如何在正确查找字符串序列时重置计数器

Question

所以我正在处理cs50 dna问题，我在计数器上遇到了困难，因为我不知道如何编码来正确计算我要查找的序列在没有其他序列的情况下重复的最高次数。例如，我正在寻找序列AAT，文本是AATDHDHDTKSDHAATAAT，因此最高数量应该是两个，因为最后两个序列是AAT，并且它们之间没有序列。

这是我的代码：

text="TCTAGTCTAGTCTAGTCTAGTCTAGACTTGTCGCTGACTCCGAGAAGATCCTAACATTAACCAATTCCCCCTAGTCTGAGGCACGGTTACCGATCGGGTTAATGGATCTCTCACCGTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAAACGTGTAACTGTAATAATCCGCCCGAAAAAACTGATCTTAGGGTTGCGGCATCTGCACGTGACAGTGTGCTACTGTTAGATAGAGGGATCAAACGAGGTTGCAAGGATTATATCTCTCCGTGCTCGATAAGACACAGCCGGTTGCGGGCTGCTTCCTCTGGATCCAATGCAGCCGTACGTACACCGTAGAGCAAATTTAGTGGTAAAGGAACTTGCTCAAACACTACGGCTTCGGGCTACTGTTGGCGCCGGTTGGGGATCCCATTCAACGCTGGCCCTTTCGCTATGGTTCGGTGATTTTACACCGAAGCGAACCTTGAACCGTGGATTTCGGGTGTCCTCCGTTTTTAGGTACTGCGTGCAGACATGGGCACCTGCCATAGTGCGATCAGCCAGAATCCATTGTATGGGAGTTGGACTCGTTTGAATTTACCGGAAACCTCATGCTTGGTCTGTAGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGAAACTGGGCGACTTGAAGTCGGCTTGCGTATTAATAGCTCTGCAATGTAACTCGGCCCTTGGCGGCGGGCAGCTTAGTATTGAACCGCGACACACCATAGGTGCGGCAAATATTAAAAGTACGCTCGAACCGGAACCTGTCTCCATGACTGGACGACCAGCCCGGCGTCTTCTACGTAACACAGGGGGCTGTCGAGGTAGGGCGTAGGAACTTCGGGGTCACTACGCCGTAACAGCACCGAATATCATATCATCCAACTTGCTTGGTACATGCCCCGTTCTGTATCAAAAGTTTACGGCCCCGGACATACCTGCTGTCAGTTGAATACCTATGCGAGTCTGAAACACGAATAGTTCAGGCGTGCAAAGACACGCTAAGCACACGCCGCAGGCAGGGGGGGTATTTTATAAGTCGTTTTTTGGAAGGGTAATGTAAACTTATCCCATAATACCCTTTGGCTTCCCCTCACTCGTGCACTTCTCATAATGATACGTCAGGGTGATTGTAGATTCACGCGTCATCAGATTGTCCCTTTCTCGAGTCTTAGTATCTTTCCTAATCCGCTCGACTCTGCGCCATGATCGAATTCCTGACAGGCTACAAGAATAAACTGCCAGCATACTCCTTACCGATTGGCGCCTACTAATTATACGCACATGGGCATCTTCGACGTCTAAACATAGGCTCTTAGTATTCCGTAGGATGTTGAGCCGACAGGAAAGTCAAACGTCGTGGGTGACCGTAGCCTGACTCGCCCGACGCAGGATTCGCTCATATGTGTGAACGGATGCTTATGTAACTTCCTAATTGCAGCGAATGGCAGTTCCGTAGTGAAGGTTCGAAACGTACGGGGTCCGGCCATGGATTAGATCTTTCAGTGCGCTAAACTCTTAACCGCAGATACTTGGCGGACCATCTTCGTGTTGCTACTATGGTATAGACCAGGCTGTCGAATCTACTTAACACAGGTGAACCCCCAGATCGGCTAGAGCCTTCGAGGCTAGACCTTTAACAATCTTTAGACACTTCCAAATCGCGGCCGGATATGTCTCGTTGGCAGCCGCAGACAAGAGAAGAGGGTCGGCAGTGTCTGCCACGCGTGACCTGTATGATCTTAGCCTTTAAGATCACACTACTGATCACAATCTATTATGATTGCCTTAGCTAACTGAGTGATGCACCCCCACAGGCTGAGAGAAATCTGTAGTTTGACGACACGCCGTCTGGCTAAAAATGTGAATCCGCCGATCCGAGACGGTGGAAGCTTGAGACCAAATGCGGGAAACCAATGACTTCATTACGGAACAAGACATAACGGCGTGAGTTGACGACTGGGATTAACCCTTTCCCGAGTCTGTACTTCTGCTACACAATGAGGATGCGAATTATCTAAGACCTTGTACTACCTAAACTAACCCTGAGGCGGGCATTGAATTCCGGCCATCTTCAGCCCAAAGAAAGACCAAATGTGAGGAAAATGAGGGATCGGTATAAGCTTTTCACGATCTCAAGGTTCACGGCCGCCAGGGCCGTAGTTGGGGCTTCATGCACATTGCCAACCCGGACATCGACAGTCGGTACCGCAGGGGTTCGAGGAATACTCCCAGCTGTGACACCTGGTCGTCGACTGGACCCAGCTGGTGGGCGGCATAGGTAGTTAATACTGAATTAAAGCCGGGAACGTCTCTCTAACTAGAAACCTTGTGATAGGATACACAGACCTAGTGCCCCGACGTTAGCATTTGAATTCATCTATCTTGGCGTCTTTTAGTAGGCCTGGGTCAACTCCGGCGTTGGCCAAAATAACCGATCTGCGTTATGTGGCCACGCATCGAGTGACAGGGTGCATACAAATTGATGGTCAAAGAGTTTAAACAAGACAