正则表达式从元组列表中捕获包含特定模式的元组

Question

I have a list of tuples:

ee = [('noise', 0.7592900395393372), ('***roice***', 0.638433039188385), ('voice', 0.7524746060371399), ('***choice***', 0.638433039188385)]

From here I want to extract only the tuples that contains the pattern which starts with ***

Expected output:

ee = [('***roice***', 0.638433039188385), ('***choice***', 0.638433039188385)]

I have tried the following regex but it only captures the words with *** but not the entire tuple, ie I also want the number present in the tuple which contains ***.

Code till now:

yy= []
for i in ee:
    t9 = re.findall("[***@*&?].*[***@*&?, ]", str(i))
#    for m in t9.finditer(t9):
#        print(m.start(), m.group())
#    
#    print(t9)
    for em in t9:
        yy.append(em)

Can someone help me fix this

Answer 1

You can try:

ee = [('noise', 0.7592900395393372), ('***roice***', 0.638433039188385), ('voice', 0.7524746060371399), ('***choice***', 0.638433039188385)]

output = []

for data in ee:
    if data[0].startswith("***")::
        output.append(data)
print(output)

Output:

[('***roice***', 0.638433039188385), ('***choice***', 0.638433039188385)]

Answer 2

I'm not sure you want a regex in this case. If all that you want to do is filtering strings that begin with "***", you can simply do:

[e for e in ee if e[0].startswith('***')]

If you still want to use a regex, you can do:

r = re.compile(r'\*\*\*.*\*\*\*')
[s for s in ee if r.match(s[0])]

Answer 3

if you need to extract the tuples which 0 element starts and ends with *** , you can try with this:

extracted = []
for item in ee:
    if item[0][:3] == '***' and item[0][-3:] == '***':
        extracted.append(item)

This doesn't use regex.