字符串在 Python 中是不可变的，这意味着一旦创建了字符串，就不能更改字符串的内容.如果您需要更改它，则会使用更改创建字符串的新实例.

考虑到这一点，我们有很多方法可以解决这个问题

1. 使用 str.replace，

   >>>"它是冰".replace("i", "")'不是'

2. 使用 str.translate，

   >>>它是冰冷的".翻译(无，我")'不是'

3. 使用正则表达式，

   >>>进口重新>>>re.sub(r'i', "", "很冷")'不是'

4. 使用理解作为过滤器，

   >>>"".join([char for char in "it is icy" if char != "i"])'t 是 cy'

5. 使用filter函数

   >>>"".join(filter(lambda char: char != "i", "它是冰的"))'不是'

时间对比

def findreplace(m_string, char):m_string = 列表(m_string)对于 m_string 中的 k:如果 k == 字符:德尔(m_string[m_string.index(k)])返回 "".join(m_string)定义替换(m_string，字符):return m_string.replace("i", "")定义翻译(m_string，字符):返回 m_string.translate(None, "i")从时间导入时间print timeit("findreplace('it is icy','i')", "from __main__ import findreplace")print timeit("replace('it is icy','i')", "from __main__ import replace")print timeit("translate('it is icy','i')", "from __main__ import translate")

结果

1.644745826720.292785882950.311302900314

str.replace 和 str.translate 方法比接受的答案快 8 到 5 倍.

注意:理解方法和过滤方法预计会更慢，对于这种情况，因为它们必须创建列表，然后必须再次遍历以构造一个字符串.而 re 对于单个字符替换来说有点矫枉过正.因此，它们都被排除在时间比较之外.

How do I delete all the instances of a character in this string? Here is my code:

def findreplace(char, string):
    place = string.index(char)
    string[place] = ''
    return string

However, if I run this, this is what happens:

>>> findreplace('i', 'it is icy')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in findreplace
TypeError: 'str' object does not support item assignment

Why is this?

Strings are immutable in Python, which means once a string is created, you cannot alter the contents of the strings. If at all, you need to change it, a new instance of the string will be created with the alterations.

Having that in mind, we have so many ways to solve this

1. Using str.replace,

   >>> "it is icy".replace("i", "")
   't s cy'
   

2. Using str.translate,

   >>> "it is icy".translate(None, "i")
   't s cy'
   

3. Using Regular Expression,

   >>> import re
   >>> re.sub(r'i', "", "it is icy")
   't s cy'
   

4. Using comprehension as a filter,

   >>> "".join([char for char in "it is icy" if char != "i"])
   't s cy'
   

5. Using filter function

   >>> "".join(filter(lambda char: char != "i", "it is icy"))
   't s cy'
   

Timing comparison

def findreplace(m_string, char):
    m_string = list(m_string)
    for k in m_string:
        if k == char:
            del(m_string[m_string.index(k)])
    return "".join(m_string)

def replace(m_string, char):
    return m_string.replace("i", "")

def translate(m_string, char):
    return m_string.translate(None, "i")

from timeit import timeit

print timeit("findreplace('it is icy','i')", "from __main__ import findreplace")
print timeit("replace('it is icy','i')", "from __main__ import replace")
print timeit("translate('it is icy','i')", "from __main__ import translate")

Result

1.64474582672
0.29278588295
0.311302900314

str.replace and str.translate methods are 8 and 5 times faster than the accepted answer.

Note: Comprehension method and filter methods are expected to be slower, for this case, since they have to create list and then they have to be traversed again to construct a string. And re is a bit overkill for a single character replacement. So, they all are excluded from the timing comparison.