' to 'b. Hence your code should be:
>>> row = "xyz'b"
>>> row = row.replace("'b", "").replace("'", "").replace('b"', '').replace('"', '')
# ^ changed here
>>> print(row.encode('ascii', errors='ignore'))
xyz
I am assuming rest of the conditions you have are the part of other task/matches that you didn't mentioned here.
If all you want is to take the string before first ', then you may just do:
row.split("'")[0]
You haven't listed this to remove 'b:
.replace("'b", '')
import ast
row = "b'James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38'"
b_string = ast.literal_eval(row)
print(b_string)
u_string = b_string.decode('utf-8')
print(u_string)
out:
b_string:b'James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38'
u_string: James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38
The real question is how to convert a string to python object.
You get a string which contains an a binary string, to convert it to python's binary string object, you should use eval(). ast.literal_eval() is more safe way to do it.
Now you get a binary string, you can convert it to unicode string which do not start with "b" by using decode()
I have tried below code to split but I am unable to split
import re
s = "abcd[00451.00]"
print str(s).strip('[]')
I need output as only number or decimal format 00451.00 this value but I am able to get output as abcd[00451.00
If you know for sure that there will be one opening and closing brackets you can do
s = "abcd[00451.00]"
print s[s.index("[") + 1:s.rindex("]")]
# 00451.00
str.index is used to get the first index of the element [ in the string, where as str.rindex is used to get the last index of the element in ]. Based on those indexes, the string is sliced.
If you want to convert that to a floating point number, then you can use float function, like this
print float(s[s.index("[") + 1:s.rindex("]")])
# 451.0
You should use re.search:
import re
s = "abcd[00451.00]"
>>> print re.search(r'\[([^\]]+)\]', s).group(1)
00451.00
You can first split on the '[' and then strip the resulting list of any ']' chars:
[p.strip(']') for p in s.split('[')]
I have a variable string with unknown length that has the important string at the left side and the unimportant things on the right side separated by a single space. How do I remove the unimportant information to the right?
I have tried rstrip, and split with no success.
Edit: I'll place the actual value that needs to be fixed.
"NPC_tester_contact() ) ntact() "
The very first space (the one left to the closed parenthesis) should have everything after including itself be marked as unimportant.
Edit: The output should be "NPC_tester_contact()"!
Look carefully at my string that I placed above. There is alot of whitespace after it as well. I assume that is what is causing the hiccup.
I have tried most of the solutions here and they either don't do anything or just produce whitespace.
repr(s) gives me.
'NPC_me_lvup_event_contact()\x00t()\x00act()\x00act()\x00ntact()\x00\x00\x00\x00
\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00
\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00
\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00
\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00
\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00
\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00
\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00
\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00
\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00
\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00'
It should be "NPC_me_lvup_event_contact()".
Thanks!
Perhaps this is a better question. Is there a way to remove all characters after the first \x00 hex that shows up in the string?
For some reason, it works sometimes and doesn't always work. The above example was done with the method that Levon posted.
Solution: Problem solved. This is more of a null byte rather than a space byte. The solution would of been any of the below using "\x00" as the identifier instead of " ".
Thank you everyone!
UPDATE based on new string data:
Assuming s contains your string:
s.split('\x00')[0]
yields
'NPC_me_lvup_event_contact()'
split() will give you a list of strings separated by the character you specify with split. If none is specified space is used, in this case we use the hex character you are interested in.
USE split(' ')[0]
>>> a = 'aaa bbb'
>>> a.split(' ')[0]
'aaa'
>>> >
>>> mystring = 'important useless'
>>> mystring[:mystring.find(' ')]
'important'
split() w/o delimiter splits by any whitespace:
>>> "asdasd xyz".split()[0]
'asdasd'
str = "important unimportant"
important = str.split(' ')[0]
try this:
lhs,rhs=s.split() #lhs is what you want.
This only works if there is really only one space.
Otherwise, you can get lhs by (but you lose rhs):
lhs=s.split()[0]
Use the split() function, and get the first item that it returns:
raw_string = 'NPC_tester_contact() ) ntact() '
important = raw_string.split(' ')[0]
Will return:
NPC_tester_contact()
try this,
will assume that your string is stored in str
print str[0:str.index(" ")]
comment if it dont work, will solve it..
here is
My code
str = "NPC_tester_contact() ) ntact() "
print str[0:str.index(" ")]
output
NPC_tester_contact()
link
http://ideone.com/i9haI
and if you want output to be have surrounded with double-quotes then
`print '"',str[0:str.index(" ")],'"'
you could use a regex type solution also. Something like:
import re
input_string = 'NPC_me_lvup_event_contact()\x00t()\x00act()\x00act()\x00ntact()\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00'
string_pat = re.compile(r'[a-zA-Z0-9\(\)_]+')
try:
first_part = string_pat.findall(input_string)[0]
except IndexError:
# There is nothing of interest for you in this string
first_part = ''