Meaning of a statement - python

I want to understand the meaning of the following line in the code mentioned below: (As in how to read that line?)
print('#%*s' % (a, '#') if a else '')
From this code:
lines = int(input("Enter number of lines for pattern: "))
for a in range(lines):
print('#%*s' % (a, '#') if a else '')

This is a conditional expression:
'#%*s' % (a, '#') if a else ''
Read it like this: if the variable a is not null and not empty format it, else the expression evaluates to the empty string ''. Now for the format part (which uses the old % syntax):
'#%*s' % (a, '#')
It says: print an #, then a number of spaces and finally one last # character. The * gets substituted with the value of a and then the format string is applied to the # character For example, if a = 5 the above expression will evaluate to this:
'#%5s' % '#'
Which we can print and see the result:
print('#%5s' % '#')
# #
Notice that the % syntax is deprecated, in modern Python the recommendation is to use str.format or even better, f-strings for Python 3.6+.

You can see some string formatting examples here. So your formatting does something like:
Format the output (your #) as a string and place an additional #
at the end of the line. Fill the line with so many spaces that the
line has a total length of a. If a is zero and so the if
statement is FALSE print and empty line.
Enter number of lines for pattern: 9
##
# #
# #
# #
# #
# #
# #
# #
A modern solution for the same output would be looking like this:
lines = int(input("Enter number of lines for pattern: "))
for a in range(lines):
print(("{:" + str(a) + "}{}").format("#", "#") if a else "")

Related

How can I implement isalnum() into this Python web scraper to remove special characters? [duplicate]