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCCCCACGCTTCTACATAGCCACACTGGAGCTAGTCCTCGTGTTAAATTTTTCGCTTGTTGCACGGTTATCATCAGAAGTGCCACTGGTATTCCTCTGTAGCTCCCGTATGCCGAAGGTTGCGGCTTAGGTACTGCTTATACACGTCTCTCAAGTTTGTCAGCCGCGTGATCTTTCTGCGGGGATAGGTGATCGTCCCTCGCTCCGGACATTGCATTAAAATTACCTAGTTGATAGGGCGGCGGAGTTGCATACCGGCGTTCAATCGCGGCTCCAGACTGGTTTGAGCTACGCGTCTGCCAGCGTGAAAAAGCTGATTTGTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCCAGGTATTATCATTTGAATCGTATGTTTTCTGCCGTACGTCAACTGCGTCGTCGGGGACTGAAATGGTCTGCCTCCAGACCCTTACCTCCCGATAAGCCATGACTAAGTATGTGAAGGATCACCTGAATTGCTGAAAGTTAACGGTAAGATATCTGAAAGAGCTCATTAGATCCAACACTTATCTACTCAAAAATTCGTCATATTTCGGTGACTTGCTAGAAAGGCTCTTGCACAGTAAGGTTATAGAGAATGCTACCGTTGAAGCACCAGCCGTTGAAGCCCGCCTTTAACCACGCGATATATCCAATTAACCAAGGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGTCGCCTTGTAATAATTACTTTGGCCCGGATTATAACGAAGGAACTCGCCATGAACTCGCAGCACGTTGTACTGGAACAATCTACTTTTTATAATATAGCGATAACTCCCAGCTTTTATGTGGGTGATATTGTCCTAGCTTTTTAAAGATACCCTCTGGCCCGGTCCAAGTAAGGTCCACATTGCCTGACGTAAGCGTACGGTCAACGGGTGCACCGGTTCCCGCTAAAGCTCGATCCTATTCTTTCAGTCGGGGGGAAATAAACTCGTATACTCTCCACCCACCCGTACGTCCCGGACTAGAATAACTACCGGGTATTTCCGGTTCGTAACACCACGCCATGACGTGTCAACATAAACGCTTCTTTTGAAAGGTGCACATGCAGATTGCACAAGCAGCAGGCACCGCCCTTATCCATATCCTGTTGAGGCCCTCGATCCTAGTGTTCCTTGTTATCAGGATATTTTCTCGCTGTACGTTATTGTCCTTTTCAAATTACAACTGACCGCTTCCTCACCCGCTAAACCCTACCTTACGCACAACCAAGGCCTTGTCCCGGATGAACCCGGCTGCTCCTATGGATAAGCAACCCAGCCCGGCAGTTTACTTCAGGTGTTATCGTCGACTGACACCCTCAGCTTTCTCCCATTACACAGCGAGTATTTTCCGCGTAGCAATGGCAGTGACTTTGAGCGCACACTCAGAAGCCGTTGGAATGGCACCGGGGACGGCCCGATTTAGCCCCGCACACCTCCTGGAATCTTAGATCGCACGGCGATCTCGGTTCAGGCACCAACCCCAAAGAGTGTTTTGAGTTTTTGGTATGGCTCGCCTCAATTATCGGTTTTCGCTGCTCTGTGCCTGTCAACTCGGCTAGCTGTCGTGTTTTGTCGATCAGTGCGTGGACACTCTCGGTCGATGGTCGTGGATGGGACTGTAGTAAGTTTCACCGAAGCAGGAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAACTTCGCTTCATATAACGTAGCCATAGTGCTGTCTGCCATCAATAAGTCTTGCTCAGTGGTGCATACGTCGGGGAGGTTTGTTCCGCCTGGTCAGAACGAGTCTAGGGCGAGCCTATAGGCCAGTCGAGAGCCAAGATTCTATGAAATTAATACGACTACTGGGTGAGAGGTCATACAATTCCCGTGGAATCTGTACCTAAGATATTTCCAGATAGGGATGGCTACTGGTTAAGTTGACAGTGTCTAGATACGTGAGAGCACCTGAGAGGACGCCACGAGTCGGAGCGTGGGCGATCACCCTTCTGAGTCATAAGTCATGTCTATATATCCCTCACTAAAAAGGGCACACGACTATACATGCTTGAGCTTTACGGTCTGGCATGTGGAATGCCCGGAGCAACCCAGTCTTACCATCCTTTACGTACATTTACCGACCCGGCAGTGGCCGGCGCGGAAACCCAGGAGAACGTCGGTCATGATACGCGCCCTCCGCCGAAAGCGTGCTCACACCTCAGGATATCAGCGCTATTACCGGACGTCCCGCGTCCACCATCTAATAATTCAGGTGCTCCTAATAAGTGGGCTGGAGAGCGAGGATTGATATACGTTGAGGAGCTCCGACGGCCCTCTCGTGCGTTTGATGTAGATTGCGTTACCGACGGAGCACGCGTTTGTCAATTTCTGTCTAGGGACGTTTATGTCCTCAATACGAATACCAGGCCTATTTTAGTGTACAAATCACTTAGCAGTCGGAATTGGAAACCTGATGGAAGCGT"
counter=0
length=len(text)
search="AGATC"
tmp=0
for i in range(length):
if text[i:i + len(search)] == search:
tmp += 1
if tmp > counter:
counter = tmp
if text[i:i + len(search)] != search:
tmp = 0

