pandas 通过引用传递pandas DataFrame
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/39783570/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Passing pandas DataFrame by reference
提问by labrynth
My question is regarding immutability of pandas DataFrame when it is passed by reference. Consider the following code:
我的问题是关于 Pandas DataFrame 在通过引用传递时的不变性。考虑以下代码：
import pandas as pd
def foo(df1, df2):
df1['B'] = 1
df1 = df1.join(df2['C'], how='inner')
return()
def main(argv = None):
# Create DataFrames.
df1 = pd.DataFrame(range(0,10,2), columns=['A'])
df2 = pd.DataFrame(range(1,11,2), columns=['C'])
foo(df1, df2) # Pass df1 and df2 by reference.
print df1
return(0)
if __name__ == '__main__':
status = main()
sys.exit(status)
The output is
输出是
A B
0 0 1
1 2 1
2 4 1
3 6 1
4 8 1
and not
并不是
A B C
0 0 1 1
1 2 1 3
2 4 1 5
3 6 1 7
4 8 1 9
In fact, if foo is defined as
事实上，如果 foo 被定义为
def foo(df1, df2):
df1 = df1.join(df2['C'], how='inner')
df1['B'] = 1
return()
(i.e. the "join" statement before the other statement) then the output is simply
（即另一个语句之前的“join”语句）那么输出就是
A
0 0
1 2
2 4
3 6
4 8
I'm intrigued as to why this is the case. Any insights would be appreciated.
我很好奇为什么会这样。任何见解将不胜感激。
回答by Jezzamon
The issue is because of this line:
问题是因为这一行：
df1 = df1.join(df2['C'], how='inner')
df1.join(df2['C'], how='inner')returns a new dataframe. After this line, df1no longer refers to the same dataframe as the argument, but a new one, because it's been reassigned to the new result. The first dataframe continues to exist, unmodified. This isn't really a pandas issue, just the general way python, and most other languages, work.
df1.join(df2['C'], how='inner')返回一个新的数据帧。在这一行之后，df1不再引用与参数相同的数据帧，而是一个新的数据帧，因为它已被重新分配给新结果。第一个数据帧继续存在，未修改。这不是真正的Pandas问题，只是 python 和大多数其他语言的一般工作方式。
Some pandas functions have an inplaceargument, which would do what you want, however the join operation doesn't. If you need to modify a dataframe, you'll have to return this new one instead and reassign it outside the function.
一些 Pandas 函数有一个inplace参数，它可以做你想要的，但是连接操作没有。如果你需要修改一个数据框，你必须返回这个新的，并在函数之外重新分配它。
回答by Ami Tavory
Python doesn't have pass by value vs. pass by reference - there are just bindings from names to objects.
Python 没有按值传递和按引用传递——只有从名称到对象的绑定。
If you change your function to
如果您将函数更改为
def foo(df1, df2):
res = df1.join(df2['C'], how='inner')
res['B'] = 1
return res
Then df1, df2, in the function, are bound to the objects you sent. The result of the join, which is a new object in this case, is bound to the name res. You can manipulate it, and return it, without affecting any of the other objects or bindings.
然后df1，df2在函数中，绑定到您发送的对象。的结果，join在这种情况下是一个新对象，绑定到名称res。您可以操作它并返回它，而不会影响任何其他对象或绑定。
In your calling code, you could just write
在你的调用代码中，你可以写
print foo(df1, df2)
