如何处理Pandas中的SettingWithCopyWarning？

Itachi2020-08-03

背景

我刚刚将Pandas从0.11升级到0.13.0rc1。现在，该应用程序弹出许多新警告。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])

古一柳叶风吹2020/08/03 14:13:37

对我来说，此问题发生在下面的> simplified <示例中。而且我也能够解决它（希望有一个正确的解决方案）：

带有警告的旧代码：

def update_old_dataframe(old_dataframe, new_dataframe):
    for new_index, new_row in new_dataframe.iterrorws():
        old_dataframe.loc[new_index] = update_row(old_dataframe.loc[new_index], new_row)

def update_row(old_row, new_row):
    for field in [list_of_columns]:
        # line with warning because of chain indexing old_dataframe[new_index][field]
        old_row[field] = new_row[field]  
    return old_row

这打印了该行的警告 old_row[field] = new_row[field]

由于update_row方法中的行实际上是type Series，因此我将行替换为：

old_row.at[field] = new_row.at[field]

即用于访问/查找的方法Series。即使两种方法都可以正常工作并且结果是相同的，但是通过这种方式，我不必禁用警告（=将其保留在其他地方的其他链索引问题中）。

我希望这可以帮助某人。

DavaidTony宝儿2020/08/03 14:13:36

这个话题确实让Pandas感到困惑。幸运的是，它有一个相对简单的解决方案。

问题在于，并不总是清楚数据过滤操作（例如loc）是否返回DataFrame的副本或视图。因此，这种过滤后的DataFrame的进一步使用可能会造成混淆。

一个简单的解决方案是（除非您需要处理非常大的数据集）：

每当需要更新任何值时，请始终确保在分配之前隐式复制DataFrame。

df  # Some DataFrame
df = df.loc[:, 0:2]  # Some filtering (unsure whether a view or copy is returned)
df = df.copy()  # Ensuring a copy is made
df[df["Name"] == "John"] = "Johny"  # Assignment can be done now (no warning)

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