pandas 使用混合列类型增加数据框的单元格值
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/19777751/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
提示：将鼠标放在中文语句上可以显示对应的英文。显示中英文
时间：2020-09-13 21:18:24 来源:igfitidea点击:
pandas increment cell value of dataframe with mixed column types
提问by will
I'd like to increment a cell of a dataframe:
我想增加数据帧的一个单元格：
from pandas import DataFrame
foo = DataFrame([[1,'a'],[2,'b'],[3,'c']],columns=['a','z'])
foo.ix[0,['a']] += 1
which gives the following error:
这给出了以下错误：
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-141-cf9b905bd544> in <module>()
1 foo = DataFrame([[1,'a'],[2,'b'],[3,'c']],columns=['a','z'])
----> 2 foo.ix[0,['a']] += 1
/home/ubuntu/anaconda/lib/python2.7/site-packages/pandas/core/indexing.pyc in __setitem__(self, key, value)
86 indexer = self._convert_to_indexer(key)
87
---> 88 self._setitem_with_indexer(indexer, value)
89
90 def _has_valid_tuple(self, key):
/home/ubuntu/anaconda/lib/python2.7/site-packages/pandas/core/indexing.pyc in _setitem_with_indexer(self, indexer, value)
156 # we have an equal len list/ndarray
157 elif len(labels) == 1 and (
--> 158 len(self.obj[labels[0]]) == len(value) or len(plane_indexer[0]) == len(value)):
159 setter(labels[0], value)
160
TypeError: object of type 'int' has no len()
Even though the following willwork:
即使以下内容有效：
foo = DataFrame([[1,4],[2,5],[3,6]],columns=['a','z'])
foo.ix[0,['a']] += 1
Which leads me to believe the issue is different column types.
这让我相信问题是不同的列类型。
How can I increment the cell value of the first dataframe?
如何增加第一个数据帧的单元格值？
回答by Jeff
In [1]: foo = DataFrame([[1,'a'],[2,'b'],[3,'c']],columns=['a','z'])
In [3]: foo
Out[3]:
a z
0 1 a
1 2 b
2 3 c
In [4]: foo.dtypes
Out[4]:
a int64
z object
dtype: object
In [5]: foo.ix[0,'a'] += 1
In [6]: foo
Out[6]:
a z
0 2 a
1 2 b
2 3 c
