蟒蛇：为什么我越来越settingwithcopywarning(Why does this code throw a SettingWithCopyWarning? [duplicate])

蟒蛇：为什么我越来越settingwithcopywarning(Why does this code throw a SettingWithCopyWarning? [duplicate])

所以我试图创建一个新的列，指示指定的条件是否为True。 我希望列简单说明“1”或“0”。

这是我的代码：

data_sub = data_orig.loc[~pd.isnull(data_orig['Last_Audit_Date']), :] data_sub.reset_index(inplace=True) data_sub['PackageLengthFlag'] = (abs(data_sub.loc['AUDIT_Primary_Length'] - data_sub.loc[:, 'PKG_SUB_Length']) > threshold)

我认为True = 1和False = 0默认情况下，如果我将它转换为整数，对不对？ （我以为我在某处说这个......）

这是我不断得到的警告：

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

我读到：

如何在Pandas中处理SettingWithCopyWarning？

在熊猫片上设置值的正确方法

熊猫SettingWithCopyWarning

但我不认为他们做我正在寻找的东西。 任何人有任何建议？ 我知道这个问题可能听起来很愚蠢，但仍然感谢任何帮助！

编辑我在创建data_sub的地方添加了两行代码。 希望有所帮助！

This question already has an answer here:

How to deal with SettingWithCopyWarning in Pandas? 13 answers

so I'm trying to create a new column that indicates whether or not the specified condition is True. I want the column to simply state "1" or "0".

Here's my code:

data_sub = data_orig.loc[~pd.isnull(data_orig['Last_Audit_Date']), :] data_sub.reset_index(inplace=True) data_sub['PackageLengthFlag'] = (abs(data_sub.loc['AUDIT_Primary_Length'] - data_sub.loc[:, 'PKG_SUB_Length']) > threshold)

I am thinking that True = 1 and False = 0 by default, if I convert it into integers, right? (thought I read somewhere saying this...)

And here's the warning that I keep getting:

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

I read into:

How to deal with SettingWithCopyWarning in Pandas?

Correct way to set value on a slice in pandas

Pandas SettingWithCopyWarning

But I don't think they do what I am looking for. Anyone has any advice? I know this question may sound painfully stupid, but still appreciate any help!

Edit I've added the 2 lines of code where I created the data_sub. Hope that helps!

最满意答案

错误出现在上面的代码中，当您尝试提取某些数据帧子切片而没有创建完整副本时，因此您拥有的引用实际上是对另一个较大数据帧的一部分的引用。

没有太多关于你想要做什么的背景，只需事先做一个副本：

data_sub = data_sub.copy() data_sub['PackageLengthFlag'] = ((data_sub['AUDIT_Primary_Length'] \ - data_sub['PKG_SUB_Length']).abs() > threshold).astype(int)

调用df.abs对整个结果执行abs函数。 abs是一种不能处理pd.Series对象的香草python方法。

最后一个astype调用将结果转换为整数值。

---

以下是你正在做的一个例子：

df A_Key B_ID C_Key D_NA 0 123 22 343 23 1 121 23 45.4 52 x = df.iloc[[0], :] x A_Key B_ID C_Key D_NA 0 123 22 343 23 x.iloc[:, 0] += 2 /Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/site-packages/pandas/core/indexing.py:517: 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

您会看到生成的错误。 但在大多数情况下，原件仍应在不影响原件的情况下进行修改。 现在，先复制：

x = x.copy() x.iloc[:, 0] += 2 # no warning

而且你看到错误被抑制了。 有趣的是，当执行垂直子模式的类似操作时，不会看到相同的行为。 我相信熊猫通过返回完整的独立副本来巧妙地处理这个问题。

The error is in the code above this, when you try extracting some dataframe subslice without making a complete copy, so the reference you have is actually a reference to part of another larger dataframe.

Without much context on what you're trying to do, just make a copy beforehand:

data_sub = data_sub.copy() data_sub['PackageLengthFlag'] = ( data_sub['AUDIT_Primary_Length'] .sub(data_sub['PKG_SUB_Length']) .abs() .gt(threshold) .astype(int)

Call df.abs to perform the abs function on the entire result. abs is a vanilla python method that cannot handle pd.Series objects.

One final astype call converts the result to integral values.

---

Here's an example of what you're doing:

df A_Key B_ID C_Key D_NA 0 123 22 343 23 1 121 23 45.4 52 x = df.iloc[[0], :] x A_Key B_ID C_Key D_NA 0 123 22 343 23 x.iloc[:, 0] += 2 /Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/site-packages/pandas/core/indexing.py:517: 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

You see the error generated. But in most cases, the original should still be modified without affecting the original. Now, copy first:

x = x.copy() x.iloc[:, 0] += 2 # no warning

And you see the error suppressed. Interestingly, the same behaviour is not seen when performing similar operations on vertical subslices. I believe pandas smartly handles this by returning a full independent copy.