def figure_filter (self, url):
index = 0
end_figure = re.findall("[^/]+(?!.*/)", url)
if end_figure:
all_figure = re.findall("\d+", end_figure[0])
for each_figure in all_figure:
if len(each_figure) < 5:
index += 1
if index == len(all_figure):
return urlthe counter I use here is to calculate a regular match of all numbers less than 5 and return this url, but I use an index count, is there a more concise way to write
although not python, you can fix
with regular matching [5-9] + if it succeeds, there is a number > = 5
if it fails, return URL
the method of @ Fractal upstairs is feasible.
all digits whose length is less than 5 are returned equivalent to only one or more digits whose length is greater than 5 are not returned
the following is the supplementary python code
import re
def figure_filter(url):
end_figure = re.findall('[^/]+(?!.*/)', url)
if end_figure:
-sharp check if there is figure with length > 5
long_figure = re.findall('\d{5,}', end_figure[0])
if not long_figure:
return url
print(figure_filter('http://www.12a123b123c.com')) -sharp http://www.12a123b123c.com
print(figure_filter('http://www.12a345678bc.com')) -sharp None
dict1 = { "system.cpu.user.pct": { "value": 12.83 }, "system.load.1": { "value": 0.33 } } dict2 = { "system": { "...
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