python - 在这种情况下，Python 是按引用传递还是按值传递，为什么？

Question

这个问题已经有答案了:

How do I clone a list so that it doesn't change unexpectedly after assignment? (24 个回答)

已关闭 3 年前。

以下代码是我的简单示例，对于每一步，我都会在注释中解释我在做什么，问题在最后。

import pandas as pd
todays_date = datetime.datetime.now().date()
index = pd.date_range(todays_date, periods=3, freq='D')
columns = ['A','B','C']
df1 = pd.DataFrame(index=index, columns=columns)
df1 = df1.fillna(1) 

# up to here, i've just create a random df1, which looks like the follow:  
#               A   B   C
# 2020-03-24    1   1   1
# 2020-03-25    1   1   1
# 2020-03-26    1   1   1

df2 = df1     # here created a copy of df1 named as df2, it should pass by value based on my knowledge 
df1 += df2.shift(1)    # this should be the same as df1 = df1 + df2.shift(1)

display(df1)  # here I print out df1 which looks like the follow: 
#               A   B   C
# 2020-03-24    NaN NaN NaN
# 2020-03-25    2.0 2.0 2.0
# 2020-03-26    2.0 2.0 2.0

display(df2)  # here I print out df2, the result surprise me because i thought df2 isn't changed from when it is defined , however it becomes the same as the new df1: 
#               A   B   C
# 2020-03-24    NaN NaN NaN
# 2020-03-25    2.0 2.0 2.0
# 2020-03-26    2.0 2.0 2.0

谁能向我解释一下为什么df2这些步骤有变化吗？我真的很困惑。

Answer 1

df2 = df1     # here created a copy of df1 named as df2

该评论不正确，可能是造成您误解的原因。

这行代码的意思是:df2 现在是当前称为 df1 的另一个名称。

因此，如果您更改名为 df1 的对象，当您引用 df2 时，您也会看到此更改。