I'm trying to remove specific characters from a string using Python. This is the code I'm using right now. Unfortunately it appears to do nothing to the string.
for char in line:
if char in " ?.!/;:":
line.replace(char,'')
How do I do this properly?
Strings in Python are immutable (can't be changed). Because of this, the effect of line.replace(...) is just to create a new string, rather than changing the old one. You need to rebind (assign) it to line in order to have that variable take the new value, with those characters removed.
Also, the way you are doing it is going to be kind of slow, relatively. It's also likely to be a bit confusing to experienced pythonators, who will see a doubly-nested structure and think for a moment that something more complicated is going on.
Starting in Python 2.6 and newer Python 2.x versions *, you can instead use str.translate, (see Python 3 answer below):
line = line.translate(None, '!##$')
or regular expression replacement with re.sub
import re
line = re.sub('[!##$]', '', line)
The characters enclosed in brackets constitute a character class. Any characters in line which are in that class are replaced with the second parameter to sub: an empty string.
Python 3 answer
In Python 3, strings are Unicode. You'll have to translate a little differently. kevpie mentions this in a comment on one of the answers, and it's noted in the documentation for str.translate.
When calling the translate method of a Unicode string, you cannot pass the second parameter that we used above. You also can't pass None as the first parameter. Instead, you pass a translation table (usually a dictionary) as the only parameter. This table maps the ordinal values of characters (i.e. the result of calling ord on them) to the ordinal values of the characters which should replace them, or—usefully to us—None to indicate that they should be deleted.
So to do the above dance with a Unicode string you would call something like
translation_table = dict.fromkeys(map(ord, '!##$'), None)
unicode_line = unicode_line.translate(translation_table)
Here dict.fromkeys and map are used to succinctly generate a dictionary containing
{ord('!'): None, ord('#'): None, ...}
Even simpler, as another answer puts it, create the translation table in place:
unicode_line = unicode_line.translate({ord(c): None for c in '!##$'})
Or, as brought up by Joseph Lee, create the same translation table with str.maketrans:
unicode_line = unicode_line.translate(str.maketrans('', '', '!##$'))
* for compatibility with earlier Pythons, you can create a "null" translation table to pass in place of None:
import string
line = line.translate(string.maketrans('', ''), '!##$')
Here string.maketrans is used to create a translation table, which is just a string containing the characters with ordinal values 0 to 255.
Am I missing the point here, or is it just the following:
string = "ab1cd1ef"
string = string.replace("1", "")
print(string)
# result: "abcdef"
Put it in a loop:
a = "a!b#c#d$"
b = "!##$"
for char in b:
a = a.replace(char, "")
print(a)
# result: "abcd"
>>> line = "abc##!?efg12;:?"
>>> ''.join( c for c in line if c not in '?:!/;' )
'abc##efg12'
With re.sub regular expression
Since Python 3.5, substitution using regular expressions re.sub became available:
import re
re.sub('\ |\?|\.|\!|\/|\;|\:', '', line)
Example
import re
line = 'Q: Do I write ;/.??? No!!!'
re.sub('\ |\?|\.|\!|\/|\;|\:', '', line)
'QDoIwriteNo'
Explanation
In regular expressions (regex), | is a logical OR and \ escapes spaces and special characters that might be actual regex commands. Whereas sub stands for substitution, in this case with the empty string ''.
The asker almost had it. Like most things in Python, the answer is simpler than you think.
>>> line = "H E?.LL!/;O:: "
>>> for char in ' ?.!/;:':
... line = line.replace(char,'')
...
>>> print line
HELLO
You don't have to do the nested if/for loop thing, but you DO need to check each character individually.
For the inverse requirement of only allowing certain characters in a string, you can use regular expressions with a set complement operator [^ABCabc]. For example, to remove everything except ascii letters, digits, and the hyphen:
>>> import string
>>> import re
>>>
>>> phrase = ' There were "nine" (9) chick-peas in my pocket!!! '
>>> allow = string.letters + string.digits + '-'
>>> re.sub('[^%s]' % allow, '', phrase)
'Therewerenine9chick-peasinmypocket'
From the python regular expression documentation:
Characters that are not within a range can be matched by complementing
the set. If the first character of the set is '^', all the characters
that are not in the set will be matched. For example, [^5] will match
any character except '5', and [^^] will match any character except
'^'. ^ has no special meaning if it’s not the first character in the
set.
line = line.translate(None, " ?.!/;:")
>>> s = 'a1b2c3'
>>> ''.join(c for c in s if c not in '123')
'abc'
Strings are immutable in Python. The replace method returns a new string after the replacement. Try:
for char in line:
if char in " ?.!/;:":
line = line.replace(char,'')
This is identical to your original code, with the addition of an assignment to line inside the loop.
Note that the string replace() method replaces all of the occurrences of the character in the string, so you can do better by using replace() for each character you want to remove, instead of looping over each character in your string.
