对 Pandas 数据框列使用条件 if/else 逻辑

Using conditional if/else logic with pandas dataframe columns

提问by Aaraeus

My dataframe called pw2looks something like this, where I have two columns, pw1 and pw2, which are probability of wins. I'd like to perform some conditional logic to create another column called WINNERbased off pw1and pw2.

我的数据pw2框看起来像这样，我有两列，pw1 和 pw2，它们是获胜的概率。我想执行一些条件逻辑来创建另一个名为WINNERbased off pw1and 的列pw2。

+-------------------------+-------------+-----------+-------------+
|          Name1          |     pw1     |   Name2   |     pw2     |
+-------------------------+-------------+-----------+-------------+
| Seaking                 | 0.517184213 | Lickitung | 0.189236181 |
| Ferrothorn              | 0.172510623 | Quagsire  | 0.260884258 |
| Thundurus Therian Forme | 0.772536272 | Hitmonlee | 0.694069408 |
| Flaaffy                 | 0.28681284  | NaN       | NaN         |
+-------------------------+-------------+-----------+-------------+

I want to do this conditionally in a function but I'm having some trouble.

我想在一个函数中有条件地做到这一点，但我遇到了一些麻烦。

• if pw1> pw2, populate with Name1
• if pw2> pw1, populate with Name2
• if pw1is populated but pw2isn't, populate with Name1
• if pw2is populated but pw1isn't, populate with Name2

• 如果pw1> pw2，填充Name1
• 如果pw2> pw1，填充Name2
• 如果pw1已填充但未pw2填充，则填充Name1
• 如果pw2已填充但未pw1填充，则填充Name2

But my function isn't working - for some reason checking if a value is null isn't working.

但是我的函数不起作用 - 出于某种原因检查值是否为空不起作用。

def final_winner(df):
    # If PW1 is missing and PW2 is populated, Pokemon 1 wins
    if df['pw1'] = None and df['pw2'] != None:
        return df['Number1']
    # If it's the same thing but the other way around, Pokemon 2 wins
    elif df['pw2'] = None and df['pw1'] != None:
        return df['Number2']
    # If pw2 is greater than pw1, then Pokemon 2 wins
    elif df['pw2'] > df['pw1']:
        return df['Number2']
    else
        return df['Number1']

pw2['Winner'] = pw2.apply(final_winner, axis=1)

回答by rafaelc

Do not use apply, which is very slow. Use np.where

不要使用apply，这很慢。用np.where

pw2 = df.pw2.fillna(-np.inf)
df['winner'] = np.where(df.pw1 > pw2, df.Name1, df.Name2)

Once NaNs always lose, can just fillna()it with -np.infto yield same logic.

一旦NaNs总是输，可以用fillna()它-np.inf来产生同样的逻辑。

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Looking at your code, we can point out several problems. First, you are comparing df['pw1'] = None, which is invalid python syntax for comparison. You usually want to compare things using ==operator. However, for None, it is recommended to use is, such as if variable is None: (...). However again, you are in a pandas/numpyenvironment, where there actually several values for null values (None, NaN, NaT, etc).

查看您的代码，我们可以指出几个问题。首先，您正在比较df['pw1'] = None，这是用于比较的无效 python 语法。您通常希望使用==运算符比较事物。但是，对于None，建议使用is，例如if variable is None: (...)。然而再次，你是在一个pandas/numpy环境中，为空值，其中居然有几个值（None，NaN，NaT等）。

So, it is preferable to check for nullability using pd.isnull()or df.isnull().

因此，最好使用pd.isnull()或检查可空性df.isnull()。

Just to illustrate, this is how your code should look like:

只是为了说明，您的代码应该如下所示：

def final_winner(df):
    if pd.isnull(df['pw1']) and not pd.isnull(df['pw2']):
        return df['Name1']
    elif pd.isnull(df['pw2']) and not pd.isnull(df['pw1']):
        return df['Name1']
    elif df['pw2'] > df['pw1']:
        return df['Name2']
    else:
        return df['Name1']

df['winner'] = df.apply(final_winner, axis=1)

But again, definitely use np.where.

但同样，绝对使用np.where.