如何在字典的值中统计字符串的出现次数

Question

我是Python的新手，但我正在帮助一个项目，以确定数据中有偏见的单词的数量。

我现在有一个密码词的列表：

male_coded_words = ['active','adventurous','aggress','ambitio','analy','assert']

我有一本职称和技能的字典：

jobsdict = {'fork lift truck driver': ['fork lift truck driv','assert], 'assistant fraud and payment risk manager': ['fraud', 'online fraud', 'fraud detect', 'payment system', 'risk manag'], 'paralegal vacancy corporate immigration (london office)': ['legal', 'microsoft offic', 'communication skil'], 'transport operator': ['transport','active'], 'year 5 primary teacher': ['newham'], 'multi agency safeguarding administrator': ['admin', 'social work', 'safeguard', 'social work admin', 'children administr', 'social work administr', 'safeguarding administr']

我想遍历字典，找出每个键在male_coded_words列表中出现的次数。

{'fork lift truck driver': "count":"1", "coded_words":["assert"].....}形式的字典形式的输出

到目前为止我的代码；

final_count = 0
final_output = {}

for k, v in jobsdict:
    final_output[k] = []
    if 'analy' in str(v):
        n = final_count + 1
    else:
        n = 0  
    final_output[k].append(n)
    final_output[k].append(v)

Answer 1

这里的一个好主意是利用Python的set对象，它充当列表的无序替代品。集合上的操作往往比列表上的等价操作快得多。为简洁起见，我还使用了一个dictionary comprehension和一个Counter对象来自动计算编码的words.The的实例数，下面的脚本应该会给出您指定的输出：

from collections import Counter

# General form of the data provided above, for reference
# male_coded_words = ['active', ...]
# jobsdict = {'fork lift truck driver': ['fork lift truck driv','assert'], ...}

result = {k: Counter(set(v) & set(male_coded_words)) for k,v in jobsdict.items()}

# result will look like {'fork lift truck driver': {"assert": 1}, ...}.
# If no coded words exist for a specific job, then its value in the 
# result dict will just be an empty set.

Answer 2

我是一个初学者，但我会研究正则表达式并对其进行计数(这是我能想到的最简单的方法)。

Answer 3

除了使用少于2的for循环之外，我想不出其他方法，一个用于迭代jobsdict，另一个用于编码的单词。另外，使用jobsdict.items()通过键和值对其进行迭代：

final_count = 0
final_output = {}

for k, v in jobsdict.items():
    count, words = 0, []
    s = ''.join(v) 
    # merge all the strings into one to avoid a third nested loop iterating over them
    for w in male_coded_words:
        c = s.count(w) 
        # can be replaced with `w in s` if you don't want to count multiple occurrences of a word each time
        if c:
            count += c
            words.append(w)
    final_count += count
    final_output[k] = [count, words]

print(final_output, final_count)

这给了我以下输出：

{'fork lift truck driver': [1, ['assert']], 'assistant fraud and payment risk manager': [0, []], 'paralegal vacancy corporate immigration (london office)': [0, []], 'transport operator': [1, ['active']], 'year 5 primary teacher': [0, []], 'multi agency safeguarding administrator': [0, []]} 2

编辑:如果希望final_output中包含字典，请将倒数第二行替换为final_output[k] = {"count":count, "words":words}