问题:查找和替换列表中的元素

我必须搜索一个列表,然后用一个元素替换所有出现的元素。到目前为止,我在代码方面的尝试使我无处可寻,做到这一点的最佳方法是什么?

例如,假设我的列表具有以下整数

>>> a = [1,2,3,4,5,1,2,3,4,5,1]

我需要将所有出现的数字1替换为值10,所以我需要的输出是

>>> a = [10, 2, 3, 4, 5, 10, 2, 3, 4, 5, 10]

因此,我的目标是将数字1的所有实例替换为数字10。

I have to search through a list and replace all occurrences of one element with another. So far my attempts in code are getting me nowhere, what is the best way to do this?

For example, suppose my list has the following integers

>>> a = [1,2,3,4,5,1,2,3,4,5,1]

and I need to replace all occurrences of the number 1 with the value 10 so the output I need is

>>> a = [10, 2, 3, 4, 5, 10, 2, 3, 4, 5, 10]

Thus my goal is to replace all instances of the number 1 with the number 10.


回答 0

>>> a= [1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1]
>>> for n, i in enumerate(a):
...   if i == 1:
...      a[n] = 10
...
>>> a
[10, 2, 3, 4, 5, 10, 2, 3, 4, 5, 10]
>>> a= [1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1]
>>> for n, i in enumerate(a):
...   if i == 1:
...      a[n] = 10
...
>>> a
[10, 2, 3, 4, 5, 10, 2, 3, 4, 5, 10]

回答 1

尝试使用列表推导三元运算符

>>> a=[1,2,3,1,3,2,1,1]
>>> [4 if x==1 else x for x in a]
[4, 2, 3, 4, 3, 2, 4, 4]

Try using a list comprehension and the ternary operator.

>>> a=[1,2,3,1,3,2,1,1]
>>> [4 if x==1 else x for x in a]
[4, 2, 3, 4, 3, 2, 4, 4]

回答 2

列表理解效果很好,使用enumerate循环可以节省一些内存(b / c操作实际上已就位)。

还有功能编程。查看地图用法:

>>> a = [1,2,3,2,3,4,3,5,6,6,5,4,5,4,3,4,3,2,1]
>>> map(lambda x: x if x != 4 else 'sss', a)
[1, 2, 3, 2, 3, 'sss', 3, 5, 6, 6, 5, 'sss', 5, 'sss', 3, 'sss', 3, 2, 1]

List comprehension works well, and looping through with enumerate can save you some memory (b/c the operation’s essentially being done in place).

There’s also functional programming. See usage of map:

>>> a = [1,2,3,2,3,4,3,5,6,6,5,4,5,4,3,4,3,2,1]
>>> map(lambda x: x if x != 4 else 'sss', a)
[1, 2, 3, 2, 3, 'sss', 3, 5, 6, 6, 5, 'sss', 5, 'sss', 3, 'sss', 3, 2, 1]

回答 3

如果您要替换多个值,则还可以使用字典:

a = [1, 2, 3, 4, 1, 5, 3, 2, 6, 1, 1]
dic = {1:10, 2:20, 3:'foo'}

print([dic.get(n, n) for n in a])

> [10, 20, 'foo', 4, 10, 5, 'foo', 20, 6, 10, 10]

If you have several values to replace, you can also use a dictionary:

a = [1, 2, 3, 4, 1, 5, 3, 2, 6, 1, 1]
dic = {1:10, 2:20, 3:'foo'}

print([dic.get(n, n) for n in a])

> [10, 20, 'foo', 4, 10, 5, 'foo', 20, 6, 10, 10]

回答 4

>>> a=[1,2,3,4,5,1,2,3,4,5,1]
>>> item_to_replace = 1
>>> replacement_value = 6
>>> indices_to_replace = [i for i,x in enumerate(a) if x==item_to_replace]
>>> indices_to_replace
[0, 5, 10]
>>> for i in indices_to_replace:
...     a[i] = replacement_value
... 
>>> a
[6, 2, 3, 4, 5, 6, 2, 3, 4, 5, 6]
>>> 
>>> a=[1,2,3,4,5,1,2,3,4,5,1]
>>> item_to_replace = 1
>>> replacement_value = 6
>>> indices_to_replace = [i for i,x in enumerate(a) if x==item_to_replace]
>>> indices_to_replace
[0, 5, 10]
>>> for i in indices_to_replace:
...     a[i] = replacement_value
... 
>>> a
[6, 2, 3, 4, 5, 6, 2, 3, 4, 5, 6]
>>> 

