本文介绍了试图在来自 DF 的切片副本上设置值的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着跟版网的小编来一起学习吧！
问题描述
我正在用 pandas 和 python 做一些事情.我有下一个代码
Im doing some stuff with pandas and python. I have the next code
df = pd.read_csv("Request.csv", keep_default_na=False)
df1 = df.loc[(df["Request Status"] == "Closed")]
df1["Request Close-Down Actual"] = pd.to_datetime(df1["Request Close-Down Actual"], errors = 'coerce' )
df3 = df1.loc[(df1["Request Close-Down Actual"] < '2016-11-01') | (df1["Request Close-Down Actual"].isnull())]
df3.set_index("Request ID", inplace = True)
df3.to_csv("Request1.csv")
问题是当我运行代码时，我会收到下一个问题
The issue is when i run the code i receive the next issue
试图在数据帧的切片副本上设置值
A value is trying to be set on a copy of a slice from a DataFrame
df1.loc[请求关闭实际"] = pd.to_datetime(df1[请求Close-Down Actual"], errors = 'coerce')
df1.loc["Request Close-Down Actual"] = pd.to_datetime(df1["Request Close-Down Actual"], errors = 'coerce' )
请有人帮我解决这个问题.谢谢
Can someone give me a hand with this please. Thanks
推荐答案
我测试了它，对我来说效果很好.
I test it and for me it works nice.
问题应该在上面一行:
df1 = df.loc[(df["Request Status"] == "Closed")]
解决方案是 copy:
And solution is copy:
#loc is not necessary
df1 = df[df["Request Status"] == "Closed"].copy()
<小时>
错误显示 loc - 如果需要选择列，请尝试删除它:
Error show loc - try remove it if need select column:
df1.loc["Request Close-Down Actual"] = pd.to_datetime(df1["Request Close-Down Actual"], errors = 'coerce' )
到:
df1["Request Close-Down Actual"] = pd.to_datetime(df1["Request Close-Down Actual"], errors = 'coerce' )
这篇关于试图在来自 DF 的切片副本上设置值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！
本站部分内容来源互联网,如果有图片或者内容侵犯了您的权益，请联系我们，我们会在确认后第一时间进行删除！