我必须在一个列表中搜索,并用另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都没有取得什么进展,最好的方法是什么?
例如,假设我的列表有以下整数
>>> 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。
我必须在一个列表中搜索,并用另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都没有取得什么进展,最好的方法是什么?
例如,假设我的列表有以下整数
>>> 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。
当前回答
我知道这是一个非常古老的问题,有无数种方法来解决它。我发现的更简单的方法是使用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)
其他回答
对于这个古老但相关的问题,答案的速度千差万别。
kxr发布的最快的解决方案。
然而,这甚至更快,否则不在这里:
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
下面是各种解决方案的时间安排。快的答案比公认答案快3倍,比最慢答案快5倍。
公平地说,所有的方法都需要对发送给函数的数组进行替换。
请参阅以下计时代码:
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)
结果是:
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% --
我可能是个笨蛋,但我会为它写一个单独的简单函数:
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
然后根据需要调用它,如下所示:
a = [1,2,3,4,5,1,2,3,4,5,1]
a_new = convertElements(a, {1: 10})
## OUTPUT: a_new=[10, 2, 3, 4, 5, 10, 2, 3, 4, 5, 10]
列表理解工作得很好,使用enumerate进行循环可以节省一些内存(b/c操作基本上是在适当的位置完成的)。
还有函数式编程。参见map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]
>>> 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]
>>>
如果你有几个值需要替换,你也可以使用字典:
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]
注意,这种方法仅在要替换的元素是可哈希的情况下才有效。这是因为字典键必须是可哈希的。