回答 5

a = [1,2,3,4,5,1,2,3,4,5,1,12]
for i in range (len(a)):
    if a[i]==2:
        a[i]=123

您可以使用for和while循环;但是,如果您知道内置的枚举功能,则建议使用枚举。1个

a = [1,2,3,4,5,1,2,3,4,5,1,12]
for i in range (len(a)):
    if a[i]==2:
        a[i]=123

You can use a for and or while loop; however if u know the builtin Enumerate function, then it is recommended to use Enumerate.1


回答 6

所有轻易更换110a = [1,2,3,4,5,1,2,3,4,5,1]一个可以使用下面的一行拉姆达+地图相结合,与“看,妈妈,没有如果或者维权!” :

# This substitutes all '1' with '10' in list 'a' and places result in list 'c':

c = list(map(lambda b: b.replace("1","10"), a))

To replace easily all 1 with 10 in a = [1,2,3,4,5,1,2,3,4,5,1]one could use the following one-line lambda+map combination, and ‘Look, Ma, no IFs or FORs!’ :

# This substitutes all '1' with '10' in list 'a' and places result in list 'c':

c = list(map(lambda b: b.replace("1","10"), a))


回答 7

以下是Python 2.x中非常直接的方法

 a = [1,2,3,4,5,1,2,3,4,5,1]        #Replacing every 1 with 10
 for i in xrange(len(a)):
   if a[i] == 1:
     a[i] = 10  
 print a

此方法有效。欢迎发表评论。希望能帮助到你 :)

也请尝试了解outisdamzam的解决方案的工作方式。列表压缩和lambda函数是有用的工具。

The following is a very straightforward method in Python 3.x

 a = [1,2,3,4,5,1,2,3,4,5,1]        #Replacing every 1 with 10
 for i in range(len(a)):
   if a[i] == 1:
     a[i] = 10  
 print(a)

This method works. Comments are welcome. Hope it helps :)

Also try understanding how outis’s and damzam’s solutions work. List compressions and lambda function are useful tools.


回答 8

我知道这是一个非常老的问题,并且有很多方法可以解决。我发现较简单的一种是使用numpy包。

import numpy

arr = numpy.asarray([1, 6, 1, 9, 8])
arr[ arr == 8 ] = 0 # change all occurrences of 8 by 0
print(arr)

I know this is a very old question and there’s a myriad of ways to do it. The simpler one I found is using numpy package.

import numpy

arr = numpy.asarray([1, 6, 1, 9, 8])
arr[ arr == 8 ] = 0 # change all occurrences of 8 by 0
print(arr)

回答 9

我的用例已替换None为一些默认值。

我已经定时提出了解决此问题的方法,包括@kxr-using str.count

使用Python 3.8.1在ipython中测试代码:

def rep1(lst, replacer = 0):
    ''' List comprehension, new list '''

    return [item if item is not None else replacer for item in lst]


def rep2(lst, replacer = 0):
    ''' List comprehension, in-place '''    
    lst[:] =  [item if item is not None else replacer for item in lst]

    return lst


def rep3(lst, replacer = 0):
    ''' enumerate() with comparison - in-place '''
    for idx, item in enumerate(lst):
        if item is None:
            lst[idx] = replacer

    return lst


def rep4(lst, replacer = 0):
    ''' Using str.index + Exception, in-place '''

    idx = -1
    # none_amount = lst.count(None)
    while True:
        try:
            idx = lst.index(None, idx+1)
        except ValueError:
            break
        else:
            lst[idx] = replacer

    return lst


def rep5(lst, replacer = 0):
    ''' Using str.index + str.count, in-place '''

    idx = -1
    for _ in range(lst.count(None)):
        idx = lst.index(None, idx+1)
        lst[idx] = replacer

    return lst


def rep6(lst, replacer = 0):
    ''' Using map, return map iterator '''

    return map(lambda item: item if item is not None else replacer, lst)


def rep7(lst, replacer = 0):
    ''' Using map, return new list '''

    return list(map(lambda item: item if item is not None else replacer, lst))


lst = [5]*10**6
# lst = [None]*10**6

%timeit rep1(lst)    
%timeit rep2(lst)    
%timeit rep3(lst)    
%timeit rep4(lst)    
%timeit rep5(lst)    
%timeit rep6(lst)    
%timeit rep7(lst)    

我得到:

26.3 ms ± 163 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
29.3 ms ± 206 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
33.8 ms ± 191 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
11.9 ms ± 37.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
11.9 ms ± 60.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
260 ns ± 1.84 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
56.5 ms ± 204 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

str.index实际上,使用内部比任何手动比较都快。

我不知道测试4中的异常是否比使用更费力str.count,差异似乎可以忽略不计。

请注意,map()(测试6)返回迭代器而不是实际列表,因此测试7。

My usecase was replacing None with some default value.

