if str(trend.name[0]) == '#'
I fail to see the problem it looks like a normal code to me I just wanted to check that this text is beginning with a hash "#"
You can use built-in starstwith string method.
if str(trend.name).startswith('#')
Related
There is one condition where I have to split my string in the manner that all the alphabetic characters should stay as one unit and everything else should be separated like the example shown below.
Example:
Some_var='12/1/20 Balance Brought Forward 150,585.80'
output_var=['12/1/20','Balance Brought Forward','150,585.80']
Yes, you could use some regex to get over this.
Some_var = '12/1/20 Balance Brought Forward 150,585.80'
match = re.split(r"([0-9\s\\\/\.,-]+|[a-zA-Z\s\\\/\.,-]+)", Some_var)
print(match)
You will get some extra spaces but you can trim that and you are good to go.
split isn't gonna cut it. You might wanna look into Regular Expressions (abbreviated regex) to accomplish this.
Here's a link to the Python docs: re module
As for a pattern, you could try using something like this:
([0-9\s\\\/\.,-]+|[a-zA-Z\s\\\/\.,-]+)
then trim each part of the output.
It's a weird problem
to_be_stripped="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120"
And two strings below:
s1="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120\\[Content_Types].xml"
s2="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120\\_rels\.rels"
When I use the command below:
s1.strip(to_be_stripped)
s2.strip(to_be_stripped)
I get these outputs:
'[Content_Types].x'
'_rels\\.'
If I use lstrip(), they will be:
'[Content_Types].xml'
'_rels\\.rels'
Which is the right outputs.
However, if we replace all Project Known with zeus_pipeline:
to_be_stripped="D:\\Users\\UserKnown\\PycharmProjects\\zeus_pipeline\\PT\\collections\\120"
And:
s2="D:\\Users\\UserKnown\\PycharmProjects\\zeus_pipeline\\PT\\collections\\120\\_rels\.rels"
s2.lstrip(to_be_stripped)will be '.rels'
If I use / instead of \\, nothing goes wrong. I am wondering why this problem happens.
strip isn't meant to remove full strings exactly. Rather, you give it a string, and every character in that string is removed from the start and of the string to be stripped.
In your case, the variable to_be_stripped contains the characters m and l, so those are stripped from the end of s1. However, it doesn't contain the character x, so the stripping stops there and no characters beyond that are removed.
Check out this question. The accepted answer is probably more extensive than you need - I like another user's suggestion of using replace instead of strip. This would look like:
s1.replace(to_be_stripped, "")
I want to use printing command bellow in many places of my script. But I need to keep replacing "Survived" with some other string.
print(df.Survived.value_counts())
Can I automate the process by formating variable the same way as string? So if I want to replace "Survived" with "different" can I use something like:
var = 'different'
text = 'df.{}.value_counts()'.format(var)
print(text)
unfortunately this prints out "df.different.value_counts()" as as a string, while I need to print the value of df.different.value_counts()
I'm pretty sure alot of IDEs, have this option that is called refactoring, and it allows you to change a similar line of code/string on every line of code to what you need it to be.
I'm aware of VSCode's way of refactoring, is by selecting a part of the code and right click to select the option called change all occurances. This will replace the exact code on every line if it exists.
But if you want to do what you proposed, then eval('df.{}.value_counts()'.format(var)) is an option, but this is very unsecured and dangerous, so a more safer approach would be importing the ast module and using it's literal_eval function which is safer. ast.literal_eval('df.{}.value_counts()'.format(var)).
if ast.literal_eval() doesn't work then try this final solution that works.
def cat():
return 1
text = locals()['df.{}.value_counts'.format(var)]()
Found the way: print(df[var].value_counts())
I want to remove all the characters from a string expect whatever character is between a certain set of characters. So for example I have the input of Grade:2/2014-2015 and I want the output of just the grade, 2.
I'm thinking that I need to use the FIND function to grab whatever is between the : and the / , this also needs to work with double characters such 10 however I believe that it would work so long as the defining values with the FIND function are correct.
Unfortunately I am totally lost on this when using the FIND function however if there is another function that would work better I could probably figure it out myself if I knew what function.
It's not particularly elegant but =MID(A1,FIND(":",A1)+1,FIND("/",A1) - FIND(":",A1) - 1) would work.
MID takes start and length,FIND returns the index of a given character.
Edit:
As pointed out, "Grade:" is fixed length so the following would work just as well:
=MID(A1,7,FIND("/",A1) - 7)
You could use LEFT() to remove "Grade:"
And then use and then use LEFTB() to remove the year.
Look at this link here. This is the way I would go about it.
=SUBSTITUTE(SUBSTITUTE(C4, "Grade:", ""), "/2014-2015", "")
where C4 is the name of your cell.
I just wrote a code with the criteria above, but it doesn't seem to work properly because I either miss a letter at the end or in the middle.
Could anyone please check out my code an tell me what I'm doing wrong. By the way I already checked other threads on this similar problem, but I'm not allowed to use regex or print function.
phrase=('my room is cold')
allSpaces=findstr(' ',phrase);
k=length(allSpaces)
acr=phrase(1:allSpaces(1):allSpaces(k)-1)
Output:
acr= mrms
Change last line to
acr = phrase([1 allSpaces+1])
That way you get the first letter, and then the first after each space.