Python视图vs复制错误希望我仅在脚本中使用.loc [英] Python view vs copy error wants me to use .loc in script only

问题描述

我正在运行一个长脚本，该脚本的数据帧为df.在脚本运行时，逐列建立和修改df，我在命令行中一遍又一遍地得到此错误:

I'm running a long script which has a dataframe df. as the script runs, building up and modifying df column by column I get this error over and over again in the command line:

试图在DataFrame的切片副本上设置一个值.尝试 使用.loc [row_indexer，col_indexer] = value代替请参见 文档: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy

A value is trying to be set on a copy of a slice from a DataFrame. Try using .loc[row_indexer,col_indexer] = value instead See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy

但是随后它将打印出引起警告的行，并且看起来不会像问题一样.如下所示的行将触发它(每行分别触发它):

But then it will print out the line that is causing the warning and it wont look like a problem. Lines such as the following will trigger it (each line triggered it separately):

df['ZIP_DENS'] = df['ZIP_DENS'].astype(str)
df['AVG_WAGE'] = df['AVG_WAGE'].astype(str).apply(lambda x:x if x != 'nan' else 'unknown')
df['TERM_BIN'] = df['TERMS'].map(terms_dict)
df['LOSS_ONE'] = 'T_'+ df['TERM'].astype(str) +'_C_'+ df['COMP'].astype(str) + df['SIZE']
# this one's inside a loop:
df[i + '_BIN'] = df[i + '_BIN'].apply(lambda x:x if x != 'nan' else 'unknown')

有一些我在数据框上进行突变的示例.现在，此警告刚刚开始出现，但我无法在解释器中重新创建此问题.当我打开终端机时，我会尝试类似的操作，但它不会给我任何警告:

There are some examples of the mutations I'm making on the dataframe. Now, this warning just started showing up but I can't recreate this problem in the interpreter. When I open a terminal I try things like this and it gives me no warnings:

import pandas as pd
df = pd.DataFrame([list('ab'),list('ef')],columns=['first','second'])
df['third'] = df[['first','second']].astype('str')

我是否缺少某些东西，对于这个警告试图告诉我的关于DataFrames的性质我不了解的东西?您是否认为我可能在脚本开始时对此数据帧进行了某些操作，然后该对象上的所有后续突变都是视图或视图副本的突变，或者类似的事情正在发生?

Is there something I'm missing, something I don't understand about the nature of DataFrames that this warning is trying to tell me? Do you think perhaps I did something to this dataframe at the beginning of the script and then all subsequent mutations on the object are mutations on a view or a copy of it or something weird like that is going on?

推荐答案

正如我在评论中提到的那样，可能的问题是，在代码上游的某个位置，您为df分配了一些其他pd.DataFrame的一部分. 这是造成混乱的常见原因，并且在为什么在使用链式索引时分配失败.

As I mentioned in my comment, the likely issue is that somewhere upstream in your code, you assigned a slice of some other pd.DataFrame to df. This is a common cause of confusion and is also explained under why-does-assignment-fail-when-using-chained-indexing in the link that the Warning mentions.

一个最小的例子:

data = pd.DataFrame({'a':range(7), 'b':list('abcccdb')})
df = data[data.a % 2 == 0]  #making a subselection of the DataFrame  
df['b'] = 'b'

/home/user/miniconda3/envs/myenv/lib/python3.6/site-packages/ipykernel_launcher.py:1: SettingWithCopyWarning:试图在一个副本上设置一个值 从DataFrame切片.尝试使用.loc [row_indexer，col_indexer] = 值代替

/home/user/miniconda3/envs/myenv/lib/python3.6/site-packages/ipykernel_launcher.py:1: SettingWithCopyWarning: A value is trying to be set on a copy of a slice from a DataFrame. Try using .loc[row_indexer,col_indexer] = value instead

请参阅文档中的警告: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy "启动IPython内核的入口点.

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy """Entry point for launching an IPython kernel.

请注意本节:

df = data[data.a % 2 == 0]  #making a subselection of the DataFrame  
df['b'] = 'b'

也可以这样重写:

data[data.a % 2 == 0]['b'] = 'b'  #obvious chained indexing  
df = data[data.a % 2 == 0]

写此位的正确方法如下:

The correct way of writing this bit is the following way:

data = pd.DataFrame({'a':range(7), 'b':list('abcccdb')})
df = data.loc[data.a % 2 == 0].copy()  #making a copy of the subselection   
df.loc[:,'b'] = 'b'

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