Reputation: 375
I know from the title you might think that this is a duplicate but it's not.
for id,row in enumerate(rows):
columns = row.findall("td")
teamName = columns[0].find("a").text, # Lag
playedGames = columns[1].text, # S
wins = columns[2].text,
draw = columns[3].text,
lost = columns[4].text,
dif = columns[6].text, # GM-IM
points = columns[7].text, # P - last column
dict[divisionName].update({id :{"teamName":teamName, "playedGames":playedGames, "wins":wins, "draw":draw, "lost":lost, "dif":dif, "points":points }})
This is how my Python code looks like. Most of the code is removed but essentially i am extracting some information from a website. And i am saving the information as a dictionary. When i print the dictionary every value has a bracket around them ["blbal"] which causes trouble in my Iphone application. I know that i can convert the variables to strings but i want to know if there is a way to get the information DIRECTLY as a string.
Upvotes: 15
Views: 149904
Reputation: 134
import re
text = "some (string) [another string] in brackets"
re.sub("\(.*?\)", "", text)
# some in brackets
# works for () and will work for [] if you replace () with [].
The \(.*?\)
format matches brackets with some text in them with an unspecified length. And the \[.*?\]
format matches also but a square brackets with some text inside the brackets.
The output will not contain brackets and texts inside of them.
If you want to match only square brackets replace square brackets with the bracket of choice and vise versa.
To match ()
and []
bracket in one go, use this format (\(.*?\)|\[.*?\]:)
joining two pattern with the |
character.
Upvotes: -1
Reputation: 2998
you can also you replace
to just replace the text/symbol that you don't want with the empty string.
text = ["blbal","test"]
strippedText = str(text).replace('[','').replace(']','').replace('\'','').replace('\"','')
print(strippedText)
Upvotes: 3
Reputation: 180401
That looks like you have a string inside a list:
["blbal"]
To get the string just index l = ["blbal"]
print(l[0]) -> "blbal"
.
If it is a string use str.strip
'["blbal"]'.strip("[]")
or slicing '["blbal"]'[1:-1]
if they are always present.
Upvotes: 39