Question

我希望能够在数据框上使用df.fillna()函数，但要根据该特定单元格的“索引和列”名称对其应用一个条件。

我正在尝试根据以下数据集（以下大型词典的歉意）创建曲棍球队友数据的热图-

linemates_toi = {
    'Player 1': {'Player 2': 0.25, 'Player 3': 7.95, 'Player 4': 0.6333, 'Player 5': 9.95, 'Player 6': 0.6333, 'Player 7': 0.8, 'Player 8': 4.2667, 'Player 9': 7.8833, 'Player 10': 0.3, 'Player 11': 11.2333, 'Player 12': 3.35, 'Player 13': 0.2167},
    'Player 10': {'Player 14': 2.3, 'Player 18': 1.2667, 'Player 2': 6.8333, 'Player 4': 5.5833, 'Player 5': 0.9, 'Player 16': 6.9167, 'Player 6': 4.9667, 'Player 7': 4.15, 'Player 15': 1.0, 'Player 8': 0.3167, 'Player 17': 5.3167, 'Player 1': 0.3, 'Player 11': 1.6167, 'Player 12': 0.6833, 'Player 13': 12.7167}, 
    'Player 12': {'Player 14': 4.5333, 'Player 18': 4.3333, 'Player 2': 3.1167, 'Player 3': 1.2333, 'Player 4': 5.7333, 'Player 5': 3.5167, 'Player 16': 3.0, 'Player 6': 3.0167, 'Player 7': 2.4, 'Player 15': 2.0167, 'Player 8': 11.6667, 'Player 17': 2.2667, 'Player 9': 0.1167, 'Player 1': 3.35, 'Player 10': 0.6833, 'Player 11': 3.35}, 
    'Player 17': {'Player 14': 4.55, 'Player 18': 1.65, 'Player 2': 0.8833, 'Player 3': 2.85, 'Player 5': 0.0333, 'Player 16': 2.9167, 'Player 6': 7.8167, 'Player 7': 6.0833, 'Player 8': 3.8, 'Player 9': 2.25, 'Player 10': 5.3167, 'Player 12': 2.2667, 'Player 13': 5.7833},
    'Player 7': {'Player 18': 0.3667, 'Player 2': 0.6667, 'Player 3': 1.55, 'Player 4': 0.3333, 'Player 5': 0.15, 'Player 16': 1.2167, 'Player 6': 6.8333, 'Player 15': 0.3333, 'Player 8': 3.0667, 'Player 17': 6.0833, 'Player 9': 1.8833, 'Player 1': 0.8, 'Player 10': 4.15, 'Player 11': 1.0, 'Player 12': 2.4, 'Player 13': 4.4333}, 
    'Player 16': {'Player 14': 2.2833, 'Player 2': 8.5333, 'Player 3': 2.7, 'Player 4': 8.0167, 'Player 5': 0.45, 'Player 6': 0.4, 'Player 7': 1.2167, 'Player 8': 2.3, 'Player 17': 2.9167, 'Player 9': 2.15, 'Player 10': 6.9167, 'Player 11': 0.1333, 'Player 12': 3.0, 'Player 13': 6.5833},
    'Player 18': {'Player 14': 10.05, 'Player 2': 0.75, 'Player 3': 5.0, 'Player 4': 3.45, 'Player 5': 0.3333, 'Player 6': 0.8333, 'Player 7': 0.3667, 'Player 15': 5.2, 'Player 8': 5.8167, 'Player 17': 1.65, 'Player 9': 4.3833, 'Player 10': 1.2667, 'Player 11': 1.5, 'Player 12': 4.3333, 'Player 13': 1.5333},
    'Player 13': {'Player 14': 3.0333, 'Player 18': 1.5333, 'Player 2': 5.9167, 'Player 3': 0.7333, 'Player 4': 4.95, 'Player 5': 0.8167, 'Player 16': 6.5833, 'Player 6': 5.1333, 'Player 7': 4.4333, 'Player 15': 1.2667, 'Player 8': 0.2833, 'Player 17': 5.7833, 'Player 1': 0.2167, 'Player 10': 12.7167, 'Player 11': 1.5333},
    'Player 5': {'Player 18': 0.3333, 'Player 2': 0.8333, 'Player 3': 8.0333, 'Player 16': 0.45, 'Player 6': 0.3333, 'Player 7': 0.15, 'Player 8': 3.0167, 'Player 17': 0.0333, 'Player 9': 6.7333, 'Player 1': 9.95, 'Player 10': 0.9, 'Player 11': 11.2333, 'Player 12': 3.5167, 'Player 13': 0.8167},
