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python - 字典字典到数据框

问题描述

我有一个函数可以计算评论中中心词和上下文词之间的共现。

def get_coocs(x):
    
    occurdict={}
    
    # Pre-processing
    tokens = nltk.word_tokenize(x)
    tokenslower = list(map(str.lower, tokens)) 
    
    # Save all the nouns in each review
    allnouns=[word for word in tokenslower if word in cent_vocab]
    
    # Save all the verbs/adjectives in each review
    allverbs_adj=Counter(word for word in tokenslower if word in cont_vocab)
    
    # Creating a dictionary of dictionaries
    for noun in allnouns:
        occurdict[noun]=dict(allverbs_adj)
        
    return occurdict

coocs=df['comments'].apply(lambda x: get_coocs(x))

我的字典看起来像这样：

{'host': {'is': 3, 'most': 1, 'amazing': 1},
{'time': {'had': 1, 'such': 1, 'great': 1},
{'room': {'very': 2, 'professional': 1},
{'way': {'is': 3, 'recommended': 1, 'provided': 2}

但是，当我尝试将其转换为数据框时，将名词作为列，动词/形容词作为索引，并具有相应的共现值，我最终得到以下结果：

def cooc_dict2df(coocs):
    
    coocdf=pd.DataFrame.from_dict({i:coocs[i] for i in coocs.keys()}, orient='index')
    
    return coocdf

数据框

我尝试了其他解决方案，但我似乎仍然无法得到我想要的。

标签： pythonpandasnltk

解决方案

你可以试试这个：

# Toy data
coocs = {
    "host": [
        {"is": 3, "most": 8, "amazing": 1},
        {"had": 5, "such": 7, "great": 9},
        {"very": 3, "recommended": 1, "provided": 2},
    ],
    "time": [
        {"is": 2, "most": 9, "amazing": 7},
        {"had": 6, "such": 6, "great": 8},
        {"very": 2, "recommended": 3, "provided": 4},
    ],
    "room": [
        {"is": 7, "most": 1, "amazing": 2},
        {"had": 7, "such": 5, "great": 8},
        {"very": 1, "recommended": 5, "provided": 4},
    ],
    "way": [
        {"is": 1, "most": 6, "amazing": 9},
        {"had": 8, "such": 4, "great": 9},
        {"very": 7, "recommended": 7, "provided": 1},
    ],
}

# Make a list of dataframes
dfs = [pd.DataFrame({colname: col}) for colname, cols in coocs.items() for col in cols]

# Merge dataframes
new_df = dfs[0]
for df in dfs[1:]:
    new_df = new_df.merge(df, how="outer", left_index=True, right_index=True)

new_df.fillna(0, inplace=True)

# Add identical columns
for name in coocs.keys():
    new_df[f"new_{name}"] = 0
    for col in new_df.columns:
        if col.startswith(name):
            new_df[f"new_{name}"] = new_df[f"new_{name}"] + new_df[col]

# Drop useless columns and rename the remaining ones
new_df = new_df.drop(columns=[col for col in new_df.columns if "new_" not in col])
new_df.columns = [col[4:] for col in new_df.columns]


print(new_df)
# Outputs
            host    time    room    way
amazing     1.0     7.0     2.0     9.0
great       9.0     8.0     8.0     9.0
had         5.0     6.0     7.0     8.0
is          3.0     2.0     7.0     1.0
most        8.0     9.0     1.0     6.0
provided    2.0     4.0     4.0     1.0
recommended 1.0     3.0     5.0     7.0
such        7.0     6.0     5.0     4.0
very        3.0     2.0     1.0     7.0

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