如何在计算后添加新列
如何在计算后添加新列
提问于 2018-07-30 05:32:11
我有这个DataFrame：
Open High Low Close AdjClose Volume
datetime
2018-07-27 28.8200 29.3350 27.7050 28.1300 28.1300 8101362
2018-07-26 28.4800 29.1200 27.5500 28.9800 28.9800 10582061
2018-07-25 30.0900 30.0900 28.4800 28.9800 28.9800 9801897
2018-07-24 30.4500 30.9400 29.9650 30.1400 30.1400 5706941
2018-07-23 31.1100 31.3500 30.6000 30.8200 30.8200 6023310 我想在下面的for循环中计算两个新列(我使用的是python 2.7.15，而且我不是程序员)。这是我的循环：
A=pd.DataFrame(df)
print (A.head(5))
Len=len(A)
print (Len)
for Rw in range(Len-1):
def adj(A):
BB=float(A.iloc[Rw,4])
CC=float(A.iloc[Rw+1,4])
#print (BB,CC)
if BB>CC:
x1=BB-CC
x2=0
print(x1,x2)
else:
x1=0
x2=CC-BB
print(x1,x2)
A.at[Rw+1,'Up']=x1 #Store calculation results in Up column
A.at[Rw+1,'Down']=x2 #Store calculation results in Down column
adj(A)
print(A.head(5))当我启用==> A.atRw+1，'up'=x1时，我得到错误：
Traceback (most recent call last):
File "D:\Yossi\eclipse-workspace_yossi\Tests\to_forum___003.py", line 85, in <module>
adj(A)
File "D:\Yossi\eclipse-workspace_yossi\Tests\to_forum___003.py", line 82, in adj
A.at[Rw+1,'Up']=x1 #Store calculation results in UpMv columns
File "C:\Python27\lib\site-packages\pandas\core\indexing.py", line 2159, in __setitem__
self.obj._set_value(*key, takeable=self._takeable)
File "C:\Python27\lib\site-packages\pandas\core\frame.py", line 2587, in _set_value
self.loc[index, col] = value
File "C:\Python27\lib\site-packages\pandas\core\indexing.py", line 189, in __setitem__
self._setitem_with_indexer(indexer, value)
File "C:\Python27\lib\site-packages\pandas\core\indexing.py", line 375, in _setitem_with_indexer
labels = index.insert(len(index), key)
File "C:\Python27\lib\site-packages\pandas\core\indexes\datetimes.py", line 2194, in insert
"cannot insert DatetimeIndex with incompatible label")
TypeError: cannot insert DatetimeIndex with incompatible label如何使用DataFrame查看'Up‘列的x1计算结果和'Down’列的x2计算结果，如下图所示？
谢谢你们所有人。
回答 1
Stack Overflow用户
回答已采纳
发布于 2018-07-30 05:43:31
从在循环中定义函数开始，原始代码中有很多地方是错误的。这是一个紧凑、高效、基于pandas/numpy的解决方案：
import numpy as np
difference = df['Close'] - df['Close'].shift()
df['Up'] = -np.minimum(difference, 0)
df['Down'] = np.maximum(difference, 0)页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/51584734
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