Question

这可能会在非常愚蠢的问题上结束，但作为python中的新手，我无法找到解决问题的好方法。

class Preprocessor:
    mPath = None;
    df = None;

    def __init__(self, path):
        self.mPath = path;


    def read(self):
        self.df = pd.read_csv(self.mPath);
        return self.df;

    def __findUniqueGenres(self):
        setOfGenres = set();
        for index, genre in self.df['genres'].iteritems():
            listOfGenreInMovie = genre.lower().split("|");
            for i, _genre in np.ndenumerate(listOfGenreInMovie):
                setOfGenres.add(_genre)
        return setOfGenres;

    def __prepareDataframe(self, genres):
        all_columns = set(["title", "movieId"]).union(genres)
        _df = pd.DataFrame(columns=all_columns)
        return _df;

    def __getRowTemplate(self, listOfColumns):
        _rowTemplate = {}
        for col in listOfColumns:
            _rowTemplate[col] = 0
        return _rowTemplate;

    def __createRow(self, rowTemplate, row):
        rowTemplate['title'] = row.title;
        rowTemplate['movieId'] = row.movieId;
        movieGenres = row.genres.lower().split("|");
        for movieGenre in movieGenres:
            rowTemplate[movieGenre] = 1;
        return rowTemplate;

    def tranformDataFrame(self):
        genres = self.__findUniqueGenres();
        print('### List of genres...', genres);
        __df = self.__prepareDataframe(genres); # Data frame with all required columns.
        rowTemplate = self.__getRowTemplate(__df.columns)
        print('### Row template looks like -->', rowTemplate)
        collection = []
        for index, row in self.df.iterrows():
        _rowToAdd=self.__createRow(rowTemplate, row);
            print('### Row looks like', _rowToAdd)
            collection.append(_rowToAdd)

        print('### Collection looks like', collection)
        return __df.append(collection)

当我尝试将_rowToAdd附加到集合时，它会结束最后一项（self.df的最后一行）的集合。

下面是相同的日志（self.df这里有3行），

### List of genres... {'mystery', 'horror', 'comedy', 'drama', 'thriller', 'children', 'adventure'}
### Row template looks like --> {'title': 0, 'horror': 0, 'comedy': 0, 'drama': 0, 'children': 0, 'mystery': 0, 'movieId': 0, 'thriller': 0, 'adventure': 0}
### Row looks like {'title': 'Big Night (1996)', 'horror': 0, 'comedy': 1, 'drama': 1, 'children': 0, 'mystery': 0, 'movieId': 994, 'thriller': 0, 'adventure': 0}
### Row looks like {'title': 'Grudge, The (2004)', 'horror': 1, 'comedy': 1, 'drama': 1, 'children': 0, 'mystery': 1, 'movieId': 8947, 'thriller': 1, 'adventure': 0}
### Row looks like {'title': 'Cheetah (1989)', 'horror': 1, 'comedy': 1, 'drama': 1, 'children': 1, 'mystery': 1, 'movieId': 2039, 'thriller': 1, 'adventure': 1}
### Collection looks like [{'title': 'Cheetah (1989)', 'horror': 1, 'comedy': 1, 'drama': 1, 'children': 1, 'mystery': 1, 'movieId': 2039, 'thriller': 1, 'adventure': 1}, {'title': 'Cheetah (1989)', 'horror': 1, 'comedy': 1, 'drama': 1, 'children': 1, 'mystery': 1, 'movieId': 2039, 'thriller': 1, 'adventure': 1}, {'title': 'Cheetah (1989)', 'horror': 1, 'comedy': 1, 'drama': 1, 'children': 1, 'mystery': 1, 'movieId': 2039, 'thriller': 1, 'adventure': 1}]

我想要我的收藏品

### [
{'title': 'Big Night (1996)', 'horror': 0, 'comedy': 1, 'drama': 1, 'children': 0, 'mystery': 0, 'movieId': 994, 'thriller': 0, 'adventure': 0},
{'title': 'Grudge, The (2004)', 'horror': 1, 'comedy': 0, 'drama': 0, 'children': 0, 'mystery': 1, 'movieId': 8947, 'thriller': 1, 'adventure': 0},
{'title': 'Cheetah (1989)', 'horror': 0, 'comedy': 0, 'drama': 0, 'children': 1, 'mystery': 0, 'movieId': 2039, 'thriller': 0, 'adventure': 1}
]

数据集 - https://grouplens.org/datasets/movielens/

Answer 1

我现在理解了这个问题，我试图改变字典对象。

def tranformDataFrame(self):
    genres = self.__findUniqueGenres();
    print('### List of genres...', genres);
    __df = self.__prepareDataframe(genres); # Data frame with all required columns.
    rowTemplate = self.__getRowTemplate(__df.columns)
    print('### Row template looks like -->', rowTemplate)
    collection = []
    for index, row in self.df.iterrows():
        # Creating the fresh copy of row template every time prevent mutation. 
        _rowToAdd = self.__createRow(self.__getRowTemplate(__df.columns), row);
        print('### Row looks like', _rowToAdd)
        collection.append(_rowToAdd)

    print('### Collection looks like', collection)
    return __df.append(collection)

虽然必须有某种方法来缓存副本并每次克隆它（而不是处理某些逻辑，并创建字典）。但是，此解决方案至少可以解决这一特定问题。

list.append仅复制最后一项

1 个答案: