Python 用熊猫系列替换NaN。地图（dict）

我正在学习pandas教程，该教程显示通过将字典传递给series.map方法来替换列中的值。以下是本教程的一个片段：

但是，当我尝试这一点时：

cols = star_wars.columns[3:9]

# Booleans for column values
answers = {
        "Star Wars: Episode I  The Phantom Menace":True, 
        "Star Wars: Episode II  Attack of the Clones":True, 
        "Star Wars: Episode III  Revenge of the Sith":True,
        "Star Wars: Episode IV  A New Hope":True,
        "Star Wars: Episode V  The Empire Strikes Back":True,
        "Star Wars: Episode VI  Return of the Jedi":True,
        NaN:False
        }

for c in cols:
    star_wars[c] = star_wars[c].map(answers) 

我得到

name错误：未定义名称“NaN”

那么我做错了什么

编辑：为了更好地解释我的目标，我有如下专栏：

我试着用假来代替南，用真来代替非南

编辑2:这是我将

NaN

更改为

np.NaN

后仍然面临的问题的图像：

---

然后，如果我重新运行mapping单元格并再次显示输出，所有False和NaN值都会触发。

很简单，Python没有内置的

NaN

名称。但是，NumPy确实如此，因此您可以使用

np.nan

使您的映射不会抛出错误。还有

math.nan

，正如乔恩指出的那样，它等于

float（'nan'）

answers = {
        "Star Wars: Episode I  The Phantom Menace":True, 
        "Star Wars: Episode II  Attack of the Clones":True, 
        "Star Wars: Episode III  Revenge of the Sith":True,
        "Star Wars: Episode IV  A New Hope":True,
        "Star Wars: Episode V  The Empire Strikes Back":True,
        "Star Wars: Episode VI  Return of the Jedi":True,
        np.nan:False
        }

但不要在这里停下来，因为这样做行不通。 另一个棘手的问题是，

nan

在技术上并不等于任何东西，因此在这样的映射中使用它将不会有效

>>> np.nan == np.nan 
False

因此，数据帧中的NaN值不会被

np.NaN

作为键拾取，而是保持NaN。有关这方面的进一步解释，请参阅。此外，我敢打赌您的

nan

值实际上就是字符串

nan

最小演示

>>> df
                                          0                                  1
0  Star Wars: Episode I  The Phantom Menace                                nan
1         Star Wars: Episode IV  A New Hope                                nan
2         Star Wars: Episode IV  A New Hope  Star Wars: Episode IV  A New Hope

>>> for c in df.columns:
        df[c] = df[c].map(answers)


>>> df
      0     1
0  True   NaN
1  True   NaN
2  True  True

# notice we're still stuck with NaN, as our nan strings weren't picked up

>>> answers = {
            "Star Wars: Episode I  The Phantom Menace",
            "Star Wars: Episode II  Attack of the Clones" 
            "Star Wars: Episode III  Revenge of the Sith",
            "Star Wars: Episode IV  A New Hope",
            "Star Wars: Episode V  The Empire Strikes Back",
            "Star Wars: Episode VI  Return of the Jedi",
            }

>>> df
                                          0                                  1
0  Star Wars: Episode I  The Phantom Menace                                nan
1         Star Wars: Episode IV  A New Hope                                nan
2         Star Wars: Episode IV  A New Hope  Star Wars: Episode IV  A New Hope

>>> df.isin(answers)

      0      1
0  True  False
1  True  False
2  True   True

更好的解决方案 话虽如此，这似乎不是一个很好的使用口述或地图-你可以只定义一组星球大战字符串，然后在你感兴趣的列的整个部分使用

answers = {
        "Star Wars: Episode I  The Phantom Menace",
        "Star Wars: Episode II  Attack of the Clones" 
        "Star Wars: Episode III  Revenge of the Sith",
        "Star Wars: Episode IV  A New Hope",
        "Star Wars: Episode V  The Empire Strikes Back",
        "Star Wars: Episode VI  Return of the Jedi",
        }

starwars.iloc[:, 3:9].isin(answers) 

最小演示

>>> df
                                          0                                  1
0  Star Wars: Episode I  The Phantom Menace                                nan
1         Star Wars: Episode IV  A New Hope                                nan
2         Star Wars: Episode IV  A New Hope  Star Wars: Episode IV  A New Hope

>>> for c in df.columns:
        df[c] = df[c].map(answers)


>>> df
      0     1
0  True   NaN
1  True   NaN
2  True  True

# notice we're still stuck with NaN, as our nan strings weren't picked up

>>> answers = {
            "Star Wars: Episode I  The Phantom Menace",
            "Star Wars: Episode II  Attack of the Clones" 
            "Star Wars: Episode III  Revenge of the Sith",
            "Star Wars: Episode IV  A New Hope",
            "Star Wars: Episode V  The Empire Strikes Back",
            "Star Wars: Episode VI  Return of the Jedi",
            }

>>> df
                                          0                                  1
0  Star Wars: Episode I  The Phantom Menace                                nan
1         Star Wars: Episode IV  A New Hope                                nan
2         Star Wars: Episode IV  A New Hope  Star Wars: Episode IV  A New Hope

>>> df.isin(answers)

      0      1
0  True  False
1  True  False
2  True   True

---

所以我对另一个解决方案的问题是，由于它的工作方式，代码在第一次运行后不会以相同的方式运行。我在一个Jupyter笔记本上工作，所以我想要一些可以运行多次的东西。我只是一名Python初学者，但以下代码似乎能够运行多次，并且只在第一次运行时更改值：

cols = star_wars.columns[3:9]

# Booleans for column values
answers = {
        "Star Wars: Episode I  The Phantom Menace":True,
        "Star Wars: Episode II  Attack of the Clones":True, 
        "Star Wars: Episode III  Revenge of the Sith":True,
        "Star Wars: Episode IV  A New Hope":True,
        "Star Wars: Episode V The Empire Strikes Back":True,
        "Star Wars: Episode VI Return of the Jedi":True,
        True:True,
        False:False,
        np.nan:False
        }

for c in cols:
    star_wars[c] = star_wars[c].map(answers)