I have a list i need to .join as string and append characters
my_list = ['3.3.3.3', '2.2.2.3', '2.2.2.2']
my_list.append(')"')
my_list.insert(0,'"(')
hostman = '|'.join('{0}'.format(w) for w in my_list)
#my_list.pop()
print(hostman)
print(my_list)
My output = "(|3.3.3.3|2.2.2.3|2.2.2.2|)"
I need the output to be = "(3.3.3.3|2.2.2.3|2.2.2.2)"
how can i strip the first and last | from the string
You are making it harder than it needs to be. You can just use join() directly with the list:
my_list = ['3.3.3.3', '2.2.2.3', '2.2.2.2']
s = '"(' + '|'.join(my_list) + ')"'
# s is "(3.3.3.3|2.2.2.3|2.2.2.2)"
# with quotes as part of the string
or if you prefer format:
s = '"({})"'.format('|'.join(my_list))
Try this :
hostman = "("+"|".join(my_list)+")"
OUTPUT :
'(3.3.3.3|2.2.2.3|2.2.2.2)'
Related
While I am writing this code lst=list(map(int,input().split().strip())) then I am getting an AttributeError 'list' object has no attribute strip
But it is working when I remove the strip() method.
My question is that list object also has no attribute split. So in this case (lst=list(map(int,input().split())) why it is not giving any error and why it is giving error in case of strip() method?
Before you read the rest of the answer: you shouldn't have to strip() after you call split() because split() will consider multiple whitespace characters as a single delimiter and automatically remove the extra whitespace. For example, this snippet evaluates to True:
s1 = "1 2 3"
s2 = "1 2 3"
s3 = " 1 2 3 "
s1.split() == s2.split() == s3.split()
split() and strip() are both attributes of string objects!
When you're confused by code that's been stuffed into one line, it often helps to unravel that code out over multiple lines to understand what it's doing
Your line of code can be unraveled like so:
user_input_str = input()
split_input_list = user_input_str.split()
stripped_input = split_input_list.strip() ### ERROR!!!
lst = list(map(int, stripped_input))
Clearly, you tried to access the strip() method of a list object, and you know that doesn't exist.
In your second example, you do
user_input_str = input()
split_input_list = user_input_str.split()
lst = list(map(int, split_input_list))
Which works perfectly fine because you don't try to access strip() on a list object
Now to fix this, you need to change the order of operations: first, you get your input. Next, strip it. This gives you back a string. Then, split this stripped string.
user_input_str = input()
stripped_input_str = user_input_str.strip() ### No error now!
split_input_list = stripped_input_str.split()
lst = list(map(int, split_input_list))
#or in one line:
lst = list(map(int, input().strip().split()))
Or, if you want to strip each element of the split input list, you will need to map the strip() function to split_input_list like so:
user_input_str = input()
split_input_list = user_input_str.split()
stripped_input_list = list(map(str.strip, split_input_list))
lst = list(map(int, stripped_input_list))
#or in one line
lst = list(map(int, map(str.strip, input().split())))
# or, create a function that calls strip and then converts to int, and map to it
def stripint(value):
return int(value.strip())
lst = list(map(stripint, input().split()))
I have a string for example "streemlocalbbv"
and I have my_function that takes this string and a string that I want to find ("loc") in the original string. And what I want to get returned is this;
my_function("streemlocalbbv", "loc")
output = ["streem","loc","albbv"]
what I did so far is
def find_split(string,find_word):
length = len(string)
find_word_start_index = string.find(find_word)
find_word_end_index = find_word_start_index + len(find_word)
string[find_word_start_index:find_word_end_index]
a = string[0:find_word_start_index]
b = string[find_word_start_index:find_word_end_index]
c = string[find_word_end_index:length]
return [a,b,c]
Trying to find the index of the string I am looking for in the original string, and then split the original string. But from here I am not sure how should I do it.
