在pycharm中使用Jupyter notebook命令时出现Key error 5问题

Question

我偶然发现了一个我不理解或找不到解决方案的问题。

我键入的代码在我的 Jupyter Notebook 中运行完美，但在我使用 PyCharm 时出现错误。

我在 PyCharm 中输入了这些代码：

df_ticker.loc[df_ticker['Red'] == True, 'Value'] = df_ticker['Open'] - df_ticker['Close']

df_ticker.loc[df_ticker['Red'] == False, 'Value'] = df_ticker['Close'] - df_ticker['Open']

单独使用这些代码并打印它们的值是成功的。H

但是，当与这些代码行匹配时：

for i in range(1, len(df_ticker)):

if df_ticker.Close[i] > df_ticker.Close[i-1] and df_ticker.Red[i] == False and \

        df_ticker.Value.round(decimals = 0)[i] >= df_ticker.Value.round(decimals = 0)[i-1] \

        and not df_ticker.Close[i-1] >= df_ticker [i-2]:

    rise.append(df_ticker.Close[i])

它返回此错误：

Traceback (most recent call last):

  File "/Users/benjaminwong/PycharmProjects/Yahoo Finance project/venv/lib/python3.8/site-packages/pandas/core/indexes/base.py", line 2889, in get_loc

    return self._engine.get_loc(casted_key)

  File "pandas/_libs/index.pyx", line 70, in pandas._libs.index.IndexEngine.get_loc

  File "pandas/_libs/index.pyx", line 97, in pandas._libs.index.IndexEngine.get_loc

  File "pandas/_libs/hashtable_class_helper.pxi", line 1675, in pandas._libs.hashtable.PyObjectHashTable.get_item

  File "pandas/_libs/hashtable_class_helper.pxi", line 1683, in pandas._libs.hashtable.PyObjectHashTable.get_item

KeyError: 5

The above exception was the direct cause of the following exception:

Traceback (most recent call last):

  File "/Users/benjaminwong/PycharmProjects/Yahoo Finance project/Rise_Finder 2.py", line 29, in <module>

    and not df_ticker.Close[i-1] >= df_ticker [i-2]:

  File "/Users/benjaminwong/PycharmProjects/Yahoo Finance project/venv/lib/python3.8/site-packages/pandas/core/frame.py", line 2899, in __getitem__

    indexer = self.columns.get_loc(key)

  File "/Users/benjaminwong/PycharmProjects/Yahoo Finance project/venv/lib/python3.8/site-packages/pandas/core/indexes/base.py", line 2891, in get_loc

    raise KeyError(key) from err

KeyError: 5

请帮忙，谢谢你的时间:)

Answer 1

and&nbsp;not&nbsp;df_ticker.Close[i-1]&nbsp;>=&nbsp;df_ticker&nbsp;[i-2]:和KeyError:&nbsp;5df_ticker.Close[&nbsp;5&nbsp;] 或 df_ticker [&nbsp;5&nbsp;] 不存在。

Answer 2

它应该是：df_ticker.Close[i]&nbsp;>&nbsp;df_ticker.Close[i-2]我错过了第二次关闭。