I was surprised that no one had yet recommended using the builtin filter function.
import operator
import string # only for the example you could use a custom string
s = "1212edjaq"
Say we want to filter out everything that isn't a number. Using the filter builtin method "...is equivalent to the generator expression (item for item in iterable if function(item))" [Python 3 Builtins: Filter]
sList = list(s)
intsList = list(string.digits)
obj = filter(lambda x: operator.contains(intsList, x), sList)))
In Python 3 this returns
>> <filter object # hex>
To get a printed string,
nums = "".join(list(obj))
print(nums)
>> "1212"
I am not sure how filter ranks in terms of efficiency but it is a good thing to know how to use when doing list comprehensions and such.
UPDATE
Logically, since filter works you could also use list comprehension and from what I have read it is supposed to be more efficient because lambdas are the wall street hedge fund managers of the programming function world. Another plus is that it is a one-liner that doesnt require any imports. For example, using the same string 's' defined above,
num = "".join([i for i in s if i.isdigit()])
That's it. The return will be a string of all the characters that are digits in the original string.
If you have a specific list of acceptable/unacceptable characters you need only adjust the 'if' part of the list comprehension.
target_chars = "".join([i for i in s if i in some_list])
or alternatively,
target_chars = "".join([i for i in s if i not in some_list])
Using filter, you'd just need one line
line = filter(lambda char: char not in " ?.!/;:", line)
This treats the string as an iterable and checks every character if the lambda returns True:
>>> help(filter)
Help on built-in function filter in module __builtin__:
filter(...)
filter(function or None, sequence) -> list, tuple, or string
Return those items of sequence for which function(item) is true. If
function is None, return the items that are true. If sequence is a tuple
or string, return the same type, else return a list.
Try this one:
def rm_char(original_str, need2rm):
''' Remove charecters in "need2rm" from "original_str" '''
return original_str.translate(str.maketrans('','',need2rm))
This method works well in Python 3
Here's some possible ways to achieve this task:
def attempt1(string):
return "".join([v for v in string if v not in ("a", "e", "i", "o", "u")])
def attempt2(string):
for v in ("a", "e", "i", "o", "u"):
string = string.replace(v, "")
return string
def attempt3(string):
import re
for v in ("a", "e", "i", "o", "u"):
string = re.sub(v, "", string)
return string
def attempt4(string):
return string.replace("a", "").replace("e", "").replace("i", "").replace("o", "").replace("u", "")
for attempt in [attempt1, attempt2, attempt3, attempt4]:
print(attempt("murcielago"))
PS: Instead using " ?.!/;:" the examples use the vowels... and yeah, "murcielago" is the Spanish word to say bat... funny word as it contains all the vowels :)
PS2: If you're interested on performance you could measure these attempts with a simple code like:
import timeit
K = 1000000
for i in range(1,5):
t = timeit.Timer(
f"attempt{i}('murcielago')",
setup=f"from __main__ import attempt{i}"
).repeat(1, K)
print(f"attempt{i}",min(t))
In my box you'd get:
attempt1 2.2334518376057244
attempt2 1.8806643818474513
attempt3 7.214925774955572
attempt4 1.7271184513757465
So it seems attempt4 is the fastest one for this particular input.
Here's my Python 2/3 compatible version. Since the translate api has changed.
def remove(str_, chars):
"""Removes each char in `chars` from `str_`.
Args:
str_: String to remove characters from
chars: String of to-be removed characters
Returns:
A copy of str_ with `chars` removed
Example:
remove("What?!?: darn;", " ?.!:;") => 'Whatdarn'
"""
try:
# Python2.x
return str_.translate(None, chars)
except TypeError:
# Python 3.x
table = {ord(char): None for char in chars}
return str_.translate(table)
#!/usr/bin/python
import re
strs = "how^ much for{} the maple syrup? $20.99? That's[] ricidulous!!!"
print strs
nstr = re.sub(r'[?|$|.|!|a|b]',r' ',strs)#i have taken special character to remove but any #character can be added here
print nstr
nestr = re.sub(r'[^a-zA-Z0-9 ]',r'',nstr)#for removing special character
print nestr
You can also use a function in order to substitute different kind of regular expression or other pattern with the use of a list. With that, you can mixed regular expression, character class, and really basic text pattern. It's really useful when you need to substitute a lot of elements like HTML ones.
*NB: works with Python 3.x
import re # Regular expression library
def string_cleanup(x, notwanted):
for item in notwanted:
x = re.sub(item, '', x)
return x
line = "<title>My example: <strong>A text %very% $clean!!</strong></title>"
print("Uncleaned: ", line)
# Get rid of html elements
html_elements = ["<title>", "</title>", "<strong>", "</strong>"]
line = string_cleanup(line, html_elements)
print("1st clean: ", line)
# Get rid of special characters
special_chars = ["[!##$]", "%"]
line = string_cleanup(line, special_chars)
print("2nd clean: ", line)
In the function string_cleanup, it takes your string x and your list notwanted as arguments. For each item in that list of elements or pattern, if a substitute is needed it will be done.
The output:
Uncleaned: <title>My example: <strong>A text %very% $clean!!</strong></title>
