Checkio--题目之Elementary（上）

下面是做CHECKIO的题目，Elementary的题目，题目很基础，可以熟悉语言特性。关键看看别人怎么写。更好的部分很好。有很多可以学习的东西。

Even the last

http://www.checkio.org/mission/even-last/

def checkio(array): """ sums even-indexes elements and multiply at the last """ if len(array) == 0: return 0 return sum(array[0::2]) * array[-1]#这里array[0::2]表示取序号为0，2，4，6....的元素，切片插座，2的参数表示步进if __name__ == '__main__': assert checkio([0, 1, 2, 3, 4, 5]) == 30, "(0+2+4)*5=30" assert checkio([1, 3, 5]) == 30, "(1+5)*5=30" assert checkio([6]) == 36, "(6)*6=36" assert checkio([]) == 0, "Empty"

Monkey Typing

http://www.checkio.org/mission/monkey-typing/

def count_words(text, words):    result = 0    #无论单词是否连在一起都可以直接判断，字符串操作好方便    for w in words:        if w in text.lower():            result = result +1            #print w    return resultif __name__ == '__main__':    assert count_words(u"How aresjfhdskfhskd you?", {u"how", u"are", u"you", u"hello"}) == 3, "Example"    assert count_words(u"Bananas, give me bananas!!!", {u"banana", u"bananas"}) == 2, "BANANAS!"    assert count_words(u"Lorem ipsum dolor sit amet, consectetuer adipiscing elit.",                       {u"sum", u"hamlet", u"infinity", u"anything"}) == 1, "Weird text"

Three words

http://www.checkio.org/mission/three-words/

判断是否有连续的三个单词，有输出TRUE，否则FALSE

def checkio(words):    cout = 3    flag = 0    word = words.split()    for w in word:        if w.isalpha() and cout:            cout -= 1            flag = 1            if cout == 0:                return True        else:            cout = 3    if not flag :        return False    return Falseif __name__ == '__main__':    assert checkio(u"Hello World hello") == True, "Hello"    assert checkio(u"He is 123 man") == False, "123 man"    assert checkio(u"1 2 3 4") == False, "Digits"    assert checkio(u"bla bla bla bla") == True, "Bla Bla"    assert checkio(u"Hi") == False, "Hi"更好的：def checkio(words):    succ = 0    for word in words.split():        succ = (succ + 1)*word.isalpha()        if succ == 3: return True    else: return False

Secret Message

http://www.checkio.org/mission/secret-message/

输出给定字符串的所有大写字母def find_message(text):    """Find a secret message"""    #lis =[]    l = ""  #字符串型初始化    lis = list(text)    flag=0    #逐个遍历字符串字符的简单方法    for w in text:        #print w        if w.isupper():            flag =1            l += w     if flag == 0:        return ''    else:        return l    return 0if __name__ == '__main__':    assert find_message(u"How are you? Eh, ok. Low or Lower? Ohhh.") == "HELLO", "hello"    assert find_message(u"hello world!") == "", "Nothing"    assert find_message(u"HELLO WORLD!!!") == "HELLOWORLD", "Capitals"更好的：find_message = lambda text: ''.join(filter(str.isupper, text))#filter是内建函数，过滤器。用法filter(条件，对象)，满足条件的返回该值，即filter返回由seq队列中使条件返回值为真的元素组成的队列。import refind_message = lambda text: re.sub(r'[a-z]|[^\w]|[\d]','',text)def find_message(text):    return ''.join(c for c in text if c.isupper())

The Most Numbers

http://www.checkio.org/mission/most-numbers/

返回列表中最大最小数的差值，列表空返回0def checkio(*args):    if not args:        return 0    return max(args) - min(args)​if __name__ == '__main__':    def almost_equal(checked, correct, significant_digits):        precision = 0.1 ** significant_digits        return correct - precision < checked < correct + precision    assert almost_equal(checkio(1, 2, 3), 2, 3), "3-1=2"    assert almost_equal(checkio(5, -5), 10, 3), "5-(-5)=10"    assert almost_equal(checkio(10.2, -2.2, 0, 1.1, 0.5), 12.4, 3), "10.2-(-2.2)=12.4"    assert almost_equal(checkio(), 0, 3), "Empty"def checkio(*args):    if not args: return 0    argslist = sorted( args )    return argslist[-1] - argslist[0]

Boolean Algebra

http://www.checkio.org/mission/boolean-algebra/

位运算OPERATION_NAMES = ("conjunction", "disjunction", "implication", "exclusive", "equivalence")def boolean(x, y, operation):    if(operation == 'conjunction'):        return (x and y)    if(operation == 'disjunction'):        return (x or y)    if(operation == 'implication'):        return (not x or y)    if(operation == 'exclusive'):        return (x ^ y)    if(operation == 'equivalence'):        return (not (x^y))    return if __name__ == '__main__':    assert boolean(1, 0, u"conjunction") == 0, "and"    assert boolean(1, 0, u"disjunction") == 1, "or"    assert boolean(1, 1, u"implication") == 1, "material"    assert boolean(0, 1, u"exclusive") == 1, "xor"    assert boolean(0, 1, u"equivalence") == 0, "same?"更好地:def boolean(x, y, operation):    if operation == "conjunction": return x & y    if operation == "disjunction": return x | y    if operation == "implication": return (1 ^ x) | y    if operation == "exclusive":   return x ^ y    if operation == "equivalence": return x ^ y ^ 1    return 0def boolean(x, y, operation):    mapp = {"conjunction": lambda a, b: a and b,             "disjunction": lambda a, b: a or b,             "implication": lambda a, b: not a or b,             "exclusive": lambda a, b: a ^ b,             "equivalence": lambda a, b: not(a ^ b)}    return mapp[operation](x, y)

