pandas 系列的真值不明确 - 调用函数时出错
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The truth value of a Series is ambiguous - Error when calling a function
提问by i.n.n.m
I know following error
我知道以下错误
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
ValueError：系列的真值不明确。使用 a.empty、a.bool()、a.item()、a.any() 或 a.all()。
has been asked a long time ago.
很久以前就被问到了。
However, I am trying to create a basic function and return a new column with df['busy']with 1or 0. My function looks like this,
但是，我正在尝试创建一个基本函数并返回一个df['busy']带有1或的新列0。我的功能看起来像这样，
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
I can execute the function, but when I call it with the DataFrame, I get the error mentioned above. I followed the following threadand another threadto create that function. I used &instead of andin my ifclause.
我可以执行该函数，但是当我使用 DataFrame 调用它时，出现上述错误。我遵循以下线程和另一个线程来创建该函数。我在我的条款中使用了&而不是。andif
Anyhow, when I do the following, I get my desired output.
无论如何，当我执行以下操作时，我会得到我想要的输出。
df['busy'] = np.where((df['hour'] >= '14:00:00') & (df['hour'] <= '23:00:00') & \
(df['week_day'] != 'Saturday') & (df['week_day'] != 'Sunday'),'1','0')
Any ideas on what mistake am I making in my hour_busfunction?
关于我在我的hour_bus函数中犯了什么错误的任何想法？
采纳答案by MSeifert
The
这
(df['hour'] >= '14:00:00') & (df['hour'] <= '23:00:00')& (df['week_day'] != 'Saturday') & (df['week_day'] != 'Sunday')
gives a boolean array, and when you index your dfwith that you'll get a (probably) smaller part of your df.
给出一个布尔数组，当你df用它索引你时，你会得到一个（可能）较小的df.
Just to illustrate what I mean:
只是为了说明我的意思：
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
However it's still a DataFrameso it's ambiguous to use it as expression that requires a truth value (in your case an if).
但是它仍然是 aDataFrame所以将它用作需要真值的表达式（在你的情况下是 an if）是不明确的。
bool(indexed_df)
# ValueError: The truth value of a DataFrame is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
You could use the np.whereyou used - or equivalently:
您可以使用np.where您使用的 - 或等效的：
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
However the np.wherewill be the better solution (it's more readable and probably faster).
然而，这np.where将是更好的解决方案（它更具可读性并且可能更快）。
