这些天我正在学习python;尝试各种挑战来改善我的观念。
我刚刚尝试过的一个迷你挑战是确定给定字符串中n个相邻重复字符的发生率。
我尝试如下：
def uniform_string(text):
text = text.lower()
i = 0
while i < (len(text)-3):
if text[i] == text[i+1] and text[i+1] == text[i+2] and text[i+2] == text[i+3]:
return True
i += 1
return False
这假设为n=4，我现在正在尝试概括它。
但另一方面，它也让我思考：有没有更有效的方法来实现这一目标？我当前的解决方案使我的查找次数比字符串的长度多4倍（这意味着，对于增加n的值，这会增加到O(n^2)）。
有什么比这更好的方法来应对这样的挑战？
答案 0 :(得分：2)
在以下函数中，n是要检查相等性的字符数，为了保持原始函数调用相同，您还可以将n的默认值设置为4
def uniform_string(text, n=4):
text = text.lower()
i = 0
while i < (len(text)-n):
if text[i:i + n] == text[i] * n:
return True
i += 1
return False
或者，您也可以使用for循环：
def uniform_string(text, n):
text = text.lower()
for i in range(len(text) - n + 1):
if text[i:i + n] == text[i] * n:
return True
return False
答案 1 :(得分：1)
您可以使用groupby对连续的相同字符进行分组，并计算每组中字符的数量。如果任何计数的阈值更高，则结果为True：
from itertools import groupby
def uniform_string(text, n):
return any(sum(1 for _ in g) >= n for _, g in groupby(text.lower()))
print(uniform_string('fofo', 1))
print(uniform_string('fofo', 2))
print(uniform_string('fofoo', 2))
输出：
True
False
True
上述时间复杂度为 O（n）。
答案 2 :(得分：1)
您可以使用slice s（some_list[start:stop]）和set来解决您的问题。
def uniform_string(text, n):
text = text.lower()
for i in range(len(text) - n + 1):
if len(set(text[i:i+n])) == 1: # all equal
return True
return False
如果使用for循环而不是while循环，您的代码也会更清晰一些。 ：）
答案 3 :(得分：1)
到目前为止发布的答案错过了Python的一个更好的迭代函数enumerate：
def uniform_string(text, n):
for i, c in enumerate(text):
if text[i:i+4] == c * n:
print( c, 'at', i, 'in', text )
我不确定这正是你所要求的，但它可能会让你有所作为。
答案 4 :(得分：1)
一个通配符条目：
优势非常快：立即分析4,000,000个字符的示例
缺点依赖于numpy;旋绕
这里是：
import numpy as np
a = "etr" + 1_000_000 * "zr" + "hhh" + 1_000_000 * "Ar"
np.max(np.diff(np.r_[-1, np.where(np.diff(np.frombuffer(a.encode('utf16'), dtype=np.uint16)[1:]))[0], len(a) - 1]))
3
工作原理：
更新：
这是一个混合版本，可以进行一些粗略的短路加上一些基本的基准测试，以找到最佳参数：
import numpy as np
from timeit import timeit
occ = 4
loc = (10, 20, 40, 80, 160, 320, 1000, 2000, 4000, 8000, 16000, 32000, 64000,
125000, 250000, 500000, 1_000_000, 2_000_000)
a = ['pafoe<03' + o * 'gr' + occ * 'x' + (2_000_000 - o) * 'u1'
+ 'leto50d-fjeoa'[occ:] for o in loc]
def brute_force(a):
np.max(np.diff(np.r_[-1, np.where(np.diff(np.frombuffer(
a.encode('utf16'), dtype=np.uint16)[1:]))[0], len(a) - 1]))
def reverse_bisect(a, chunk, encode_all=True):
j = 0
i = chunk
n = len(a)
if encode_all:
av = np.frombuffer(a.encode('utf16'), dtype=np.uint16)[1:]
while j<n:
if encode_all:
s = av[j : j + chunk]
else:
s = np.frombuffer(a[j:j+chunk].encode('utf16'), dtype=np.uint16)[1:]
if np.max(np.diff(np.r_[-1, np.where(np.diff(s))[0], len(s)-1])) >= occ:
return True
j += chunk - occ + 1
chunk *= 2
return False
leave_out = 2
out = []
print('first repeat at', loc[:-leave_out])
print('brute force {}'.format(
(timeit('[f(a) for a in A]', number=100, globals={
'f': brute_force, 'A': a[:-leave_out]}))))
print('hybrid (reverse bisect)')
for chunk in 2**np.arange(2, 18):
out.append(timeit('[f(a,c,e) for a in A]', number=100, globals={
'f': reverse_bisect, 'A': a[:-leave_out], 'c': chunk, 'e': True}))
out.append(timeit('[f(a,c,e) for a in A]', number=100, globals={
'f': reverse_bisect, 'A': a[:-leave_out], 'c': chunk, 'e': False}))
print('chunk: {}, timings: encode all {} -- encode chunks {}'.format(
