麻省理工学院公开课：计算机科学及编程导论习题3下

有一个字符串“ATGACATGCA”，我想搜索“ATGC”，可以在(5, 15)处找到。

如果把“ATGC”分成“A”、“T”和"GC"；假设不知道"T”，只查“A”和"GC"，那么分别n = (0, 3, 5, 9)和m = (7,)处找到。

“A”的长度1，在n和m的范围中选择合适的数字，5+1+1=7。

“A”所在的位置有四个，如果隔一个字母后要是“GC”的话，那“A”所在位置必须在加1才是“GC”的位置。

习题3：

将一个字符串分成三个部分，第一个子字符串，丢失的部分（长度为1），第二个子字符串。

写出函数constrainedMatchPair。此函数有三个参数：第一个子字符串起始位置的元组；第二个子字符串起始位置的元组；第一个子字符串的长度。

函数能够返回所有第一个组元中的所有数据（n），此数据能够和第二个组元中的数据（k），满足n+m+1 = k，m为第一个字字符串的长度。

函数按如下定义，

def constrainedMatchPair(firstMatch,secondMatch,length):

def constrainedMatchPair(firstMatch, secondMatch, length):
	m = length
	match = ()
	for n in firstMatch:
		for k in secondMatch:
			if n + m + 1 == k:
				match = match + (n,)
	return match

然后通过subStringMatchOneSub来验证,此函数要用到之前的subStringMatchExact

通过bing就能很容易搜到ps3_template.py：

所有代码如下：

from string import find

def subStringMatchExact(target, key):
	tuple = ()
	order = 0
	while find(target, key, order) != -1:
		tuple = tuple + (find(target, key, order),)
		order = find(target, key, order) + 1
	return tuple




def constrainedMatchPair(firstMatch, secondMatch, length):
	m = length
	match = ()
	for n in firstMatch:
		for k in secondMatch:
			if n + m + 1 == k:
				match = match + (n,)
	return match





# this is a code file that you can use as a template for submitting your
# solutions


# these are some example strings for use in testing your code

#  target strings

target1 = 'atgacatgcacaagtatgcat'
target2 = 'atgaatgcatggatgtaaatgcag'

# key strings

key10 = 'a'
key11 = 'atg'
key12 = 'atgc'
key13 = 'atgca'



### the following procedure you will use in Problem 3


def subStringMatchOneSub(key,target):
    """search for all locations of key in target, with one substitution"""
    allAnswers = ()
    for miss in range(0,len(key)):
        # miss picks location for missing element
        # key1 and key2 are substrings to match
        key1 = key[:miss]
        key2 = key[miss+1:]
        print 'breaking key',key,'into',key1,key2
        # match1 and match2 are tuples of locations of start of matches
        # for each substring in target
        match1 = subStringMatchExact(target,key1)
        match2 = subStringMatchExact(target,key2)
        # when we get here, we have two tuples of start points
        # need to filter pairs to decide which are correct
        filtered = constrainedMatchPair(match1,match2,len(key1))
        allAnswers = allAnswers + filtered
        print 'match1',match1
        print 'match2',match2
        print 'possible matches for',key1,key2,'start at',filtered
    return allAnswers





for key in (key10, key11, key12, key13):
	for target in (target1, target2):
		print subStringMatchOneSub(key,target)

习题4：

写一个函数subStringMatchExactlyOneSub，有两个参数：目标字符串和关键词字符串。返回关键词中有一个不同在目标中的起始位置的元组。

def subStringMatchExactlyOneSub(target,key):

def subStringMatchExactlyOneSub(target,key):
	possible = subStringMatchOneSub(key,target)
	certain = subStringMatchExact(target, key)
	one = ()
	for p in possible:
		for c in certain:
			if p == c:
				break
			elif c == certain[-1]:
				one = one + (p,)
	return one