熊猫如何不适用于整列

Question

self.df['Regular Price'] = self.df['Regular Price'].apply(
            lambda x: int(round(x)) if isinstance(
                x, (int, float)) else None
        )

只要在数据帧中遇到非数字值，上面的代码就会将 None 分配给字段 Regular Price 的每个值。我只想将None 分配给它的非数字值的那个单元格。

谢谢

Answer 1

首先不可能将NaNs 与integers 一起返回，因为NaNs 在设计上是float。

如果mixed 类型为数字，strings 则您的解决方案有效：

df = pd.DataFrame({
    'Regular Price': ['a',1,2.3,'a',7],
    'B': list(range(5))
})
print (df)
   B Regular Price
0  0             a
1  1             1
2  2           2.3
3  3             a
4  4             7

df['Regular Price'] = df['Regular Price'].apply(
            lambda x: int(round(x)) if isinstance(
                x, (int, float)) else None
        )

print (df)
   B  Regular Price
0  0            NaN
1  1            1.0
2  2            2.0
3  3            NaN
4  4            7.0

但如果所有数据都是字符串，则需要 to_numeric 和 errors='coerce' 将非数字转换为 NaNs：

df = pd.DataFrame({
    'Regular Price': ['a','1','2.3','a','7'],
    'B': list(range(5))
})
print (df)
   B Regular Price
0  0             a
1  1             1
2  2           2.3
3  3             a
4  4             7

df['Regular Price'] = pd.to_numeric(df['Regular Price'], errors='coerce').round()
print (df)
   B  Regular Price
0  0            NaN
1  1            1.0
2  2            2.0
3  3            NaN
4  4            7.0

编辑：

我还需要删除浮点并仅使用int

可以将NaNs 转换为None 并转换为int：

df['Regular Price'] = pd.to_numeric(df['Regular Price'],
                                    errors='coerce').round()

df['Regular Price'] = np.where(df['Regular Price'].isnull(), 
                               None,
                               df['Regular Price'].fillna(0).astype(int))

print (df)
   B Regular Price
0  0          None
1  1             1
2  2             2
3  3          None
4  4             7


print (df['Regular Price'].apply(type))
0    <class 'NoneType'>
1         <class 'int'>
2         <class 'int'>
3    <class 'NoneType'>
4         <class 'int'>
Name: Regular Price, dtype: object

但它会降低性能，所以最好不要使用它。还应该有另一个问题 - soe 功能失败，所以最好是 floats 如果使用 NaNs：

在50k rows DataFrame 中测试diff 之类的函数：

df = pd.DataFrame({
    'Regular Price': ['a','1','2.3','a','7'],
    'B': list(range(5))
})
df = pd.concat([df]*10000).reset_index(drop=True)
print (df)

df['Regular Price'] = pd.to_numeric(df['Regular Price'], errors='coerce').round()

df['Regular Price1'] = np.where(df['Regular Price'].isnull(), 
                               None,
                               df['Regular Price'].fillna(0).astype(int))

---

In [252]: %timeit df['Regular Price2'] = df['Regular Price1'].diff()

TypeError: 不支持的操作数类型 -: 'int' 和 'NoneType'

In [274]: %timeit df['Regular Price3'] = df['Regular Price'].diff()
1000 loops, best of 3: 301 µs per loop

---

In [272]: %timeit df['Regular Price2'] = df['Regular Price1'] * 1000
100 loops, best of 3: 4.48 ms per loop

In [273]: %timeit df['Regular Price3'] = df['Regular Price'] * 1000
1000 loops, best of 3: 469 µs per loop

---

编辑：

df = pd.DataFrame({
    'Regular Price': ['a','1','2.3','a','7'],
    'B': list(range(5))
})
print (df)
   B Regular Price
0  0             a
1  1             1
2  2           2.3
3  3             a
4  4             7

df['Regular Price'] = pd.to_numeric(df['Regular Price'], errors='coerce').round()
print (df)
   B  Regular Price
0  0            NaN
1  1            1.0
2  2            2.0
3  3            NaN
4  4            7.0

首先可以逐列删除NaNs 行Regular Price，然后转换为int。

df1 = df.dropna(subset=['Regular Price']).copy()
df1['Regular Price']  = df1['Regular Price'].astype(int)
print (df1)
   B  Regular Price
1  1              1
2  2              2
4  4              7

处理您需要的内容，但不要更改索引。

#e.g. some process 
df1['Regular Price']  = df1['Regular Price'] * 100

最后一个combine_first - 它将NaN 添加到Regular Price 列。

df2 = df1.combine_first(df)
print (df2)
     B  Regular Price
0  0.0            NaN
1  1.0          100.0
2  2.0          200.0
3  3.0            NaN
4  4.0          700.0