如何处理Pandas中的SettingWithCopyWarning

Question

背景
我刚刚将熊猫从0.11升级到0.13.0rc1。现在，该应用程序会弹出许多新警告。其中之一是这样的：

E:\FinReporter\FM_EXT.py:449: SettingWithCopyWarning: A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_index,col_indexer] = value instead
  quote_df['TVol']   = quote_df['TVol']/TVOL_SCALE

我想知道这到底是什么意思？我需要改变什么吗？

如果我坚持使用该如何警告quote_df[‘TVol’] = quote_df[‘TVol’]/TVOL_SCALE？

产生错误的功能

def _decode_stock_quote(list_of_150_stk_str):
    """decode the webpage and return dataframe"""

    from cStringIO import StringIO

    str_of_all = "".join(list_of_150_stk_str)

    quote_df = pd.read_csv(StringIO(str_of_all), sep=',', names=list('ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefg')) #dtype={'A': object, 'B': object, 'C': np.float64}
    quote_df.rename(columns={'A':'STK', 'B':'TOpen', 'C':'TPCLOSE', 'D':'TPrice', 'E':'THigh', 'F':'TLow', 'I':'TVol', 'J':'TAmt', 'e':'TDate', 'f':'TTime'}, inplace=True)
    quote_df = quote_df.ix[:,[0,3,2,1,4,5,8,9,30,31]]
    quote_df['TClose'] = quote_df['TPrice']
    quote_df['RT']     = 100 * (quote_df['TPrice']/quote_df['TPCLOSE'] - 1)
    quote_df['TVol']   = quote_df['TVol']/TVOL_SCALE
    quote_df['TAmt']   = quote_df['TAmt']/TAMT_SCALE
    quote_df['STK_ID'] = quote_df['STK'].str.slice(13,19)
    quote_df['STK_Name'] = quote_df['STK'].str.slice(21,30)#.decode('gb2312')
    quote_df['TDate']  = quote_df.TDate.map(lambda x: x[0:4]+x[5:7]+x[8:10])

    return quote_df

更多错误讯息

E:\FinReporter\FM_EXT.py:449: SettingWithCopyWarning: A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_index,col_indexer] = value instead
  quote_df['TVol']   = quote_df['TVol']/TVOL_SCALE
E:\FinReporter\FM_EXT.py:450: SettingWithCopyWarning: A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_index,col_indexer] = value instead
  quote_df['TAmt']   = quote_df['TAmt']/TAMT_SCALE
E:\FinReporter\FM_EXT.py:453: SettingWithCopyWarning: A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_index,col_indexer] = value instead
  quote_df['TDate']  = quote_df.TDate.map(lambda x: x[0:4]+x[5:7]+x[8:10])

Answer 1

在SettingWithCopyWarning被创造的标志可能造成混淆的“链接”的任务，比如下面这并不总是按预期方式工作，特别是当第一选择返回一个副本。

df[df['A'] > 2]['B'] = new_val  # new_val not set in df

该警告提出了如下重写建议：

df.loc[df['A'] > 2, 'B'] = new_val

但是，这不适合您的用法，相当于：

df = df[df['A'] > 2]
df['B'] = new_val

显然，您不关心将其写回到原始框架的操作（因为您正在覆盖对它的引用），但是不幸的是，这种模式无法与第一个链式分配示例区分开。因此，（误报）警告。如果您想进一步阅读，可能会在建立索引的文档中解决误报的可能性。您可以通过以下分配安全地禁用此新警告。

import pandas as pd
pd.options.mode.chained_assignment = None  # default='warn'