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I want to select the files, whose names ends with _90.jpeg|_180.jpeg|_270.jpeg|_90.jpg|_180.jpg|_270.jpg.
currently I am using the following approach
pattern = re.compile('_90.jpeg|_180.jpeg|_270.jpeg|_90.jpg|_180.jpg|_270.jpg')
pattern.search(filename)
Is there any cleaner way to represent the _xxx.yyyy in regular expression.
You can use:
pattern = re.compile('(_9|_18|_27)0\.jpe?g$')
If any digits are fine:
pattern = re.compile('_\d+\.jpe?g$')
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import ast
ast.literal_eval(df['cyberbullying_type'])
error
I was expecting to return back the string values as a list
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I am getting this error time data 2020-07-09T00:00:00+05:30 does not match format yyyy-MM-dd'T'HH:mm:ssZ please help
datetime.datetime.fromisoformat('2020-07-09T00:00:00+05:30')
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While passing the parameters mp_neuron.fit(), it asks for a third parameter.
Can anyone explain why self is not being recognized
You haven't new a object for MPNeuron, it should be
mp_neuron = MPNeuron()
You're not instantiating the object.
mp_neuron = MPNeuron
This should be
mp_neuron = MPNeuron()
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I want to ask that if we can use comprehension in print statement in python.
Just like:
print(i for i in range(10))
This gives me an error. Is there any way that we can compress our code by comprehensive printing ??
Your code has no error, it returns a generator. Instead, make it a list:
print([i for i in range(10)])
or better, with the list function:
print(list(range(10)))
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The regex expression I'm trying to convert is
my $host =~m/([^.])(\.swiss\.ch)/)
But I'm not getting the desired result
Supposing you want to be able to match *.swiss.ch domains and addresses:
import re
preg = re.compile(r'([^\.]+)\.swiss\.ch')