系列的真值不明确-调用函数时出错 [英] The truth value of a Series is ambiguous - Error when calling a function

问题描述

我知道以下错误

ValueError：系列的真值不明确。使用a.empty，a.bool（），a.item（），a.any（）或a.all（）。

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

<很久以前有人问过。

has been asked a long time ago.

但是，我试图创建一个基本函数并使用 df ['busy'] 返回一个新列 1 或 0 。我的函数看起来像这样，

However, I am trying to create a basic function and return a new column with df['busy'] with 1 or 0. My function looks like this,

def hour_bus(df):
    if df[(df['hour'] >= '14:00:00') & (df['hour'] <= '23:00:00')&\
             (df['week_day'] != 'Saturday') & (df['week_day'] != 'Sunday')]:
         return df['busy'] == 1
     else:
         return df['busy'] == 0 

我可以执行该函数，但是当我用DataFrame调用它时，出现了上面提到的错误。我遵循以下线程和另一个线程创建该功能。我在 if 子句中使用了& 而不是和 。

I can execute the function, but when I call it with the DataFrame, I get the error mentioned above. I followed the following thread and another thread to create that function. I used & instead of and in my if clause.

无论如何，当我执行以下操作时，我会得到所需的输出。

Anyhow, when I do the following, I get my desired output.

df['busy'] = np.where((df['hour'] >= '14:00:00') & (df['hour'] <= '23:00:00') & \
                        (df['week_day'] != 'Saturday') & (df['week_day'] != 'Sunday'),'1','0')

关于我在 hour_bus 中犯什么错误的任何想法

Any ideas on what mistake am I making in my hour_bus function?

推荐答案

The

(df['hour'] >= '14:00:00') & (df['hour'] <= '23:00:00')& (df['week_day'] != 'Saturday') & (df['week_day'] != 'Sunday')

给出一个布尔数组，当您为您的 df 编制索引，您将获得（可能）较小的 df 部分。

gives a boolean array, and when you index your df with that you'll get a (probably) smaller part of your df.

仅说明我的意思：

import pandas as pd

df = pd.DataFrame({'a': [1,2,3,4]})
mask = df['a'] > 2
print(mask)
# 0    False
# 1    False
# 2     True
# 3     True
# Name: a, dtype: bool
indexed_df = df[mask]
print(indexed_df)
#    a
# 2  3
# 3  4

但是它仍然是 DataFrame ，因此将其用作需要真值的表达式是模棱两可的（在您的 if ）。

However it's still a DataFrame so it's ambiguous to use it as expression that requires a truth value (in your case an if).

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

您可以使用 np.where 二手-或等效方式：

You could use the np.where you used - or equivalently:

def hour_bus(df):
    mask = (df['hour'] >= '14:00:00') & (df['hour'] <= '23:00:00')& (df['week_day'] != 'Saturday') & (df['week_day'] != 'Sunday')
    res = df['busy'] == 0                             
    res[mask] = (df['busy'] == 1)[mask]  # replace the values where the mask is True
    return res

但是 np.where 是更好的解决方案（可读性更好，而且速度可能更快）。

However the np.where will be the better solution (it's more readable and probably faster).

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