1st clean: My example: A text %very% $clean!!
2nd clean: My example: A text very clean
My method I'd use probably wouldn't work as efficiently, but it is massively simple. I can remove multiple characters at different positions all at once, using slicing and formatting.
Here's an example:
words = "things"
removed = "%s%s" % (words[:3], words[-1:])
This will result in 'removed' holding the word 'this'.
Formatting can be very helpful for printing variables midway through a print string. It can insert any data type using a % followed by the variable's data type; all data types can use %s, and floats (aka decimals) and integers can use %d.
Slicing can be used for intricate control over strings. When I put words[:3], it allows me to select all the characters in the string from the beginning (the colon is before the number, this will mean 'from the beginning to') to the 4th character (it includes the 4th character). The reason 3 equals till the 4th position is because Python starts at 0. Then, when I put word[-1:], it means the 2nd last character to the end (the colon is behind the number). Putting -1 will make Python count from the last character, rather than the first. Again, Python will start at 0. So, word[-1:] basically means 'from the second last character to the end of the string.
So, by cutting off the characters before the character I want to remove and the characters after and sandwiching them together, I can remove the unwanted character. Think of it like a sausage. In the middle it's dirty, so I want to get rid of it. I simply cut off the two ends I want then put them together without the unwanted part in the middle.
If I want to remove multiple consecutive characters, I simply shift the numbers around in the [] (slicing part). Or if I want to remove multiple characters from different positions, I can simply sandwich together multiple slices at once.
Examples:
words = "control"
removed = "%s%s" % (words[:2], words[-2:])
removed equals 'cool'.
words = "impacts"
removed = "%s%s%s" % (words[1], words[3:5], words[-1])
removed equals 'macs'.
In this case, [3:5] means character at position 3 through character at position 5 (excluding the character at the final position).
Remember, Python starts counting at 0, so you will need to as well.
In Python 3.5
e.g.,
os.rename(file_name, file_name.translate({ord(c): None for c in '0123456789'}))
To remove all the number from the string
How about this:
def text_cleanup(text):
new = ""
for i in text:
if i not in " ?.!/;:":
new += i
return new
Below one.. with out using regular expression concept..
ipstring ="text with symbols!##$^&*( ends here"
opstring=''
for i in ipstring:
if i.isalnum()==1 or i==' ':
opstring+=i
pass
print opstring
Recursive split:
s=string ; chars=chars to remove
def strip(s,chars):
if len(s)==1:
return "" if s in chars else s
return strip(s[0:int(len(s)/2)],chars) + strip(s[int(len(s)/2):len(s)],chars)
example:
print(strip("Hello!","lo")) #He!
You could use the re module's regular expression replacement. Using the ^ expression allows you to pick exactly what you want from your string.
import re
text = "This is absurd!"
text = re.sub("[^a-zA-Z]","",text) # Keeps only Alphabets
print(text)
Output to this would be "Thisisabsurd". Only things specified after the ^ symbol will appear.
# for each file on a directory, rename filename
file_list = os.listdir (r"D:\Dev\Python")
for file_name in file_list:
os.rename(file_name, re.sub(r'\d+','',file_name))
Even the below approach works
line = "a,b,c,d,e"
alpha = list(line)
while ',' in alpha:
alpha.remove(',')
finalString = ''.join(alpha)
print(finalString)
output: abcde
The string method replace does not modify the original string. It leaves the original alone and returns a modified copy.
What you want is something like: line = line.replace(char,'')
def replace_all(line, )for char in line:
if char in " ?.!/;:":
line = line.replace(char,'')
return line
However, creating a new string each and every time that a character is removed is very inefficient. I recommend the following instead:
def replace_all(line, baddies, *):
"""
The following is documentation on how to use the class,
without reference to the implementation details:
For implementation notes, please see comments begining with `#`
in the source file.
[*crickets chirp*]
"""
is_bad = lambda ch, baddies=baddies: return ch in baddies
filter_baddies = lambda ch, *, is_bad=is_bad: "" if is_bad(ch) else ch
mahp = replace_all.map(filter_baddies, line)
return replace_all.join('', join(mahp))
# -------------------------------------------------
# WHY `baddies=baddies`?!?
# `is_bad=is_bad`
# -------------------------------------------------
# Default arguments to a lambda function are evaluated
# at the same time as when a lambda function is
# **defined**.
#
# global variables of a lambda function
# are evaluated when the lambda function is
# **called**
#
# The following prints "as yellow as snow"
#
# fleece_color = "white"
# little_lamb = lambda end: return "as " + fleece_color + end
#
# # sometime later...
#
# fleece_color = "yellow"
# print(little_lamb(" as snow"))
# --------------------------------------------------
replace_all.map = map
replace_all.join = str.join
If you want your string to be just allowed characters by using ASCII codes, you can use this piece of code:
for char in s:
if ord(char) < 96 or ord(char) > 123:
s = s.replace(char, "")
It will remove all the characters beyond a....z even upper cases.

