pandas :类型错误:'>'在日期列上选择时，“int"和“str"实例之间不支持 [英] Pandas: TypeError: '>' not supported between instances of 'int' and 'str' when selecting on date column

问题描述

我有一个带有时间戳列的 Pandas DataFrame.我可以从此列中选择日期范围.但是在我对 DataFrame 中的其他列进行更改后，我无法再收到错误消息TypeError: '>' not supported between 'int' and 'str'".

I have a Pandas DataFrame with a column with TimeStamps. I can select date ranges from this column. But after I make change to other columns in the DataFrame, I can no longer and I get the error "TypeError: '>' not supported between instances of 'int' and 'str'".

以下代码重现问题:

• 用一些随机数生成一个 DataFrame
• 添加带有日期的列

• 在日期列上选择

• Generate a DataFrame with some random numbers
• Add a column with dates

• Select on the date column

df = pd.DataFrame(np.random.random((200,3)))
df['date'] = pd.date_range('2000-1-1', periods=200, freq='D')
mask = (df['date'] > '2000-6-1') & (df['date'] <= '2000-6-10')
print(df.loc[mask])

一切顺利:

            0         1         2       date
153  0.280575  0.810817  0.534509 2000-06-02
154  0.490319  0.873906  0.465698 2000-06-03
155  0.070790  0.898340  0.390777 2000-06-04
156  0.896007  0.824134  0.134484 2000-06-05
157  0.539633  0.814883  0.976257 2000-06-06
158  0.772454  0.420732  0.499719 2000-06-07
159  0.498020  0.495946  0.546043 2000-06-08
160  0.562385  0.460190  0.480170 2000-06-09
161  0.924412  0.611929  0.459360 2000-06-10

但是，现在我将第 0 列设置为 0，如果它超过 0.7 并重复:

However, now I set column 0 to 0 if it exceeds 0.7 and repeat:

df[df[0] > 0.7] = 0
mask = (df['date'] > '2000-6-1') & (df['date'] <= '2000-6-10')

这给出了错误:

TypeError: '>' not supported between instances of 'int' and 'str'

为什么会发生这种情况，我该如何避免?

Why does this happen and how do I avoid it?

推荐答案

您可以将时间戳 (Timestamp('2000-01-01 00:00:00')) 与字符串进行比较，pandas 会为您将字符串转换为 Timestamp.但是一旦将值设置为 0，就无法将 int 与 str 进行比较.

You can compare a timestamp (Timestamp('2000-01-01 00:00:00')) to a string, pandas will convert the string to Timestamp for you. But once you set the value to 0, you cannot compare an int to a str.

解决此问题的另一种方法是更改​​操作顺序.

Another way to go around this is to change order of your operations.

filters = df[0] > 0.7
mask = (df['date'] > '2000-6-1') & (df['date'] <= '2000-6-10')

df[filters] = 0
print(df.loc[mask & filters])

此外，您提到您希望将第 0 列设置为 0，如果它超过 0.7，所以 df[df[0]>0.7] = 0 不这样做正是您想要的:它将整行设置为 0.相反:

Also, you mentioned you want to set column 0 to 0 if it exceeds 0.7, so df[df[0]>0.7] = 0 does not do exactly what you want: it sets the entire rows to 0. Instead:

df.loc[df[0] > 0.7, 0] = 0

那你用原来的掩码应该没有任何问题.

Then you should not have any problem with the original mask.

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