使用包含空格的列名查询 Pandas DataFrame 或使用包含空格的列名使用 drop 方法
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Querying Pandas DataFrame with column name that contains a space or using the drop method with a column name that contains a space
提问by iNoob
I am looking to use pandasto drop rows based on the column name (contains a space) and the cell value. I have tried various ways to achieve this (drop and query methods) but it seems I'm failing due to the space in the name. Is there a way to query the data using the name that has a space in it or do I need to clean all spaces first?
我希望使用pandas基于列名（包含空格）和单元格值来删除行。我尝试了各种方法来实现这一点（删除和查询方法），但由于名称中的空格，我似乎失败了。有没有办法使用其中有空格的名称查询数据，或者我是否需要先清理所有空格？
data in form of a csv file
csv文件形式的数据
Date,"price","Sale Item"
2012-06-11,1600.20,item1
2012-06-12,1610.02,item2
2012-06-13,1618.07,item3
2012-06-14,1624.40,item4
2012-06-15,1626.15,item5
2012-06-16,1626.15,item6
2012-06-17,1626.15,item7
Attempt Examples
尝试示例
df.drop(['Sale Item'] != 'Item1')
df.drop('Sale Item' != 'Item1')
df.drop("'Sale Item'] != 'Item1'")
df.query('Sale Item' != 'Item1')
df.query(['Sale Item'] != 'Item1')
df.query("'Sale Item'] != 'Item1'")
Error received in most cases
大多数情况下收到错误
ImportError: 'numexpr' not found. Cannot use engine='numexpr' for query/eval if 'numexpr' is not installed
回答by Fabio Lamanna
If I understood correctly your issue, maybe you can just apply a filter like:
如果我正确理解了您的问题，也许您可以应用以下过滤器：
df = df[df['Sale Item'] != 'item1']
which returns:
返回：
Date price Sale Item
1 2012-06-12 1610.02 item2
2 2012-06-13 1618.07 item3
3 2012-06-14 1624.40 item4
4 2012-06-15 1626.15 item5
5 2012-06-16 1626.15 item6
6 2012-06-17 1626.15 item7
回答by Anand S Kumar
As you can see from the documentation-
DataFrame.drop(labels, axis=0, level=None, inplace=False, errors='raise')
Return new object with labels in requested axis removed
DataFrame.drop(labels, axis=0, level=None, inplace=False, errors='raise')
返回移除了请求轴中标签的新对象
DataFrame.drop()takes the indexof the rows to drop, not the condition. Hence you would most probably need something like -
DataFrame.drop()需要index删除的行，而不是条件。因此，您很可能需要类似的东西 -
df.drop(df.ix[df['Sale Item'] != 'item1'].index)
Please note, this drops the rows that meet the condition, so the result would be the rows that don't meet the condition, if you want the opposite you can use ~operator before your condition to negate it.
请注意，这会删除满足条件的行，因此结果将是不满足条件的行，如果您想要相反的行，您可以~在条件之前使用运算符来否定它。
But this seems a bit too much, it would be easier to just use Boolean indexing to get the rows you want (as indicated in the other answer) .
但这似乎有点太多了，仅使用布尔索引来获取所需的行会更容易（如其他答案中所示）。
Demo -
演示 -
In [20]: df
Out[20]:
Date price Sale Item
0 2012-06-11 1600.20 item1
1 2012-06-12 1610.02 item2
2 2012-06-13 1618.07 item3
3 2012-06-14 1624.40 item4
4 2012-06-15 1626.15 item5
5 2012-06-16 1626.15 item6
6 2012-06-17 1626.15 item7
In [21]: df.drop(df.ix[df['Sale Item'] != 'item1'].index)
Out[21]:
Date price Sale Item
0 2012-06-11 1600.2 item1