Right to Left

http://www.checkio.org/mission/right-to-left/

字符串替换,将right替换位leftdef left_join(phrases):    s =",".join(phrases)    st = s.replace("right","left")    return stif __name__ == '__main__':    assert left_join(("left", "right", "left", "stop")) == "left,left,left,stop", "All to left"    assert left_join(("bright aright", "ok")) == "bleft aleft,ok", "Bright Left"    assert left_join(("brightness wright",)) == "bleftness wleft", "One phrase"    assert left_join(("enough", "jokes")) == "enough,jokes", "Nothing to replace"更好地:def left_join(phrases):    return (",".join(phrases)).replace("right","left")

Digits Multiplication

http://www.checkio.org/mission/digits-multiplication/

输出给定数字每个位相乘结果，忽略0def checkio(number):    num = 1    for x in str(number):        if x != '0':           num *= int(x)     return numif __name__ == '__main__':    assert checkio(123405) == 120    assert checkio(999) == 729    assert checkio(1000) == 1    assert checkio(1111) == 1更好地：def checkio(number):    total = 1    for i in str(number).replace('0', '1'):        total *= int(i)    return total

Count Inversions

http://www.checkio.org/mission/count-inversions/

统计多少个逆序数不在自己的顺序上def count_inversion(sequence):    """        Count inversions in a sequence of numbers    """    count = 0    for i in range(len(sequence)):        for j in range(i+1,len(sequence)):            if sequence[i] > sequence[j]:                count += 1    return countif __name__ == '__main__':    assert count_inversion((1, 2, 5, 3, 4, 7, 6)) == 3, "Example"    assert count_inversion((0, 1, 2, 3)) == 0, "Sorted"    assert count_inversion((99, -99)) == 1, "Two numbers"    assert count_inversion((5, 3, 2, 1, 0)) == 10, "Reversed"更好地：def count_inversion(sequence):    count = 0    for i, x in enumerate(sequence[:-1]):        for y in sequence[i+1:]:            if x > y: count += 1    return count

The end of other

http://www.checkio.org/mission/end-of-other/

练习十二：判断给定多个字符串，是否有后缀def checkio(words_set):    w = words_set    for a in w:        for b in w:            if a == b:                continue            elif a.endswith(b):                return True    return Falseif __name__ == '__main__':    assert checkio({u"hello", u"lo", u"he"}) == True, "helLO"    assert checkio({u"hello", u"la", u"hellow", u"cow"}) == False, "hellow la cow"    assert checkio({u"walk", u"duckwalk"}) == True, "duck to walk"    assert checkio({u"one"}) == False, "Only One"    assert checkio({u"helicopter", u"li", u"he"}) == False, "Only end"更巧妙：def checkio(s):    return any(map(lambda x: any(map(x.endswith,s-set([x]))),s))

Days Between

http://www.checkio.org/mission/days-diff/

#计算日期差#日期初始化实例：from datetime import datet = (2010, 2, 3) #元组形式 调用date函数转换date(t[0], t[1], t[2])# ordate(*t)#程序主体：import timefrom datetime import datedef days_diff(date1, date2):    """        Find absolute diff in days between dates    """    f = date(*date1)    e = date(*date2)    ans = abs((f-e).days)    return ansif __name__ == '__main__':    assert days_diff((1982, 4, 19), (1982, 4, 22)) == 3    assert days_diff((2014, 1, 1), (2014, 8, 27)) == 238    assert days_diff((2014, 8, 27), (2014, 1, 1)) == 238更好地：from datetime import datetimedef days_diff(date1, date2):    return abs((datetime(*date1)-datetime(*date2)).days)

Binary count

http://www.checkio.org/mission/binary-count/

#把数字变成二进制，并且计算1的有多少位def checkio(number):    return bin(number).count('1')if __name__ == '__main__':    assert checkio(4) == 1    assert checkio(15) == 4    assert checkio(1) == 1    assert checkio(1022) == 9

Number Base

http://www.checkio.org/mission/number-radix/

#给定一个数字（字符串）和进制，输出他的intdef checkio(str_number, radix):    try:        return int(str_number,radix)    except ValueError:        return -1if __name__ == '__main__':    assert checkio(u"AF", 16) == 175, "Hex"    assert checkio(u"101", 2) == 5, "Bin"    assert checkio(u"101", 5) == 26, "5 base"    assert checkio(u"Z", 36) == 35, "Z base"    assert checkio(u"AB", 10) == -1, "B > A > 10"更好地：def checkio(*a):    try: return int(*a)    except ValueError: return -1

Common-words

http://www.checkio.org/mission/common-words/

找出两个字符串中相同的字符串，并排序输出体会set用法，def checkio(first, second):    a = set(first.split(','))    b = set(second.split(','))    x = a.intersection(b)  # 这个要转换为set直接返回a,b相同的元素列表    x = sorted(x)    return ','.join(x)if __name__ == '__main__':    assert checkio(u"hello,world", u"hello,earth") == "hello", "Hello"    assert checkio(u"one,two,three", u"four,five,six") == "", "Too different"    assert checkio(u"one,two,three", u"four,five,one,two,six,three") == "one,three,two", "1 2 3"更好地：def checkio(first, second):    """    set data type has useful methods.    """    first_set, second_set = set(first.split(",")), set(second.split(","))    common = first_set.intersection(second_set)    return ",".join(sorted(common))def checkio(first, second):    first_set = set(first.split(","))    second_set = set(second.split(","))    return ",".join(sorted(first_set & second_set))

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