Python re.findall regex and text processing

I'm looking to find and modify some sql syntax around the convert function. I want basically any convert(A,B) or CONVERT(A,B) in all my files to be selected and converted to B::A.
So far I tried selecting them with re.findall(r"\bconvert\b\(.*?,.*\)", l, re.IGNORECASE) But it's only returning a small selection out of what I want and I also have trouble actually manipulating the A/B I mentioned.
For example, a sample line (note the nested structure here is irrelevant, I'm only getting the outer layer working if possible)
convert(varchar, '/' || convert(nvarchar, es.Item_ID) || ':' || convert(nvarchar, o.Option_Number) || '/') as LocPath
...should become...
'/' || es.Item_ID::nvarchar || ':' || o.Option_Number::nvarchar || '/' :: varchar as LocPath
Example2:
SELECT LocationID AS ItemId, convert(bigint, -1),
...should become...
SELECT LocationID AS ItemId, -1::bigint,
I think this should be possible with some kind of re.sub with groups and currently have a code structure inside a for each loop where line is the each line in the file:
matchConvert = ["convert(", "CONVERT("]
a = next((a for a in matchConvert if a in line), False)
if a:
print("convert() line")
#line = re.sub(re.escape(a) + r'', '', line)
Edit: In the end I went with a non re solution and handled each line by identifying each block and manipulate them accordingly.
This may be an X/Y problem, meaning you’re asking how to do something with Regex that may be better solved with parsing (meaning using/modifying/writing a SQL parser). An indication that this is the case is the fact that “convert” calls can be nested. I’m guessing Regex is going to be more of a headache than it’s worth here in the long run if you’re working with a lot of files and they’re at all complicated.
The task:
Swap the parameters of all the 'convert' functions in this given. Parameters can contain any character, including nested 'convert' functions.
A solution:
def convert_py(s):
#capturing start:
left=s.index('convert')
start=s[:left]
#capturing part_1:
c=0
line=''
for n1,i in enumerate(s[left+8:],start=len(start)+8):
if i==',' and c==0:
part_1=line
break
if i==')':
c-=1
if i=='(':
c+=1
line+=i
#capturing part_2:
c=0
line=''
for n2,i in enumerate(s[n1+1:],start=n1+1):
if i==')':
c-=1
if i=='(':
c+=1
if c<0:
part_2=line
break
line+=i
#capturing end:
end=s[n2+1:]
#capturing result:
result=start+part_2.lstrip()+' :: '+part_1+end
return result
def multi_convert_py(s):
converts=s.count('convert')
for n in range(converts):
s=convert_py(s)
return s
Notes:
Unlike the solution based on the re module, which is presented in another answer - this version should not fail if there are more than two parameters in the 'convert' function in the given string. However, it will swap them only once, for example: convert(a,b, c) --> b, c : a
I am afraid that unforeseen cases may arise that will lead to failure. Please tell if you find any flaws
The task:
Swap the parameters of all the 'convert' functions in the given string. Parameters can contain any character, including nested 'convert' functions.
A solution based on the re module:
def convert_re(s):
import re
start,part_1,part_2,end=re.search(r'''
(.*?)
convert\(
([^,)(]+\(.+?\)[^,)(]*|[^,)(]+)
,
([^,)(]+\(.+?\)[^,)(]*|[^,)(]+)
\)
(.*)
''',s,re.X).groups()
result=start+part_2.lstrip()+' :: '+part_1+end
return result
def multi_convert_re(s):
converts=s.count('convert')
for n in range(converts):
s=convert_re(s)
return s
Discription of the 'convert_re' function:
Regular expression:
start is the first group with what comes before 'convert'
Then follows convert\() which has no group and contains the name of the function and the opening '('
part_1 is the second group ([^,)(]+\(.+?\)[^,)(]*|[^,)(]+). This should match the first parameter. It can be anything except - ,)(, or a function preceded by anything except ,)(, optionally followed by anything except ,)( and with anything inside (except a new line)
Then follows a comma ,, which has no group
part_2 is the third group and it acts like the second, but should catch everything what's left inside the external function
Then follows ), which has no group
end is the fourth group (.*) with what's left before the new line.
The resulting string is then created by swapping part_1 and part_2, putting ' :: ' between them, removing spaces on the left from part_2 and adding start to the beginning and end to the end.
Description of the 'multi_convert_re' function
Repeatedly calls 'convert_re' function until there are no "convert" left.
Notes:
N.B.: The code implies that the 'convert' function in the string has exactly two parameters.
The code works on the given examples, but I'm afraid there may still be unforeseen flaws when it comes to other examples. Please tell, if you find any flaws.
I have provided another solution presented in another answer that is not based on the re module. It may turn out that the results will be different.
Here's my solution based on #Иван-Балван's code. Breaking this structure into blocks makes further specification a lot easier than I previously thought and I'll be using this method for a lot of other operations as well.
# Check for balanced brackets
def checkBracket(my_string):
count = 0
for c in my_string:
if c == "(":
count+=1
elif c == ")":
count-=1
return count
# Modify the first convert in line
# Based on suggestions from stackoverflow.com/questions/73040953
def modifyConvert(l):
# find the location of convert()
count = l.index('convert(')
# select the group before convert() call
before = l[:count]
group=""
n1=0
n2=0
A=""
B=""
operate = False
operators = ["|", "<", ">", "="]
# look for A group before comma
for n1, i in enumerate(l[count+8:], start=len(before)+8):
# find current position in l
checkIndex = checkBracket(l[count+8:][:n1-len(before)-8])
if i == ',' and checkIndex == 0:
A = group
break
group += i
# look for B group after comma
group = ""
for n2, i in enumerate(l[n1+1:], start=n1+1):
checkIndex = checkBracket(l[count+n1-len(before):][:n2-n1+1])
if i == ',' and checkIndex == 0:
return l
elif checkIndex < 0:
B = group
break
group += i
# mark operators
if i in operators:
operate = True
# select the group after convert() call
after = l[n2+1:]
# (B) if it contains operators
if operate:
return before + "(" + B.lstrip() + ') :: ' + A + after
else:
return before + B.lstrip() + '::' + A + after
# Modify cast syntax with convert(a,b). return line.
def convertCast(l):
# Call helper for nested cases
i = l.count('convert(')
while i>0:
i -= 1
l = modifyConvert(l)
return l

