Question

ADVANCED = False
def count_unique(string1):
    """Count unique words in a string.
    This function should take a single string and return a dictionary
    that has all of the distinct words as keys, and the number of times
    that word appears in the string.
    For example:
        >>> print_dict(count_unique("each word appears once"))
        {'appears': 1, 'each': 1, 'once': 1, 'word': 1}
    Words that appear more than once should be counted each time:
        >>> print_dict(count_unique("rose is a rose is a rose"))
        {'a': 2, 'is': 2, 'rose': 3}
    It's fine to consider punctuation part of a word (e.g., a comma
    at the end of a word can be counted as part of that word) and
    to consider differently-capitalized words as different:
        >>> print_dict(count_unique("Porcupine see, porcupine do."))
        {'Porcupine': 1, 'do.': 1, 'porcupine': 1, 'see,': 1}
    """
    unique_words = {}
    words_list = string1.split()

    for word in words_list:
        if word not in unique_words:
            unique_words[word] = 1
        else:
            unique_words[word] += 1

    return unique_words
##############################################################################
# You can ignore everything after here

def print_dict(d):
    # This method is just used to print dictionaries in key-alphabetical
    # order, and is only used for our documentation tests. You can ignore it.
    if isinstance(d, dict):
        print "{" + ", ".join("%r: %r" % (k, d[k]) for k in sorted(d)) + "}"
    else:
        print d


def sort_pairs(l):
    # Print sorted list of pairs where the pairs are sorted. This is used only
    # for documentation tests. You can ignore it.
    return sorted(sorted(pair) for pair in l)

if __name__ == "__main__":
    print
    import doctest
    for k, v in globals().items():
        if k[0].isalpha():
            if k.startswith('adv_') and not ADVANCED:
                continue
            a = doctest.run_docstring_examples(v, globals(), name=k)
    print "** END OF TEST OUTPUT"

我正在使用Python 2.7.6

ValueError: line 8 of the docstring for count_unique has inconsistent leading whitespace: ' Words that appear more than once should be counted each time:'

Answer 1

与文档字符串一致：

def count_unique(string1):
    """Count unique words in a string.
    This function should take a single string and return a dictionary
    that has all of the distinct words as keys, and the number of times
    that word appears in the string.
    For example:
    >>> print_dict(count_unique("each word appears once"))
    {'appears': 1, 'each': 1, 'once': 1, 'word': 1}
    Words that appear more than once should be counted each time:
    >>> print_dict(count_unique("rose is a rose is a rose"))
    {'a': 2, 'is': 2, 'rose': 3}
    It's fine to consider punctuation part of a word (e.g., a comma
    at the end of a word can be counted as part of that word) and
    to consider differently-capitalized words as different:
    >>> print_dict(count_unique("Porcupine see, porcupine do."))
    {'Porcupine': 1, 'do.': 1, 'porcupine': 1, 'see,': 1}
    """

输出：

**********************************************************************
File "/home/padraic/test.py", line 30, in count_unique
Failed example:
    print_dict(count_unique("each word appears once"))
Expected:
    {'appears': 1, 'each': 1, 'once': 1, 'word': 1}
    Words that appear more than once should be counted each time:
Got:
    {'appears': 1, 'each': 1, 'once': 1, 'word': 1}
**********************************************************************
File "/home/padraic/test.py", line 33, in count_unique
Failed example:
    print_dict(count_unique("rose is a rose is a rose"))
Expected:
    {'a': 2, 'is': 2, 'rose': 3}
    It's fine to consider punctuation part of a word (e.g., a comma
    at the end of a word can be counted as part of that word) and
    to consider differently-capitalized words as different:
Got:
    {'a': 2, 'is': 2, 'rose': 3}
** END OF TEST OUTPUT

或缩进字符串：

def count_unique(string1):
    """Count unique words in a string.
    This function should take a single string and return a dictionary
    that has all of the distinct words as keys, and the number of times
    that word appears in the string.
    For example:
        >>> print_dict(count_unique("each word appears once"))
        {'appears': 1, 'each': 1, 'once': 1, 'word': 1}
        Words that appear more than once should be counted each time:
        >>> print_dict(count_unique("rose is a rose is a rose"))
        {'a': 2, 'is': 2, 'rose': 3}
        It's fine to consider punctuation part of a word (e.g., a comma
        at the end of a word can be counted as part of that word) and
        to consider differently-capitalized words as different:
        >>> print_dict(count_unique("Porcupine see, porcupine do."))
        {'Porcupine': 1, 'do.': 1, 'porcupine': 1, 'see,': 1}

Answer 2

不确定，但也许你混合空格和制表符缩进？选择一个并坚持下去。如果使用选项卡，请将编辑器设置为4个空格宽度。如果选择空格，则每个缩进级别也应使用4个空格。

为什么我运行这个文件时会出现doctest错误？

2 个答案: