Python：字符串操作的聰明方法

我是Python的新手，目前正在閱讀「潛入Python」中有關字符串操作的章節。Python：字符串操作的聰明方法

我想知道什麼是一些最好（或最聰明/創意）的方式來做到以下幾點：

1）從這個字符串中提取出來：「stackoverflow.com/questions/ask」字「的問題「。我做了string.split（/）[0] - 但那不是很聰明。

2）找到在一個給定的數字或字符串

3最長迴文）與給定的字開始（即「貓」） - 找到所有可能的方式從去另一個三字母詞（「狗」），一次更換一個字母，使得每個字母的改變形成一個新的有效的單詞。

對於example--貓，嬰兒牀，圓點，狗

來源

2011-01-19 Parseltongue

這是三個獨立的問題，有沒有什麼共同點。一般來說，你似乎沒有問題。 – SilentGhost 2011-01-19 16:25:58

問題是他對語言很陌生，已經在尋找'最瘋狂的方式':-D ..順便說一句，爲什麼string.split（'/'）[1]不聰明？ – 2011-01-19 16:28:11

@Parsetongue你有什麼？ – OscarRyz 2011-01-19 16:40:56

隨着個人的運動，這裏是你的，（希望）有很好的註釋代碼一些提示。

#!/usr/bin/env python2 

# Let's take this string: 
a = "palindnilddafa" 
# I surround with a try/catch block, explanation following 
try: 
    # In this loop I go from length of a minus 1 to 0. 
    # range can take 3 params: start, end, increment 
    # This way I start from the thow longest subsring, 
    # the one without the first char and without the last 
    # and go on this way 
    for i in range(len(a)-1, 0, -1): 
    # In this loop I want to know how many 
    # Palidnrome of i length I can do, that 
    # is len(a) - i, and I take all 
    # I start from the end to find the largest first 
    for j in range(len(a) - i): 
     # this is a little triky. 
     # string[start:end] is the slice operator 
     # as string are like arrays (but unmutable). 
     # So I take from j to j+i, all the offsets 
     # The result of "foo"[1:3] is "oo", to be clear. 
     # with string[::-1] you take all elements but in the 
     # reverse order 
     # The check string1 in string2 checks if string1 is a 
     # substring of string2 
     if a[j:j+i][::-1] in a: 
     # If it is I cannot break, 'couse I'll go on on the first 
     # cycle, so I rise an exception passing as argument the substring 
     # found 
     raise Exception(a[j:j+i][::-1]) 

# And then I catch the exception, carrying the message 
# Which is the palindrome, and I print some info 
except Exception as e: 
    # You can pass many things comma-separated to print (this is python2!) 
    print e, "is the longest palindrome of", a 
    # Or you can use printf formatting style 
    print "It's %d long and start from %d" % (len(str(e)), a.index(str(e)))

經過討論，我很抱歉，如果它ot。我已經編寫了另一個迴文搜索器的實現，如果sberry2A可以，我想知道一些基準測試的結果！

請注意，關於指針和困難的「+1 -1」問題有很多錯誤（我猜），但這個想法很清楚。從中間開始，然後展開，直到可以。

下面的代碼：

#!/usr/bin/env python2 


def check(s, i): 
    mid = s[i] 
    j = 1 
    try: 
    while s[i-j] == s[i+j]: 
     j += 1 
    except: 
    pass 
    return s[i-j+1:i+j] 

def do_all(a): 
    pals = [] 
    mlen = 0 
    for i in range(len(a)/2): 
    #print "check for", i 
    left = check(a, len(a)/2 + i) 
    mlen = max(mlen, len(left)) 
    pals.append(left) 

    right = check(a, len(a)/2 - i) 
    mlen = max(mlen, len(right)) 
    pals.append(right) 

    if mlen > max(2, i*2-1): 
     return left if len(left) > len(right) else right 

string = "palindnilddafa" 

print do_all(string)

來源

2011-01-19 17:33:05