Reading a text until matched string and print it as a single line

I am trying to write a program as a python beginner.
This function scans the given text. Whenever I come across the symbols “#, &, % or $”, I have to take the text until that point and print it as a single line. Also, I have to skip the symbol, start the next line from the letter right after the symbol and print it until I come across another symbol.
I hope I told it well.
symbols= "#, &, %, $”"
for symbol in text:
if text.startswith (symbols):
print (text)
I know it's not correct but that was all I could think. Any help is appreciated.
IIUC, you need to split the string by each of the delimiters, so you could do the following:
symbols = "#, &, %, $".split(', ')
print(symbols) # this is a list
text = "The first # The second & and here you have % and finally $"
# make a copy of text
replaced = text[:]
# unify delimiters
for symbol in symbols:
replaced = replaced.replace(symbol, '#')
print(replaced) # now the string only have # in the place of other symbols
for chunk in replaced.split('#'):
if chunk: # avoid printing empty strings
print(chunk)
Output
['#', '&', '%', '$'] # print(symbols)
The first # The second # and here you have # and finally # # print(replaced)
The first
The second
and here you have
and finally
The first step:
symbols = "#, &, %, $".split(', ')
print(symbols) # this is a list
converts your string to a list. The second step replaces, using replace, all symbols with only one because str.split only works with a single string:
# unify delimiters
for symbol in symbols:
replaced = replaced.replace(symbol, '#')
The third and final step is to split the string by the chosen symbol (i.e #):
for chunk in replaced.split('#'):
if chunk: # avoid printing empty strings
print(chunk)
If I follow the instructions given to you literally, I believe it translates into nothing fancier than just replacing each of those 4 special characters in the text by a newline character:
text = """This is a line with # and & in it.
This line has % and $ in it.
This line has nothing intersting in it.
This line has just # in it.
The end.
"""
text = text.replace('#', '\n').replace('&', '\n').replace('%', '\n').replace('$', '\n')
print(text)
Prints:
This is a line with
and
in it.
This line has
and
in it.
This line has nothing intersting in it.
This line has just
in it.

