根据索引在熊猫数据框中查找特定列的值[重复]
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【中文标题】根据索引在熊猫数据框中查找特定列的值[重复]【英文标题】:Finding values in pandas dataframe for a particular column based on index [duplicate] 【发布时间】:2019-08-05 01:16:51 【问题描述】:我正在使用 Python - 3.6 和 pandas - 0.24.1
我有一个熊猫数据框df1
:
col1 col2
0 8388611 3.9386
1 8388612 1.9386
我需要在特定索引上找到col1
的值
print(df1['col1'][1])
错误:
Traceback (most recent call last):
File "/home/runner/.site-packages/pandas/core/indexes/base.py", line 2656, inget_loc
return self._engine.get_loc(key)
File "pandas/_libs/index.pyx", line 108, in pandas._libs.index.IndexEngine.get_loc
File "pandas/_libs/index.pyx", line 132, in pandas._libs.index.IndexEngine.get_loc
File "pandas/_libs/hashtable_class_helper.pxi", line 1601, in pandas._libs.hashtable.PyObjectHashTable.get_item
File "pandas/_libs/hashtable_class_helper.pxi", line 1608, in pandas._libs.hashtable.PyObjectHashTable.get_item
KeyError: 1
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "main.py", line 24, in <module>
print(df1['col1'][1])
File "/home/runner/.site-packages/pandas/core/frame.py", line 2927, in __getitem__
indexer = self.columns.get_loc(key)
File "/home/runner/.site-packages/pandas/core/indexes/multi.py", line 2397, in get_loc
loc = self._get_level_indexer(key, level=0)
File "/home/runner/.site-packages/pandas/core/indexes/multi.py", line 2652, in _get_level_indexer
code = level_index.get_loc(key)
File "/home/runner/.site-packages/pandas/core/indexes/base.py", line 2658, inget_loc
return self._engine.get_loc(self._maybe_cast_indexer(key))
File "pandas/_libs/index.pyx", line 108, in pandas._libs.index.IndexEngine.get_loc
File "pandas/_libs/index.pyx", line 132, in pandas._libs.index.IndexEngine.get_loc
File "pandas/_libs/hashtable_class_helper.pxi", line 1601, in pandas._libs.hashtable.PyObjectHashTable.get_item
File "pandas/_libs/hashtable_class_helper.pxi", line 1608, in pandas._libs.hashtable.PyObjectHashTable.get_item
KeyError: 1
在 pandas 中迭代特定列的所有元素的理想方法是什么?
【问题讨论】:
使用print(df1.loc[1, 'col1'])
,是骗子
【参考方案1】:
你可以使用 iloc
df["column"].iloc[index]
【讨论】:
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