如果和语句未在Python中输出准确的结果

Question

df.loc[(df['Platform'] == 'A321') and (df['Days'] <= 14), 'Mods Applicable'] = 'CFM56 9th Stage Duct'

我正在尝试使A321和Days的Platforms的输出小于或等于14，以打印出以下“适用的Mods”，但并没有说Platforms and the Days。

输出结果：

IF和输出：

Answer 1

尝试使用“＆”代替“ and”：

df = pd.DataFrame({'Platform': ['A123', 'A321', 'A321', 'B123'],
                   'Days': [10, 13, 20, 5]})

df.loc[(df['Platform'] == 'A321') & (df['Days'] <= 14), 'Mods Applicable'] = 'CFM56 9th Stage Duct'

print(df)

      Platform  Days       Mods Applicable
    0     A123    10                   NaN
    1     A321    13  CFM56 9th Stage Duct
    2     A321    20                   NaN
    3     B123     5                   NaN

我可以通过使用“和”而不是“＆”来重现相同的错误：

Traceback (most recent call last):

  File "<ipython-input-128-7957a219684b>", line 1, in <module>
    df.loc[(df['Platform'] == 'A321') and (df['Days'] <= 14), 'Mods Applicable'] = 'CFM56 9th Stage Duct'

  File "/Users/nathanielgates/anaconda3/lib/python3.6/site-packages/pandas/core/generic.py", line 1479, in __nonzero__
    .format(self.__class__.__name__))

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

Answer 2

这是因为将Series与值进行比较也会返回Series。这意味着(df['Platform'] == 'A321')和(df['Days'] <= 14)的结果是Series值的true/false，默认情况下不会解析为单个true或false。所以你不能and！

您可以将and替换为&。它将变成：

df.loc[(df['Platform'] == 'A321') & (df['Days'] <= 14), 'Mods Applicable'] = 'CFM56 9th Stage Duct'