Question

我需要检查是否可以使用字符列表创建字符串并返回True或False。

我在list.count或collections.Counter中使用不同的解决方案。

我也在使用这个解决方案，我不需要阅读字符列表：

def check(message, characters):
    try:
        [list(characters).remove(m) for m in message]
        return True
    except:
        return False

有最快的方法吗？对于一个非常非常大的角色列表。计数器和列表计数似乎较慢。不知道是否有快速的pythonic方法来做到这一点。

示例：

message = "hello"
characters = "hheellooasdadsfgfdgfdhgfdlkgkfd"

check(message, characters) # this should return True or False
# characters can be a veeeeery long string

重复事项，例如字符=＆＃34; hheloo＆＃34;不适用于message =＆＃34;你好＆＃34;

Answer 1

您可以使用collections.Counter()。只需构建两个计数器并使用subtract()方法检查是否存在任何负数：

>>> c1 = Counter(characters)
>>> c2 = Counter(message)
>>> c1.subtract(c2)
>>> all(v >= 0 for v in c1.values())
False

这应该在线性时间内起作用。

Answer 2

这在线性时间内是不可行的，因为两个字符串的长度很重要，并且需要为每个字符迭代它们。在没有检查其实际实现的情况下，我认为remove()是对数的。

def check(msg, chars):
    c = list(chars)  # Creates a copy
    try:
        for m in msg:
            c.remove(m)
    except ValueError:
        return False
    return True

if __name__ == '__main__':
    print(check('hello', 'ehlo'))
    print(check('hello', 'ehlol'))
    print(check('hello', 'ehloijin2oinscubnosinal'))

Answer 3

与eugene的解决方案和jbndlr的解决方案相比，这是另一种解决方案。

def test1(input_word, alphabet):
    alp_set = set(list(alphabet))
    in_set = set(list(input_word))
    return in_set.issubset(alp_set)

def test2(input_word, alphabet):
    c1 = collections.Counter(alphabet)
    c2 = collections.Counter(input_word)
    c1.subtract(c2)
    return all(v >= 0 for v in c1.values())

def check(msg, chars):
    c = list(chars)  # Creates a copy
    try:
        for m in msg:
            c.remove(m)
    except ValueError:
        return False
    return True

input_word = "hello"
alphabet = "hheellooasdadsfgfdgfdhgfdlkgkfd"


start_time = time.time()
for i in range(10000):
    test1(input_word,alphabet)
print("--- %s seconds ---" % (time.time() - start_time))

start_time = time.time()
for i in range(10000):
    test2(input_word,alphabet)
print("--- %s seconds ---" % (time.time() - start_time))

start_time = time.time()
   for i in range(10000):
       check(input_word,alphabet)
   print("--- %s seconds ---" % (time.time() - start_time))

>> --- 0.03100299835205078 seconds ---
>> --- 0.24402451515197754 seconds ---
>> --- 0.022002220153808594 seconds ---

⇒jbndlr的解决方案是最快的 - 对于这个测试用例。

另一个测试用例：

input_word = "hellohellohellohellohellohellohellohellohellohellohellohellohello"
alphabet =

“hheellooasdadsfgfdgfdhgfdlkgkfdhheellooasdadsfgfdgfdhgfdlkgkfdhheellooasdadsfgfdgfdhgfdlkgkfdhheellooasdadsfgfdgfdhgfdlkgkfdhheellooasdadsfgfdgfdhgfdlkgkfdhheellooasdadsfgfdgfdhgfdlkgkfdhheellooasdadsfgfdgfdhgfdlkgkfdhheellooasdadsfgfdgfdhgfdlkgkfdhheellooasdadsfgfdgfdhgfdlkgkfdhheellooasdadsfgfdgfdhgfdlkgkfdhheellooasdadsfgfdgfdhgfdlkgkfdhheellooasdadsfgfdgfdhgfdlkgkfdhheellooasdadsfgfdgfdhgfdlkgkfdhheellooasdadsfgfdgfdhgfdlkgkfdhheellooasdadsfgfdgfdhgfdlkgkfdhheellooasdadsfgfdgfdhgfdlkgkfdhheellooasdadsfgfdgfdhgfdlkgkfdhheellooasdadsfgfdgfdhgfdlkgkfd”

>> --- 0.21964788436889648 seconds ---
>> --- 0.518169641494751 seconds ---
>> --- 1.3148927688598633 seconds ---

⇒test1是最快的

Answer 4

有一种更快的方法可以做到这一点，显然是由于创建all（）生成器（Why is Python's 'all' function so slow?）的成本，或许for循环更快，扩展@eugene y的回答：

from collections import Counter
import time

message = "hello"
characters = "hheeooasdadsfgfdgfdhgfdlkgkfd"

def check1(message,characters):
    c1 = Counter(characters)
    c2 = Counter(message)
    c1.subtract(c2)
    return all(v > -1 for v in c1.values())

def check2(message,characters):
    c1 = Counter(characters)
    c2 = Counter(message)
    c1.subtract(c2)
    for v in c1.values():
        if v < 0:
            return False
    return True

st = time.time()
for i in range(350000):
    check1(message,characters)
end = time.time()
print ("all(): "+str(end-st))

st = time.time()
for i in range(350000):
    check2(message,characters)
end = time.time()
print ("for loop: "+str(end-st))

结果：

all(): 5.201688051223755
for loop: 4.864434719085693

检查是否可以使用python中的字符列表创建字符串的最快方法

4 个答案: