Question

所以基本上，我试图编写一个过滤器，我可以从字典中分隔数字和字母，并在它们之间留一个空格。例如12346 S 12346 Q。

def check_if_works():
    dict_info = {}
    dict_info['1234'] = "test"
    dict_info['12456s'] = "test"
    dict_info['12456q'] = "test"
    dict_info['12456b'] = "test"
    dict_info['123456'] = "test"
    dict_info['asdftes'] = "test"
    dict_info['asdftess'] = "test"
    dict_info['asdftessd'] = "test"
    arr = []
    for key,value in dict_info.iteritems():
        if key.isalpha() or key.isdigit():
            pass
        #print key
        else:
            print key

Answer 1

不需要字典列表，您想直接填写并转储new_dict。

此外，不需要try / except（由于str.isalpha()永远不会抛出异常，因此不起作用），您应该组合检查以测试它只包含字母或数字（isalnum()）并且它不仅包含字母（isalpha()）或仅包含数字（isdigit()）。

然后，您也不会在每个迭代步骤中对原始字典执行一次文件，但最后只能翻译一次以转储整个结果字典。

new_dict = {}

for key,value in dict_info.iteritems():
    check = key.isalnum() and not (key.isalpha() or key.isdigit())

    new_dict[insert_spaces(key)] = value
    print (key, ":", value)

with open('output.json', 'wb') as outfile:
    json.dump(new_dict, outfile, indent=4)

您需要在数字+字母或字母+数字之间插入空格的函数insert_spaces如下所示：

import re
def insert_spaces(key):
    return re.sub(r'(\D)(\d)', r"\1 \2", re.sub(r'(\d)(\D)', r"\1 \2", key))

或者你可以用字典理解替换整个代码（自Python 2.7起）：

with open('output.json', 'wb') as outfile:
    json.dump({ insert_spaces(key): dict_info[key] for key in dict_info if key.isalnum() and not (key.isalpha() or key.isdigit()) }, outfile, indent=4)

或者通过将检查条件提取到lambda函数来格式化：

check_key = lambda key: key.isalnum() and not (key.isalpha() or key.isdigit())
with open('output.json', 'wb') as outfile:
    json.dump({ insert_spaces(k): dict_info[k] for k in dict_info if check_key(k) }, 
              outfile, indent=4)

字母和数字之间的间距

1 个答案: