Python - 检測字符串之间的包括

检測字符串之间的包括

本文地址: http://blog.csdn.net/caroline_wendy/article/details/27048955

Python中, 能够检測字符串之间的包括问题.

containsAny, 仅仅要包括不论什么一个字符就可以;

containsOnly, 所有字符都包括在内;

containsAll, 包括所有;

代码:

# -*- coding: utf-8 -*-

'''
Created on 2014.5.25

@author: C.L.Wang
'''

'''存在不论什么'''
def containsAny(seq, aset):
    for c in seq:
        if c in aset: return True
    return False

import itertools
def containsAny2(seq, aset):
    for item in itertools.ifilter(aset.__contains__, seq) :
        return True
    return False

def containsAny3(seq, aset):
    return bool(set(aset).intersection(seq))

'''所有存在'''
def containsOnly(seq, aset):
    for c in seq:
        if c not in aset: return False
    return True

'''包括所有'''
def containsAll(seq, aset):
    #print(set(aset).difference(seq))
    return not set(aset).difference(seq)

import string
notrans = string.maketrans('', '')
def containsAny4(astr, strset):
    return len(strset) != len(strset.translate(notrans, astr))

def containsAll2(astr, strset):
    return not strset.translate(notrans, astr)

if __name__ == '__main__':
    L1 = [1, 2, 3, 4]
    L2 = [5, 6, 7, 8]
    L3 = [1, 4, 7, 10]
    print("L1 constains any in L2 : " + str(containsAny(L1, L2)))
    print("L1 constains any in L3 : " + str(containsAny(L1, L3)))
    
    print("L1 constains any in L2 (2) : " + str(containsAny2(L1, L2)))
    print("L1 constains any in L3 (2) : " + str(containsAny2(L1, L3)))
    
    print("L1 constains any in L2 (3) : " + str(containsAny3(L1, L2)))
    print("L1 constains any in L3 (3) : " + str(containsAny3(L1, L3)))
    
    L4 = [1, 1, 2, 2, 3, 4]
    L5 = [1, 1, 2, 2, 3, 4, 5]

    print("L1 constains only in L4 : " + str(containsOnly(L1, L4)))
    print("L1 constains only in L5 : " + str(containsOnly(L1, L5)))
    
    print("L1 constains all in L4 (2) : " + str(containsAll(L1, L4)))
    print("L1 constains all in L5 (2) : " + str(containsAll(L1, L5)))


    pass

输出:

L1 constains any in L2 : False
L1 constains any in L3 : True
L1 constains any in L2 (2) : False
L1 constains any in L3 (2) : True
L1 constains any in L2 (3) : False
L1 constains any in L3 (3) : True
L1 constains only in L4 : True
L1 constains only in L5 : True
L1 constains all in L4 (2) : True
L1 constains all in L5 (2) : False