有條件地格式化一個字符串 - Python

Question

我想要一個字符串並對其進行格式化，以便我可以控制更改的數量。例如..

「這是一個真棒字符串」替換法「@」的「一」能給我...

「這是@n @wesome字符串」但是，我想說的替換1「a」帶「@」，剩下的只剩下其中一個..

「這是@n真棒字符串」放置允許隨機，但最重要的是我說明了我替換的特定種類的數量。有任何想法嗎？

Answer 1

的字符串替換函數有一個可選的count參數來控制替換的最大數量，使

"This is an awesome string".replace("a","@") # "This is @n @wesome string" 

"This is an awesome string".replace("a","@",1) # "This is @n awesome string" 

---

如果需要隨意做的，我們可以寫一個函數來做到這一點

import random 
def randreplace(str,c,c_replace,maxnum=0): 
    if maxnum >= str.count(c) or maxnum<1: 
     return str.replace(c,c_replace) 
    indices = [i for i,x in enumerate(str) if x==c] 
    replacements = random.sample(indices,maxnum) 
    st_pieces = (x if not i in replacements else c_replace for i,x in enumerate(str)) 
    return "".join(st_pieces) 

此函數接受字符串替換，要替換的字符，要替換的字符以及替換的最大數量（全部爲0他們）並返回隨機完成所需數量替換的字符串。

random.seed(100) 
randreplace("This is an awesome string","a","@",1) # "This is @n awesome string" 
randreplace("This is an awesome string","a","@",1) # "This is an @wesome string" 
randreplace("This is an awesome string","a","@",2) # "This is @n @wesome string" 
randreplace("This is an awesome string","a","@") # "This is @n @wesome string" 

Answer 2

以下功能將讓你改變一個匹配的字符：

def replace_a_char(text, x, y, n): 
    matched = 0 
    for index, c in enumerate(text): 
     if c == x: 
      matched += 1 
      if matched == n: 
       return text[:index] + y + text[index+1:] 

    return text 

text = "This is an awesome string and has lot of characters" 

for n in xrange(1, 10): 
    print replace_a_char(text, 'a', '@', n) 

給你以下的輸出：

This is @n awesome string and has lot of characters 
This is an @wesome string and has lot of characters 
This is an awesome string @nd has lot of characters 
This is an awesome string and h@s lot of characters 
This is an awesome string and has lot of ch@racters 
This is an awesome string and has lot of char@cters 
This is an awesome string and has lot of characters 
This is an awesome string and has lot of characters 
This is an awesome string and has lot of characters 

Answer 3

@Andrew梅斯說，「我一直在尋找使它隨機...「

import random 

target = "a" 

replacement = "@" 

string = "This is an awesome string" 

indicies = [index for index, character in enumerate(string) if character == target] 

index = random.choice(indicies) 

string = string[:index] + replacement + string[index + 1:] 

讓我們把它變成一個函數，它可以選擇製作多少個隨機替換，並返回修改過的字符串和實際替換次數（例如，你可能會要求太多）

def random_replace(string, target, replacement, instances): 
    indicies = [index for index, character in enumerate(string) if character == target] 

    replacements = min(instances, len(indicies)) 

    random_indicies = random.sample(indicies, replacements) 

    for index in random_indicies: 
     string = string[:index] + replacement + string[index + 1:] 

    return string, replacements 

一些有用的例子：

>>> print(random_replace(string, "a", "@", 3)) 
('This is @n awesome string @nd has lot of char@cters', 3) 
>>> print(random_replace(string, "a", "@", 10)) 
('This is @n @wesome string @nd h@s lot of ch@r@cters', 6) 
>>> print(random_replace(string, "a", "@", 0)) 
('This is an awesome string and has lot of characters', 0)