I’ve timed approaches to this problem that were presented here, including the one by @kxr – using str.count.

Test code in ipython with Python 3.8.1:

def rep1(lst, replacer = 0):
    ''' List comprehension, new list '''

    return [item if item is not None else replacer for item in lst]


def rep2(lst, replacer = 0):
    ''' List comprehension, in-place '''    
    lst[:] =  [item if item is not None else replacer for item in lst]

    return lst


def rep3(lst, replacer = 0):
    ''' enumerate() with comparison - in-place '''
    for idx, item in enumerate(lst):
        if item is None:
            lst[idx] = replacer

    return lst


def rep4(lst, replacer = 0):
    ''' Using str.index + Exception, in-place '''

    idx = -1
    # none_amount = lst.count(None)
    while True:
        try:
            idx = lst.index(None, idx+1)
        except ValueError:
            break
        else:
            lst[idx] = replacer

    return lst


def rep5(lst, replacer = 0):
    ''' Using str.index + str.count, in-place '''

    idx = -1
    for _ in range(lst.count(None)):
        idx = lst.index(None, idx+1)
        lst[idx] = replacer

    return lst


def rep6(lst, replacer = 0):
    ''' Using map, return map iterator '''

    return map(lambda item: item if item is not None else replacer, lst)


def rep7(lst, replacer = 0):
    ''' Using map, return new list '''

    return list(map(lambda item: item if item is not None else replacer, lst))


lst = [5]*10**6
# lst = [None]*10**6

%timeit rep1(lst)    
%timeit rep2(lst)    
%timeit rep3(lst)    
%timeit rep4(lst)    
%timeit rep5(lst)    
%timeit rep6(lst)    
%timeit rep7(lst)    

I get:

26.3 ms ± 163 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
29.3 ms ± 206 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
33.8 ms ± 191 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
11.9 ms ± 37.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
11.9 ms ± 60.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
260 ns ± 1.84 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
56.5 ms ± 204 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Using the internal str.index is in fact faster than any manual comparison.

I didn’t know if the exception in test 4 would be more laborious than using str.count, the difference seems negligible.

Note that map() (test 6) returns an iterator and not an actual list, thus test 7.


回答 10

list.index()与其他答案中提出的单步迭代方法相比,在长列表和罕见情况下,使用它的速度大约快3倍。

def list_replace(lst, old=1, new=10):
    """replace list elements (inplace)"""
    i = -1
    try:
        while 1:
            i = lst.index(old, i + 1)
            lst[i] = new
    except ValueError:
        pass

On long lists and rare occurrences its about 3x faster using list.index() – compared to single step iteration methods presented in the other answers.

def list_replace(lst, old=1, new=10):
    """replace list elements (inplace)"""
    i = -1
    try:
        while 1:
            i = lst.index(old, i + 1)
            lst[i] = new
    except ValueError:
        pass

回答 11

您可以在python中简单地使用列表理解:

def replace_element(YOUR_LIST, set_to=NEW_VALUE):
    return [i
            if SOME_CONDITION
            else NEW_VALUE
            for i in YOUR_LIST]

对于您的情况,要将所有出现的1替换为10,代码片段将如下所示:

def replace_element(YOUR_LIST, set_to=10):
    return [i
            if i != 1  # keeps all elements not equal to one
            else set_to  # replaces 1 with 10
            for i in YOUR_LIST]

You can simply use list comprehension in python:

def replace_element(YOUR_LIST, set_to=NEW_VALUE):
    return [i
            if SOME_CONDITION
            else NEW_VALUE
            for i in YOUR_LIST]

for your case, where you want to replace all occurrences of 1 with 10, the code snippet will be like this:

def replace_element(YOUR_LIST, set_to=10):
    return [i
            if i != 1  # keeps all elements not equal to one
            else set_to  # replaces 1 with 10
            for i in YOUR_LIST]

回答 12

仅查找和替换一项

ur_list = [1,2,1]     # replace the first 1 wiz 11

loc = ur_list.index(1)
ur_list.remove(1)
ur_list.insert(loc, 11)

----------
[11,2,1]

Find & replace just one item

your_list = [1,2,1]     # replace the first 1 with 11

loc = your_list.index(1)
your_list.remove(1)
your_list.insert(loc, 11)

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