Hopefully this post goes better..
So I am stuck on this feature of this program that will return the whole word where a certain keyword is specified.
ie - If I tell it to look for the word "I=" in the string "blah blah blah blah I=1mV blah blah etc?", that it returns the whole word where it is found, so in this case, it would return I=1mV.
I have tried a bunch of different approaches, such as,
text = "One of the values, I=1mV is used"
print(re.split('I=', text))
However, this returns the same String without I in it, so it would return
['One of the values, ', '1mV is used']
If I try regex solutions, I run into the problem where the number could possibly be more then 1 digit, and so this bottom piece of code only works if the number is 1 digit. If I=10mV was that value, it would only return one, but if i have [/0-9] in twice, the code no longer works with only 1 value.
text = "One of the values, I=1mV is used"
print(re.findall("I=[/0-9]", text))
['I=1']
When I tried using re.match,
text = "One of the values, I=1mV is used"
print(re.search("I=", text))
<_sre.SRE_Match object at 0x02408BF0>
What is a good way to retrieve the word (In this case, I want to retrieve I=1mV) and cut out the rest of the string?
A better way would be to split the text into words first:
>>> text = "One of the values, I=1mV is used"
>>> words = text.split()
>>> words
['One', 'of', 'the', 'values,', 'I=1mV', 'is', 'used']
And then filter the words to find the one you need:
>>> [w for w in words if 'I=' in w]
['I=1mV']
This returns a list of all words with I=
in them. We can then just take the first element found:
>>> [w for w in words if 'I=' in w][0]
'I=1mV'
Done! What we can do to clean this up a bit is to just look for the first match, rather then checking every word. We can use a generator expression for that:
>>> next(w for w in words if 'I=' in w)
'I=1mV'
Of course you could adapt the if
condition to fit your needs better, you could for example use str.startswith()
to check if the words starts with a certain string or re.match()
to check if the word matches a pattern.
For the record, your attempt to split the string in two halves, using I=
as the separator, was nearly correct. Instead of using str.split()
, which discards the separator, you could have used str.partition()
, which keeps it.
>>> my_text = "Loadflow current was I=30.63kA"
>>> my_text.partition("I=")
('Loadflow current was ', 'I=', '30.63kA')
A more flexible and robust solution is to use a regular expression:
>>> import re
>>> pattern = r"""
... I= # specific string "I="
... \s* # Possible whitespace
... -? # possible minus sign
... \s* # possible whitespace
... \d+ # at least one digit
... (\.\d+)? # possible decimal part
... """
>>> m = re.search(pattern, my_text, re.VERBOSE)
>>> m
<_sre.SRE_Match object at 0x044CCFA0>
>>> m.group()
'I=30.63'
This accounts for a lot more possibilities (negative numbers, integer or decimal numbers).
Note the use of:
a*
- zero or more a
sa+
- at least one a
a?
- "optional" - one or zero a
sre.VERBOSE
flag) with comments - much easier to understand the pattern above than the non-verbose equivalent, I=\s?-?\s?\d+(\.\d+)
.r"..."
instead of plain strings "..."
- means that literal backslashes don't have to be escaped. Not required here because our pattern doesn't use backslashes, but one day you'll need to match C:\Program Files\...
and on that day you will need raw strings.Exercise 1: How do you extend this so that it can match the unit as well? And how do you extend this so that it can match the unit as either mA
, A
, or kA
? Hint: "Alternation operator".
Exercise 2: How do you extend this so that it can match numbers in engineering notation, i.e. "1.00e3", or "-3.141e-4"?
import re
text = "One of the values, I=1mV is used"
l = (re.split('I=', text))
print str(l[1]).split(' ') [0]
if you have more than one I=
do the above for each odd index in l sice 0 is the first one.
that is a good way since one can write "One of the values, I= 1mV is used" and I guess you want to get that I is 1mv.
BTW I is current and its units are Ampers and not Volts :)
With your re.findall attempt you would want to add a +
which means one or more.
Here are some examples:
import re
test = "This is a test with I=1mV, I=1.414mv, I=10mv and I=1.618mv."
result = re.findall(r'I=[\d\.]+m[vV]', test)
print(result)
test = "One of the values, I=1mV is used"
result = re.search(r'I=([\d\.]+m[vV])', test)
print(result.group(1))
The first print is: ['I=1mV', 'I=1.414mv', 'I=10mv', 'I=1.618mv']
I've grouped everything other than I=
in the re.search example,
so the second print is: 1mV
incase you are interested in extracting that.