print("done")
print(counter)

Answer 1

尝试这个

import re
sequence = "AATDHDHDTKSDHAATAAT"
matches = re.findall(r'(?:AAT)+', sequence)
largest = max(matches, key=len)
print(len(largest)//len('AAT'))

基本上，这种方法会找到字符串中的子字符串列表，然后选择最大的子字符串。子串的出现次数将是最大的长度除以子串的长度

Answer 2

首先，regex解决方案是Python解决此问题的方法。但是，如果您想修复您的代码。。。

代码的问题在于，索引无法确认您找到了匹配项。您无法识别连续出现的事件。

考虑这样一种情况，您已经找到了三重匹配的开始，AATAATAAT。到达第一个A，识别AAT并递增tmp。进入下一个循环迭代，现在i指向第二个A。您可以看到这里是而不是AAT(它是ATA，跨越前两次出现(，因此您记录一个实例并重置所有状态变量。

相反，你必须跳到第一场比赛的最后，寻找第二场比赛。由于您的索引不会以1的增量平滑移动，因此您将需要while循环。

请学会在变量具有任何意义。如果它所做的一切都是管理您的循环，那么i就可以了。像一旦你把它用于其他用途，就给它一个真实的名字。类似地，CCD_ 12和CCD_ 13确实需要更换。

snip_size = len(search)
pos = 0      # position in the genetic sequence
rep = 0      # number of consecutive repetitions
max_rep = 0  # longest repetition sequence found
while pos < length:
if text[pos:pos + snip_size] == search:
rep += 1
pos += snip_size
else:
max_rep = max(max_rep, rep)
rep = 0
pos += 1
print(max_rep, "repetitions found")

输出：

15 repetitions found