replace multi words using dictionary [closed] - python

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import re
dic={}
dic1={}
s="ASK NOT WHAT YOUR COUNTRY CAN DO FOR YOU ASK WHAT YOU CAN DO FOR YOUR `COUNTRY"
sentence=". ASK NOT WHAT YOUR COUNTRY CAN DO FOR YOU"
sentence0=". ASK WHAT YOU CAN DO FOR YOUR COUNTRY"
sentence2=sentence.split()
sentence1=sentence.split()
for position,char in enumerate(sentence1):
dic[(char)]=(position)
for position,char in enumerate(sentence2):
dic1[(char)]=(position)
dic.update(dic1)
del dic["."]
print(dic)
pattern = re.compile(r'\b(' + '|'.join(dic.keys()) + r')\b')
result = pattern.sub(lambda x: dic[x.group()], s)
print(result)
When I run the program I have these error TypeError: sequence item 0: expected str instance, int found.How can I solve it?


You need a string for the sub function, to fix that you can either change the dictionary's values to strings:
dic1[(char)]=str(position)
or change the lambda function to:
result = pattern.sub(lambda x: str(dic[x.group()]), s)

Related

Coverting a python list into string with comma separated in efficient way [closed]

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I have a list of acc nos:
list1 = ['1234','3456','2345','5543','1344','5679','6433','3243','0089']
Output I need is a string:
print(output): '1234','3456','2345','5543','1344','5679','6433','3243','0089'

You can join all the values with ',' and then adding a ' before and after the string like this:
"'{0}'".format("','".join(list1))

>>> list1 = ['1234','3456','2345','5543','1344','5679','6433','3243','0089']
>>> print(','.join(["'{0}'".format(s) for s in list1]))
'1234','3456','2345','5543','1344','5679','6433','3243','0089'

Return part of string between specific character pair, in string with multiple character pairs [closed]

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Text file or string:
SomeText1/SomeText2/SomeText3/SomeText4/SomeText5
#What I am looking for:
split_func(3, "/")
>>> SomeText3

Try:
s = "SomeText1/SomeText2/SomeText3/SomeText4/SomeText5"
# s.split("/") returns a list of strings, split at the "/"
# I.e. ["SomeText1", "SomeText2", "SomeText3", "SomeText4", "SomeText5"]
# Then take the second element (remembering that the count starts at 0
result = s.split("/")[2]

select the characters in after "):" and print it in between? [closed]

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I have a file like this
(E:0.13228,((D:0.08440,A:0.14071):0.29270,(H:0.30329,(B:0.06928,
(F:0.00236,G:0.00010):0.00010):0.44531):0.06201):0.57269,C:0.19183);
I need to generate it as follows:
(E:0.13228,((D:0.08440,A:0.14071)0.29270:0.29270,(H:0.30329,(B:0.06928,
(F:0.00236,G:0.00010)0.00010:0.00010)0.44531:0.44531)0.06201:0.06201)0.57269:0.57269,C:0.19183);

What you should use in this cases are the regular expressions. Check Python re
match_pattern = r'\):(\d(\.\d+)?)'
output_pattern = r')\1:\1'
input_str = """(E:0.13228,((D:0.08440,A:0.14071):0.29270,(H:0.30329,(B:0.06928,
(F:0.00236,G:0.00010):0.00010):0.44531):0.06201):0.57269,C:0.19183);"""
output_str = re.sub(match_pattern, output_pattern, input_str)
print(output_str)
And the result is:
(E:0.13228,((D:0.08440,A:0.14071)0.29270:0.29270,(H:0.30329,(B:0.06928,
(F:0.00236,G:0.00010)0.00010:0.00010)0.44531:0.44531)0.06201:0.06201)0.57269:0.57269,C:0.19183);

I guess you could go with some string manipulation like so if you're not using regular expressions
data = "(E:0.13228,((D:0.08440,A:0.14071):0.29270,(H:0.30329,(B:0.06928,
(F:0.00236,G:0.00010):0.00010):0.44531):0.06201):0.57269,C:0.19183);"
dataParts = data.split("):")
correctedData = dataParts[0]
for dataPart in dataParts[1:]:
number = dataPart[:7]
correctedData = "){}:".format(number).join([correctedData, dataPart])
but that's not clean...

python format string by length [closed]

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I have some strings below:
123123|00|992|1111
222222|2222|19|922
997997|3333|922|77
How can I format like this (sort by len):
123123|1111|00|992
222222|2222|19|922
997997|3333|77|922

Using str.split & str.join with sorted
Demo:
s = """123123|00|992|1111
222222|2222|19|922
997997|3333|922|77"""
data = ["|".join(sorted(i.split("|"), key=len, reverse=True)) for i in s.split("\n")]
print( "\n".join(data) )
Output:
123123|1111|992|00
222222|2222|922|19
997997|3333|922|77

Try this
a='''123123|00|992|1111
222222|2222|19|922
997997|3333|922|77'''
'\n'.join(['|'.join(sorted(e.split("|"),key=len,reverse=True)) for e in a.split("\n")])
Output:
123123|1111|992|00
222222|2222|922|19
997997|3333|922|77

The OPs sort order isn't based on string length, as the length of 2 comes before the length of 3 (the very last two segments)
for that purpose, a custom sort mapping is needed
ranks = {6:0, 4:1, 2:2, 3:3}
text = """123123|00|992|1111
222222|2222|19|922
997997|3333|922|77"""
result = '\n'.join(['|'.join(sorted(line.split('|'),key=lambda x: ranks[len(x)]))
for line in text.split('\n')])
print(result)
# outputs:
123123|1111|00|992
222222|2222|19|922
997997|3333|77|922

How to replace a dot with a string in a python list [closed]

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my_list = ['b','.','.']
expected_list = ['b','.','w']
May be simple, moving into python recently so any suggestions would be fine

You can do this by using list comprehension also
l = ['w' if i == '.' else i for i in my_list]

If you are replacing every occurrence of '.' with 'w' then I would just suggest:
for n, i in enumerate(my_list):
if i == '.':
my_list[n] = 'w'
Here is the documentation for how to use the enumerate() function: https://docs.python.org/2/library/functions.html#enumerate

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