Question

我需要一些关于下面代码的帮助，我有样本输入和预期输出，当前它没有打印任何东西..请提供你的输入，

基本上我的代码是解析num_ids.txt中的值并检查每个值是否大于提供的base_num值，如果该值不在“num_ignore”列表中，然后（在满足前2个条件之后），它会尝试匹配numrefs列表并在numrefs ...

中打印匹配的值

EXPECTEDOUTPUT:-
nums/39/205739/2

import os
import subprocess
def p4 (base_num):
    numrefs = ['nums/89/202089/4', 'nums/39/205739/2', 'nums/94/203455/6']
    num_ignore = [150362, 147117]
    '''
        num_ids.txt
        202089
        205739
        147117
        203455
    '''
    with open('./num_ids.txt', 'rb') as f:
        # Iterate over the file itself
        for line in f:
            num = int(line)
            if num > base_num and num not in num_ignore and line in numrefs:
                print numrefs
def main():
    base_num=203456
    p4(base_num)

    if __name__ == '__main__':
        main()

Answer 1

您可以使用理解将numrefs列表转换为名为numrefs_index的词典。这样您就可以使用in运算符并更快地访问引用。

def p4(base_num):
    numrefs = ['nums/89/202089/4', 'nums/39/205739/2', 'nums/94/203455/6']
    num_ignore = [150362, 147117]
    numrefs_index = dict((int(x.split('/')[2]), x) for x in numrefs)
    for line in file("num_ids.txt"):
        num = int(line)
        if num > base_num and num not in num_ignore and num in numrefs_index:
            print numrefs_index[num]

if __name__ == "__main__":
    p4(203456)

# prints:
# nums/39/205739/2

numrefs_index行构建了这个词：

{202089: 'nums/89/202089/4',
 203455: 'nums/94/203455/6',
 205739: 'nums/39/205739/2'}

将变量与列表和打印列表值匹配

1 个答案: