用字典交叉 dataframe 列“包含列表”

Question

我正在嘗試創建 dataframe。 來自 dataframe 與交叉字典的組合，如下所示

在此處輸入圖像描述

dataframe 包含多列“N個列x，y，z，a，b，c......等超過100列”

df = pd.DataFrame({'ID':['EF407412','KM043272']
               , 'x': ['[2788, 3140, 4836]','[539, 906, 1494, 1932, 2029,7001]']
               , 'y': ['[1408, 1572, 2277]','[]']
               # dataframe contains multiple columns "N numbers of columns x,y,z,a,b,c ......etc more than 100 columns " 
               })

字典名稱是比例，它的項目（鍵值）是可定制的，從輸入 dataframe 到 output dataframe 的轉換規則在下面的評論中提到

scale = ("500-10000", {

# Key= Scales and value = Weights, both Customizable

    500: 7000,   #  key is 500 and value is compared with List as  List items >= 7000
2500: 3000,  #  key is 2500 and value is compared with List as  List 7000 > items >= 3000
5000: 1000,  #  key is 5000 and value is compared with List as  List 3000 > items >= 1000
7500: 400,   #  key is 7500 and value is compared with List as  List 1000 > items >= 400
10000:250    #  key is 10000 and value is compared with List as  List 400 > items >= 250
             #  any others List items < 250 will be neglected 
             #  any others List items < 250 will be neglected }) 

重要 ps >>> 如果輸入的列表項包含冗余數據，則在 output 中將其視為單獨的值。 例如 x 列包含列表 [4836, 4836, 4836] output 在 x_2500 的列內將是 [4836, 4836, 4836]

Answer 1

使用您的df和scale對象...

def make_new_columns(series: pd.Series) -> pd.DataFrame:
    """Given column, make new columns using `scale`."""
    # convert str representation of list to literal list
    series = series.apply(ast.literal_eval)
    
    scale_dict = scale[1]
    
    frames = []
    for k, v in scale_dict.items():
        k_frame = pd.DataFrame({f"{series.name}_{k}": series.apply(lambda x: [i for i in x if i >= v])})
        frames.append(k_frame)
        
    frame = pd.concat(frames, axis="columns")
    
    cols = frame.columns[frame.columns.str.startswith(f"{series.name}_")]
    
    for col0, col1 in zip(cols, cols[1:]):
        frame[f"{col1}_"] = frame[[col0, col1]].applymap(set).apply(lambda x: x[col1].difference(x[col0]), axis=1)
    
    # the first `x_...` col is `x_500` and will not change -- remove others
    frame = frame.drop(columns=cols[1:])
    
    frame.columns = frame.columns.str.strip("_")
    
    frame[cols] = frame[cols].applymap(lambda x: [0] if not len(x) else list(x))
    
    return frame

# apply `make_new_columns` to x, y, z, a, b, c, ...
cols_to_apply = df.loc[:, "x":].columns

to_join = []
for col in cols_to_apply:
    new = make_new_columns(df[col])
    to_join.append(new)

df = df[["ID"]].join(to_join)

df

Answer 2

嘗試：

from ast import literal_eval

df["x"] = df["x"].apply(literal_eval)
df["y"] = df["y"].apply(literal_eval)

x = df.set_index("ID").stack().to_frame().explode(0).dropna()

x["name"] = pd.cut(
    x[0],
    list(scale[1].values())[::-1] + [float("inf")],
    right=False,
    labels=list(scale[1])[::-1],
)
x["tmp"] = x.index.get_level_values(1)

x = x.pivot_table(
    index=pd.Grouper(level=0),
    columns=["tmp", "name"],
    values=0,
    aggfunc=list,
)

idx = pd.MultiIndex.from_product(
    [set(x.columns.get_level_values(0)), scale[1].keys()]
)

x = x.reindex(idx, axis=1)
x.columns = [f"{a}_{b}" for a, b in x.columns]
x = x.apply(lambda s: s.fillna({i: [0] for i in x.index}))

print(
    x[
        sorted(x.columns, key=lambda x: (x.split("_")[0], int(x.split("_")[1])))
    ].reset_index()
)

印刷：

         ID   x_500        x_2500              x_5000      x_7500 x_10000 y_500 y_2500              y_5000 y_7500 y_10000
0  EF407412     [0]  [3140, 4836]              [2788]         [0]     [0]   [0]    [0]  [1408, 1572, 2277]    [0]     [0]
1  KM043272  [7001]           [0]  [1494, 1932, 2029]  [539, 906]     [0]   [0]    [0]                 [0]    [0]     [0]