"查找和替换列表中的元素"

Question

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.

Answer 1

Try using a list comprehension and a conditional expression .

>>> 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]

Answer 2

You can use the built-in enumerate to get both index and value while iterating the list. Then, use the value to test for a condition and the index to replace that value in the original list:

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

Answer 3

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]
replacements = {1:10, 2:20, 3:'foo'}
replacer = replacements.get  # For faster gets.

print([replacer(n, n) for n in a])

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

Note that this approach works only if the elements to be replaced are hashable. This is because dict keys are required to be hashable.

Answer 4

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

>>> 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]

Answer 5

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.

Answer 6

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

Answer 7

>>> 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]
>>> 

Answer 8

On long lists and rare occurrences its about 3x faster using list.index()<\/code> - 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 True:
            i = lst.index(old, i + 1)
            lst[i] = new
    except ValueError:
        pass

Answer 9

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))

Answer 10

一个班轮和最快的方法来做到这一点：

[10 if x==1 else x for x in a]

Answer 11

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.

Answer 12

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)

Answer 13

My usecase was replacing None<\/code> with some default value.

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)    

Answer 14

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]

Answer 15

The answers for this old but relevant question are wildly variable in speed.

The fastest of the solution posted by kxr.

However, this is even faster and otherwise not here:

def f1(arr, find, replace):
    # fast and readable
    base=0
    for cnt in range(arr.count(find)):
        offset=arr.index(find, base)
        arr[offset]=replace
        base=offset+1

Here is timing for the various solutions. The faster ones are 3X faster than accepted answer and 5X faster than the slowest answer here.

To be fair, all methods needed to do inlace replacement of the array sent to the function.

Please see timing code below:

def f1(arr, find, replace):
    # fast and readable
    base=0
    for cnt in range(arr.count(find)):
        offset=arr.index(find, base)
        arr[offset]=replace
        base=offset+1
        
def f2(arr,find,replace):
    # accepted answer
    for i,e in enumerate(arr):
        if e==find: 
            arr[i]=replace
        
def f3(arr,find,replace):
    # in place list comprehension
    arr[:]=[replace if e==find else e for e in arr]
    
def f4(arr,find,replace):
    # in place map and lambda -- SLOW
    arr[:]=list(map(lambda x: x if x != find else replace, arr))
    
def f5(arr,find,replace):
    # find index with comprehension
    for i in [i for i, e in enumerate(arr) if e==find]:
        arr[i]=replace
        
def f6(arr,find,replace):
    # FASTEST but a little les clear
    try:
        while True:
            arr[arr.index(find)]=replace
    except ValueError:
        pass    

def f7(lst, old, new):
    """replace list elements (inplace)"""
    i = -1
    try:
        while 1:
            i = lst.index(old, i + 1)
            lst[i] = new
    except ValueError:
        pass
    
    
import time     

def cmpthese(funcs, args=(), cnt=1000, rate=True, micro=True):
    """Generate a Perl style function benchmark"""                   
    def pprint_table(table):
        """Perl style table output"""
        def format_field(field, fmt='{:,.0f}'):
            if type(field) is str: return field
            if type(field) is tuple: return field[1].format(field[0])
            return fmt.format(field)     

        def get_max_col_w(table, index):
            return max([len(format_field(row[index])) for row in table])         

        col_paddings=[get_max_col_w(table, i) for i in range(len(table[0]))]
        for i,row in enumerate(table):
            # left col
            row_tab=[row[0].ljust(col_paddings[0])]
            # rest of the cols
            row_tab+=[format_field(row[j]).rjust(col_paddings[j]) for j in range(1,len(row))]
            print(' '.join(row_tab))                

    results={}
    for i in range(cnt):
        for f in funcs:
            start=time.perf_counter_ns()
            f(*args)
            stop=time.perf_counter_ns()
            results.setdefault(f.__name__, []).append(stop-start)
    results={k:float(sum(v))/len(v) for k,v in results.items()}     
    fastest=sorted(results,key=results.get, reverse=True)
    table=[['']]
    if rate: table[0].append('rate/sec')
    if micro: table[0].append('\u03bcsec/pass')
    table[0].extend(fastest)
    for e in fastest:
        tmp=[e]
        if rate:
            tmp.append('{:,}'.format(int(round(float(cnt)*1000000.0/results[e]))))

        if micro:
            tmp.append('{:,.1f}'.format(results[e]/float(cnt)))

        for x in fastest:
            if x==e: tmp.append('--')
            else: tmp.append('{:.1%}'.format((results[x]-results[e])/results[e]))
        table.append(tmp) 

    pprint_table(table)                    



if __name__=='__main__':
    import sys
    import time 
    print(sys.version)
    cases=(
        ('small, found', 9, 100),
        ('small, not found', 99, 100),
        ('large, found', 9, 1000),
        ('large, not found', 99, 1000)
    )
    for txt, tgt, mul in cases:
        print(f'\n{txt}:')
        arr=[1,2,3,4,5,6,7,8,9,0]*mul 
        args=(arr,tgt,'X')
        cmpthese([f1,f2,f3, f4, f5, f6, f7],args)   