You can use str.partition which does exactly what you want:
>>> "streemlocalbbv".partition("loc")
('streem', 'loc', 'albbv')
Use the split function:
def find_split(string,find_word):
ends = string.split(find_word)
return [ends[0], find_word, ends[1]]
Use the split, index and insert function to solve this
def my_function(word,split_by):
l = word.split(split_by)
l.insert(l.index(word[:word.find(split_by)])+1,split_by)
return l
print(my_function("streemlocalbbv", "loc"))
#['str', 'eem', 'localbbv']
I've read some switch MAC address table into a file and for some reason the MAC address if formatted as such:
'aabb.eeff.hhii'
This is not what a MAC address should be, it should follow: 'aa:bb:cc:dd:ee:ff'
I've had a look at the top rated suggestions while writing this and found an answer that may fit my needs but it doesn't work
satomacoto's answer
The MACs are in a list, so when I run for loop I can see them all as such:
Current Output
['8424.aa21.4er9','fa2']
['94f1.3002.c43a','fa1']
I just want to append ':' at every 2nd nth character, I can just remove the '.' with a simple replace so don't worry about that
Desired output
['84:24:aa:21:4e:r9','fa2']
['94:f1:30:02:c4:3a','fa1']
My code
info = []
newinfo = []
file = open('switchoutput')
newfile = file.read().split('switch')
macaddtable = newfile[3].split('\\r')
for x in macaddtable:
if '\\n' in x:
x = x.replace('\\n', '')
if carriage in x:
x = x.replace(carriage, '')
if '_#' in x:
x = x.replace('_#', '')
x.split('/r')
info.append(x)
for x in info:
if "Dynamic" in x:
x = x.replace('Dynamic', '')
if 'SVL' in x:
x = x.replace('SVL', '')
newinfo.append(x.split(' '))
for x in newinfo:
for x in x[:1]:
if '.' in x:
x = x.replace('.', '')
print(x)
Borrowing from the solution that you linked, you can achieve this as follows:
macs = [['8424.aa21.4er9','fa2'], ['94f1.3002.c43a','fa1']]
macs_fixed = [(":".join(map(''.join, zip(*[iter(m[0].replace(".", ""))]*2))), m[1]) for m in macs]
Which yields:
[('84:24:aa:21:4e:r9', 'fa2'), ('94:f1:30:02:c4:3a', 'fa1')]
If you like regular expressions:
import re
dotted = '1234.3456.5678'
re.sub('(..)\.?(?!$)', '\\1:', dotted)
# '12:34:34:56:56:78'
The template string looks for two arbitrary characters '(..)' and assigns them to group 1. It then allows for 0 or 1 dots to follow '\.?' and makes sure that at the very end there is no match '(?!$)'. Every match is then replaced with its group 1 plus a colon.
This uses the fact that re.sub operates on nonoverlapping matches.
x = '8424.aa21.4er9'.replace('.','')
print(':'.join(x[y:y+2] for y in range(0, len(x) - 1, 2)))
>> 84:24:aa:21:4e:r9
Just iterate through the string once you've cleaned it, and grab 2 string each time you loop through the string. Using range() third optional argument you can loop through every second elements. Using join() to add the : in between the two elements you are iterating.
You can use re module to achieve your desired output.
import re
s = '8424.aa21.4er9'
s = s.replace('.','')
groups = re.findall(r'([a-zA-Z0-9]{2})', s)
mac = ":".join(groups)
#'84:24:aa:21:4e:r9'
Regex Explanation
[a-zA-Z0-9]: Match any alphabets or number
{2}: Match at most 2 characters.
This way you can get groups of two and then join them on : to achieve your desired mac address format
wrong_mac = '8424.aa21.4er9'
correct_mac = ''.join(wrong_mac.split('.'))
correct_mac = ':'.join(correct_mac[i:i+2] for i in range(0, len(correct_mac), 2))
print(correct_mac)
i'm new to python and i'm having a select statement like following help_category_id, name, what is the most effective way to convert this string to this:
'help_category_id', 'name'
i've currently done this, which works fine, but is there a nicer and more clean way to do the same:
test_string = 'help_category_id, name'
column_sort_list = []
if test_string is not None:
for col in test_string.split(','):
column = "'{column}'".format(column=col)
column_sort_list.append(column)
column_sort = ','.join(column_sort_list)
print(column_sort)
Simple one liner using looping constructs:
result = ", ".join(["'" + i + "'" for i.strip() in myString.split(",")])
What we are doing here is we are creating a list that contains all substrings of your original string, with the quotes added. Then, using join, we make that list into a comma delimited string.
Deconstructed, the looping construct looks like this:
resultList = []
for i in myString.split(","):
resultList.append("'" + i.strip() + "'")
Note the call to i.strip(), which removes extraneous spaces around each substring.
Note: You can use format syntax to make this code even cleaner:
New syntax:
result = ", ".join(["'{}'".format(i.strip()) for i in myString.split(",")])
Old syntax:
result = ", ".join(["'%s'" % i.strip() for i in myString.split(",")])
it can be achieved by this also.
','.join("'{}'".format(value) for value in map(lambda text: text.strip(), test_string.split(",")))
I have to strip whitespace for extracted strings, one string at a time for which I'm using split(). The split() function returns list after removing white spaces. I want to store this in my own dynamic list since I have to aggregate all of the strings.
The snippet of my code:
while rec_id = "ffff"
output = procs.run_cmd("get sensor info", command)
sdr_li = []
if output:
byte_str = output[0]
str_1 = byte_str.split(' ')
for byte in str_1:
sdr_li.append(byte)
rec_id = get_rec_id()
Output = ['23 0a 06 01 52 2D 12']
str_1 = ['23','0a','06','01','52','2D','12']
This does not look very elegant, transferring from one list to another. Is there another way to achieve this.
list.extend():
sdr_li.extend(str_1)
str.split() returns you a list so just add your list's items to the main list. Use extend https://docs.python.org/2/tutorial/datastructures.html
so rewriting your data into something legible and properly indented you'd get:
my_list = list
while rec_id = "ffff"
output = procs.run_cmd("get sensor info", command)
if output:
result_string = output[0]
# extend my_list with the list resulting from the whitespace
# seperated tokens of the output
my_list.extend( result_string.split() )
pass # end if
rec_id = get_rec_id()
...
pass # end while