如何计算代码中关键字的出现次数但忽略comment / docstring中的关键字?

时间:2015-05-14 08:14:16

标签: python file python-2.7 count keyword

我是Python的新手。我想在下面的代码中找到Python关键字['def','in', 'if'...]的出现位置。但是,需要忽略代码中任何字符串常量中找到的关键字。 如何计算关键字出现次数而不计算字符串中的关键字?

def grade(result):
    '''
    if if (<--- example to test if the word "if" will be ignored in the counts)
    :param result: none
    :return:none
    '''

    if result >= 80:
        grade = "HD"
    elif 70 <= result:
        grade = "DI"
    elif 60 <= result:
        grade = "CR"
    elif 50 <= result:
        grade = "PA"
    else:
    #else (ignore this word)
        grade = "NN"
    return grade

result = float(raw_input("Enter a final result: "))

while result < 0 or result > 100:
    print "Invalid result. Result must be between 0 and 100."
    result = float(raw_input("Re-enter final result: "))

print "The corresponding grade is", grade(result)

1 个答案:

答案 0 :(得分:2)

使用tokenizekeywordcollections模块。

  

tokenize.generate_tokens(readline)

     

generate_tokens()生成器   需要一个参数readline,它必须是一个可调用的对象   提供与内置文件的readline()方法相同的接口   对象(请参阅文件对象部分)。每次调用函数都应该   将一行输入作为字符串返回。或者,readline可以是a   通过提高StopIteration来表示完成的可调用对象。

     

生成器使用这些成员生成5元组:令牌类型;   令牌字符串;指定行的2元组(srow,scol)   和令牌在源中开始的列;一个2元组(erow,   ecol)ofts指定令牌结束的行和列   来源;以及找到令牌的行。这条线路过去了   (最后一个元组项)是逻辑行;延续线是   包括在内。

     

2.2版中的新功能。

import tokenize
with open('source.py') as f:
    print list(tokenize.generate_tokens(f.readline))

部分输出:

[(1, 'def', (1, 0), (1, 3), 'def grade(result):\n'),
 (1, 'grade', (1, 4), (1, 9), 'def grade(result):\n'),
 (51, '(', (1, 9), (1, 10), 'def grade(result):\n'),
 (1, 'result', (1, 10), (1, 16), 'def grade(result):\n'),
 (51, ')', (1, 16), (1, 17), 'def grade(result):\n'),
 (51, ':', (1, 17), (1, 18), 'def grade(result):\n'),
 (4, '\n', (1, 18), (1, 19), 'def grade(result):\n'),
 (5, '    ', (2, 0), (2, 4), "    '''\n"),
 (3,
  '\'\'\'\n    if if (<--- example to test if the word "if" will be ignored in the counts)\n    :param result: none\n    :return:none\n    \'\'\'',
  (2, 4),
  (6, 7),
  '    \'\'\'\n    if if (<--- example to test if the word "if" will be ignored in the counts)\n    :param result: none\n    :return:none\n    \'\'\'\n'),
 (4, '\n', (6, 7), (6, 8), "    '''\n"),
 (54, '\n', (7, 0), (7, 1), '\n'),
 (1, 'if', (8, 4), (8, 6), '    if result >= 80:\n'),

您可以从keyword模块中检索关键字列表:

import keyword
print keyword.kwlist
print keyword.iskeyword('def')

集合解决方案与collections.Counter:

import tokenize
import keyword
import collections 
with open('source.py') as f:
    # tokens is lazy generator
    tokens = (token for _, token, _, _, _ in tokenize.generate_tokens(f.readline))
    c = collections.Counter(token for token in tokens if keyword.iskeyword(token))

print c  # Counter({'elif': 3, 'print': 2, 'return': 1, 'else': 1, 'while': 1, 'or': 1, 'def': 1, 'if': 1})