    'Player 15': {'Player 14': 4.5667, 'Player 18': 5.2, 'Player 2': 0.4667, 'Player 3': 2.35, 'Player 6': 0.1667, 'Player 7': 0.3333, 'Player 8': 2.0167, 'Player 9': 2.0833, 'Player 10': 1.0, 'Player 12': 2.0167, 'Player 13': 1.2667},
    'Player 2': {'Player 18': 0.75, 'Player 3': 2.65, 'Player 4': 8.6, 'Player 5': 0.8333, 'Player 16': 8.5333, 'Player 6': 0.8333, 'Player 7': 0.6667, 'Player 15': 0.4667, 'Player 8': 2.3333, 'Player 17': 0.8833, 'Player 9': 1.9167, 'Player 1': 0.25, 'Player 10': 6.8333, 'Player 11': 1.6167, 'Player 12': 3.1167, 'Player 13': 5.9167},
    'Player 8': {'Player 14': 5.8333, 'Player 18': 5.8167, 'Player 2': 2.3333, 'Player 3': 1.1167, 'Player 4': 5.6833, 'Player 5': 3.0167, 'Player 16': 2.3, 'Player 6': 4.2667, 'Player 7': 3.0667, 'Player 15': 2.0167, 'Player 17': 3.8, 'Player 9': 1.1333, 'Player 1': 4.2667, 'Player 10': 0.3167, 'Player 11': 3.8167, 'Player 12': 11.6667, 'Player 13': 0.2833},
    'Player 4': {'Player 14': 3.2833, 'Player 18': 3.45, 'Player 2': 8.6, 'Player 3': 2.0667, 'Player 16': 8.0167, 'Player 6': 0.8333, 'Player 7': 0.3333, 'Player 8': 5.6833, 'Player 9': 1.85, 'Player 1': 0.6333, 'Player 10': 5.5833, 'Player 11': 0.85, 'Player 12': 5.7333, 'Player 13': 4.95},
    'Player 9': {'Player 14': 4.5167, 'Player 18': 4.3833, 'Player 2': 1.9167, 'Player 3': 14.35, 'Player 4': 1.85, 'Player 5': 6.7333, 'Player 16': 2.15, 'Player 6': 0.8833, 'Player 7': 1.8833, 'Player 15': 2.0833, 'Player 8': 1.1333, 'Player 17': 2.25, 'Player 1': 7.8833, 'Player 11': 9.0667, 'Player 12': 0.1167},
    'Player 14': {'Player 18': 10.05, 'Player 3': 5.7167, 'Player 4': 3.2833, 'Player 16': 2.2833, 'Player 6': 1.8833, 'Player 15': 4.5667, 'Player 8': 5.8333, 'Player 17': 4.55, 'Player 9': 4.5167, 'Player 10': 2.3, 'Player 11': 0.9833, 'Player 12': 4.5333, 'Player 13': 3.0333},
    'Player 11': {'Player 14': 0.9833, 'Player 18': 1.5, 'Player 2': 1.6167, 'Player 3': 9.7667, 'Player 4': 0.85, 'Player 5': 11.2333, 'Player 16': 0.1333, 'Player 6': 0.5, 'Player 7': 1.0, 'Player 8': 3.8167, 'Player 9': 9.0667, 'Player 1': 11.2333, 'Player 10': 1.6167, 'Player 12': 3.35, 'Player 13': 1.5333},
    'Player 6': {'Player 14': 1.8833, 'Player 18': 0.8333, 'Player 2': 0.8333, 'Player 3': 1.1333, 'Player 4': 0.8333, 'Player 5': 0.3333, 'Player 16': 0.4, 'Player 7': 6.8333, 'Player 15': 0.1667, 'Player 8': 4.2667, 'Player 17': 7.8167, 'Player 9': 0.8833, 'Player 1': 0.6333, 'Player 10': 4.9667, 'Player 11': 0.5, 'Player 12': 3.0167, 'Player 13': 5.1333},
    'Player 3': {'Player 14': 5.7167, 'Player 18': 5.0, 'Player 2': 2.65, 'Player 4': 2.0667, 'Player 5': 8.0333, 'Player 16': 2.7, 'Player 6': 1.1333, 'Player 7': 1.55, 'Player 15': 2.35, 'Player 8': 1.1167, 'Player 17': 2.85, 'Player 9': 14.35, 'Player 1': 7.95, 'Player 11': 9.7667, 'Player 12': 1.2333, 'Player 13': 0.7333}
}