chunk, out[-2], out[-1]))
样品运行：
first repeat at (10, 20, 40, 80, 160, 320, 1000, 2000, 4000, 8000, 16000, 32000, 64000, 125000, 250000, 500000)
brute force 90.26514193788171
hybrid (reverse bisect)
chunk: 4, timings: encode all 5.257935176836327 -- encode chunks 2.3392367498017848
chunk: 8, timings: encode all 5.210895746946335 -- encode chunks 2.288218504982069
chunk: 16, timings: encode all 5.268893962958828 -- encode chunks 2.2223802611697465
chunk: 32, timings: encode all 5.109196993988007 -- encode chunks 2.1715646600350738
chunk: 64, timings: encode all 5.05742059298791 -- encode chunks 2.1255820950027555
chunk: 128, timings: encode all 5.110778157133609 -- encode chunks 2.100305920932442
chunk: 256, timings: encode all 5.058305847924203 -- encode chunks 2.153960411902517
chunk: 512, timings: encode all 5.108077083015814 -- encode chunks 2.056686638854444
chunk: 1024, timings: encode all 4.969490061048418 -- encode chunks 2.0368234540801495
chunk: 2048, timings: encode all 5.153041162993759 -- encode chunks 2.465495347045362
chunk: 4096, timings: encode all 5.28073402796872 -- encode chunks 2.173405918991193
chunk: 8192, timings: encode all 5.044360157102346 -- encode chunks 2.1234876308590174
chunk: 16384, timings: encode all 5.294338152976707 -- encode chunks 2.334656815044582
chunk: 32768, timings: encode all 5.7856643970590085 -- encode chunks 2.877617093967274
chunk: 65536, timings: encode all 7.04935942706652 -- encode chunks 4.1559580829925835
chunk: 131072, timings: encode all 7.516369879012927 -- encode chunks 4.553452031919733
first repeat at (10, 20, 40)
brute force 16.363576064119115
hybrid (reverse bisect)
chunk: 4, timings: encode all 0.6122389689553529 -- encode chunks 0.045893668895587325
chunk: 8, timings: encode all 0.5982049370650202 -- encode chunks 0.03538667503744364
chunk: 16, timings: encode all 0.5907809699419886 -- encode chunks 0.025738760828971863
chunk: 32, timings: encode all 0.5741697370540351 -- encode chunks 0.01634934707544744
chunk: 64, timings: encode all 0.5719085780438036 -- encode chunks 0.013115004170686007
chunk: 128, timings: encode all 0.5666680270805955 -- encode chunks 0.011037093820050359
chunk: 256, timings: encode all 0.5664500128477812 -- encode chunks 0.010536623885855079
chunk: 512, timings: encode all 0.5695593091659248 -- encode chunks 0.01133729494176805
chunk: 1024, timings: encode all 0.5688401609659195 -- encode chunks 0.012476094998419285
chunk: 2048, timings: encode all 0.5702746720053256 -- encode chunks 0.014690137933939695
chunk: 4096, timings: encode all 0.5782928131520748 -- encode chunks 0.01891179382801056
chunk: 8192, timings: encode all 0.5943365979474038 -- encode chunks 0.0272749038413167
chunk: 16384, timings: encode all 0.609349318081513 -- encode chunks 0.04354232898913324
chunk: 32768, timings: encode all 0.6489383969455957 -- encode chunks 0.07695812894962728
chunk: 65536, timings: encode all 0.7388215309474617 -- encode chunks 0.14061277196742594
chunk: 131072, timings: encode all 0.8899400909431279 -- encode chunks 0.2977339250501245
答案 5 :(得分：0)
如果这是为了告诉您每个字符都相同，则可以执行以下操作：
def uniform_string(text):
text = text.lower()
if text.count(text[0]) == len(text):
return True
return False