Python test if string matches a template value

I am trying to iterate through a list of strings, keeping only those that match a naming template I have specified. I want to accept any list entry that matches the template exactly, other than having an integer in a variable <SCENARIO> field.
The check needs to be general. Specifically, the string structure could change such that there is no guarantee <SCENARIO> always shows up at character X (to use list comprehensions, for example).
The code below shows an approach that works using split, but there must be a better way to make this string comparison. Could I use regular expressions here?
template = 'name_is_here_<SCENARIO>_20131204.txt'
testList = ['name_is_here_100_20131204.txt', # should accept
'name_is_here_100_20131204.txt.NEW', # should reject
'other_name.txt'] # should reject
acceptList = []
for name in testList:
print name
acceptFlag = True
splitTemplate = template.split('_')
splitName = name.split('_')
# if lengths do not match, name cannot possibly match template
if len(splitTemplate) == len(splitName):
print zip(splitTemplate, splitName)
# compare records in the split
for t, n in zip(splitTemplate, splitName):
if t!=n and not t=='<SCENARIO>':
#reject if any of the "other" fields are not identical
#(would also check that '<SCENARIO>' field is numeric - not shown here)
print 'reject: ' + name
acceptFlag = False
else:
acceptFlag = False
# keep name if it passed checks
if acceptFlag == True:
acceptList.append(name)
print acceptList
# correctly prints --> ['name_is_here_100_20131204.txt']
Try with the re module for regular expressions in Python:
import re
template = re.compile(r'^name_is_here_(\d+)_20131204.txt$')
testList = ['name_is_here_100_20131204.txt', #accepted
'name_is_here_100_20131204.txt.NEW', #rejected!
'name_is_here_aabs2352_20131204.txt', #rejected!
'other_name.txt'] #rejected!
acceptList = [item for item in testList if template.match(item)]
This should do, I understand that name_is_here is just a placeholder for alphanumeric characters?
import re
testList = ['name_is_here_100_20131204.txt', # should accept
'name_is_here_100_20131204.txt.NEW', # should reject
'other_name.txt',
'name_is_44ere_100_20131204.txt',
'name_is_here_100_2013120499.txt',
'name_is_here_100_something_2013120499.txt',
'name_is_here_100_something_20131204.txt']
def find(scenario):
begin = '[a-z_]+100_' # any combinations of chars and underscores followd by 100
end = '_[0-9]{8}.txt$' #exactly eight digits followed by .txt at the end
pattern = re.compile("".join([begin,scenario,end]))
result = []
for word in testList:
if pattern.match(word):
result.append(word)
return result
find('something') # returns ['name_is_here_100_something_20131204.txt']
EDIT: scenario in separate variable, regex now only matches characters followed by 100, then scenarion, then eight digits followed by .txt.

Python: Remove First Character of each Word in String

I am trying to figure out how to remove the first character of a words in a string.
My program reads in a string.
Suppose the input is :
this is demo
My intention is to remove the first character of each word of the string, that is
tid, leaving his s emo.
I have tried
Using a for loop and traversing the string
Checking for space in the string using isspace() function.
Storing the index of the letter which is encountered after the
space, i = char + 1, where char is the index of space.
Then, trying to remove the empty space using str_replaced = str[i:].
But it removed the entire string except the last one.
List comprehensions is your friend. This is the most basic version, in just one line
str = "this is demo";
print " ".join([x[1:] for x in str.split(" ")]);
output:
his s emo
In case the input string can have not only spaces, but also newlines or tabs, I'd use regex.
In [1]: inp = '''Suppose we have a
...: multiline input...'''
In [2]: import re
In [3]: print re.sub(r'(?<=\b)\w', '', inp)
uppose e ave
ultiline nput...
You can simply using python comprehension
str = 'this is demo'
mstr = ' '.join([s[1:] for s in str.split(' ')])
then mstr variable will contains these values 'his s emo'
This method is a bit long, but easy to understand. The flag variable stores if the character is a space. If it is, the next letter must be removed
s = "alpha beta charlie"
t = ""
flag = 0
for x in range(1,len(s)):
if(flag==0):
t+=s[x]
else:
flag = 0
if(s[x]==" "):
flag = 1
print(t)
output
lpha eta harlie

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