df = pd.DataFrame(linemates_toi)

我现在想要实现的是使用df.fillna(0)并应用一个条件，因此唯一被替换的NaN是当索引和列名称不匹配时，因为我希望这些单元格能够保留NaN，这样当我将它们绘制到热图中时，它们在Matplotlib所应用的cmap中就没有任何颜色。

如果我正在编写伪代码，它将看起来像这样-

df.fillna(0, df.cell.Index.Name != df.cell.Column.Name)

谢谢！

Answer 1

使用一些广播和NaN-掩码

mask = df.index.to_numpy() == df.columns.to_numpy()[:, None]

df.fillna(0).mask(mask)

>>> df.head()

           Player 1  Player 10  Player 12  Player 17  Player 7  Player 16  \
Player 1        NaN     0.3000     3.3500     0.0000    0.8000     0.0000   
Player 10    0.3000        NaN     0.6833     5.3167    4.1500     6.9167   
Player 11   11.2333     1.6167     3.3500     0.0000    1.0000     0.1333   
Player 12    3.3500     0.6833        NaN     2.2667    2.4000     3.0000   
Player 13    0.2167    12.7167     0.0000     5.7833    4.4333     6.5833   

           Player 18  Player 13  Player 5  Player 15  Player 2  Player 8  \
Player 1      0.0000     0.2167    9.9500     0.0000    0.2500    4.2667   
Player 10     1.2667    12.7167    0.9000     1.0000    6.8333    0.3167   
Player 11     1.5000     1.5333   11.2333     0.0000    1.6167    3.8167   
Player 12     4.3333     0.0000    3.5167     2.0167    3.1167   11.6667   
Player 13     1.5333        NaN    0.8167     1.2667    5.9167    0.2833   

           Player 4  Player 9  Player 14  Player 11  Player 6  Player 3  
Player 1     0.6333    7.8833     0.0000    11.2333    0.6333    7.9500  
Player 10    5.5833    0.0000     2.3000     1.6167    4.9667    0.0000  
Player 11    0.8500    9.0667     0.9833        NaN    0.5000    9.7667  
Player 12    5.7333    0.1167     4.5333     3.3500    3.0167    1.2333  
Player 13    4.9500    0.0000     3.0333     1.5333    5.1333    0.7333

Answer 2

您也可以执行以下操作：

df.fillna(0, inplace=True)
for col in df:
    df.loc[col, col] = np.nan

说明：

填写NaNs
设置行名称等于列名称的NaN

Answer 3

使用df.apply在每个列上映射一个lambda：

df = df.apply(lambda col: col.where((col.name == col.index) | col.notnull(), 0))

如果col为true，则

col.where(condition, value_if_false)返回condition中的原始值。否则返回value_if_false

基于索引和列名称的条件的数据框填充

3 个答案: