字符串列表中的Python搜索模式

Question

I have a list of strings like the following in Python:

         SS = ['T', 'Q', 'T', 'D', 'Q', 'D', 'D', 'Q', 'T', 'D']

Is there anyway I could check how many Ts are directly followed by D, so in this case, there are 2 Ts (second and the last T) that meet the requirement.

I tried this but it was not working though, any advice? thx!

            if ['T','Q'] in SS:
                 print ("yes")

Answer 1

You could do this iteratively checking if the next character was a D and implementing some sort of counter. This is (probably) the tersest, although maybe not the best way.

SS = ['T', 'Q', 'T', 'D', 'Q', 'D', 'D', 'Q', 'T', 'D']
print("".join(SS).count("TD"))

Result:

2

Answer 2

SS = ['T', 'Q', 'T', 'D', 'Q', 'D', 'D', 'Q', 'T', 'D']
print zip(SS, SS[1:]).count(('T', 'D'))

Answer 3

A sequence of single-character strings can be expressed either as your given list or as a multi-character string, the main differences being list methods vs str methods (there's also the tuple type), and mutability.

>>> SS = ['T', 'Q', 'T', 'D', 'Q', 'D', 'D', 'Q', 'T', 'D']
>>> SS_joined = ''.join(SS)
>>> 'TQ' in SS_joined
True

Answer 4

You can always you re.

>>> import re
>>> SS = ['T', 'Q', 'T', 'D', 'Q', 'D', 'D', 'Q', 'T', 'D']
>>> new = ''.join(SS)
>>> len(re.findall('TD',new))
2