python - Pandas 用 None 替换单元格匹配条件

Question

采用以下示例数据框:

df = pd.DataFrame([['de', None, None], 
                   ['de ditos', 2, 3], 
                   [4, None, None], 
                   [None, None, 9],
                   ['de', 4, 6]])

看起来像

          0    1    2
0        de  NaN  NaN
1  de ditos  2.0  3.0
2         4  NaN  NaN
3      None  NaN  9.0
4        de  4.0  6.0

我想用 None 替换第 0 列中单元格值为“de”的所有值，以便数据框像这样结束:

          0    1    2
0      None  NaN  NaN
1  de ditos  2.0  3.0
2         4  NaN  NaN
3      None  NaN  9.0
4      None  4.0  6.0

我试过:

df[0].where(df[0] == 'de') = None

返回 SyntaxError: can't assign to function call

我也试过:

def erasedes(x):
    if x == 'de':
        return None
    else: pass
df[0] = df[0].apply(lambda x: erasedes(x))

但这会将每个值替换为 None

Answer 1

这应该有效:

df[0]= df[0].replace({'de': None})