And the results:

3.9.1 (default, Feb  3 2021, 07:38:02) 
[Clang 12.0.0 (clang-1200.0.32.29)]

small, found:
   rate/sec μsec/pass     f4     f3     f5     f2     f6     f7     f1
f4  133,982       7.5     -- -38.8% -49.0% -52.5% -78.5% -78.6% -82.9%
f3  219,090       4.6  63.5%     -- -16.6% -22.4% -64.8% -65.0% -72.0%
f5  262,801       3.8  96.1%  20.0%     --  -6.9% -57.8% -58.0% -66.4%
f2  282,259       3.5 110.7%  28.8%   7.4%     -- -54.6% -54.9% -63.9%
f6  622,122       1.6 364.3% 184.0% 136.7% 120.4%     --  -0.7% -20.5%
f7  626,367       1.6 367.5% 185.9% 138.3% 121.9%   0.7%     -- -19.9%
f1  782,307       1.3 483.9% 257.1% 197.7% 177.2%  25.7%  24.9%     --

small, not found:
   rate/sec μsec/pass     f4     f5     f2     f3     f6     f7     f1
f4   13,846      72.2     -- -40.3% -41.4% -47.8% -85.2% -85.4% -86.2%
f5   23,186      43.1  67.5%     --  -1.9% -12.5% -75.2% -75.5% -76.9%
f2   23,646      42.3  70.8%   2.0%     -- -10.8% -74.8% -75.0% -76.4%
f3   26,512      37.7  91.5%  14.3%  12.1%     -- -71.7% -72.0% -73.5%
f6   93,656      10.7 576.4% 303.9% 296.1% 253.3%     --  -1.0%  -6.5%
f7   94,594      10.6 583.2% 308.0% 300.0% 256.8%   1.0%     --  -5.6%
f1  100,206      10.0 623.7% 332.2% 323.8% 278.0%   7.0%   5.9%     --

large, found:
   rate/sec μsec/pass     f4     f2     f5     f3     f6     f7     f1
f4      145   6,889.4     -- -33.3% -34.8% -48.6% -85.3% -85.4% -85.8%
f2      218   4,593.5  50.0%     --  -2.2% -22.8% -78.0% -78.1% -78.6%
f5      223   4,492.4  53.4%   2.3%     -- -21.1% -77.5% -77.6% -78.2%
f3      282   3,544.0  94.4%  29.6%  26.8%     -- -71.5% -71.6% -72.3%
f6      991   1,009.5 582.4% 355.0% 345.0% 251.1%     --  -0.4%  -2.8%
f7      995   1,005.4 585.2% 356.9% 346.8% 252.5%   0.4%     --  -2.4%
f1    1,019     981.3 602.1% 368.1% 357.8% 261.2%   2.9%   2.5%     --

large, not found:
   rate/sec μsec/pass     f4     f5     f2     f3     f6     f7     f1
f4      147   6,812.0     -- -35.0% -36.4% -48.9% -85.7% -85.8% -86.1%
f5      226   4,424.8  54.0%     --  -2.0% -21.3% -78.0% -78.1% -78.6%
f2      231   4,334.9  57.1%   2.1%     -- -19.6% -77.6% -77.7% -78.2%
f3      287   3,484.0  95.5%  27.0%  24.4%     -- -72.1% -72.2% -72.8%
f6    1,028     972.3 600.6% 355.1% 345.8% 258.3%     --  -0.4%  -2.7%
f7    1,033     968.2 603.6% 357.0% 347.7% 259.8%   0.4%     --  -2.3%
f1    1,057     946.2 619.9% 367.6% 358.1% 268.2%   2.8%   2.3%     --

Answer 16

I might be a dumb-dumb, but I would write a separate, simple function for this:

def convertElements( oldlist, convert_dict ):
  newlist = []
  for e in oldlist:
    if e in convert_dict:
      newlist.append(convert_dict[e])
    else:
      newlist.append(e)
  return newlist

Answer 17

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.