复杂曲线方程式在np.where使用中给出错误 [英] Complex curve equation giving error in np.where usage
问题描述
我已经查看了其他有关Operand错误的答案，但似乎没有一个适合这个示例.数学/方程式可以用X值编码，也可以从DataFrame导入.在np.where表达式中使用相同的方程会导致操作数错误.
I have looked through the other answers for Operand errors and none seem to fit this example. The mathematics/equation works, either coding in X values or importing from the DataFrame. Using the same equation in an np.where expression causes the operand error.
import csv
import pandas as pd
from pandas import DataFrame
import numpy as np
data= pd.read_csv('miniDF.csv')
df=pd.DataFrame(data, columns=['X','Z'])
df['y']=df['Z']*0.01
df['y']=(14.6413819224756*(df['X']**0.5)+64.4092780704338*(np.log(df['X'])**-2)
+1675.7498523727*(np.exp(-df['X']))+3.07221083927051*np.cos(df['X']))
print(df)
df['y']=np.where(df['Z']>=(14.6413819224756*(df['X']**0.5)+64.4092780704338*(np.log(df['X'])**-2)
+1675.7498523727*(np.exp(-df['X']))+3.07221083927051*np.cos(df['X']),8,9))
print(df)
我的数据框中的值，第一个print(df)的输出和错误如下.
The values in my Dataframe, the output from the first print(df) and the error are as follows.
X Z y
0 1.4 1 999.999293
1 2.0 2000 380.275104
2 3.0 3 159.114194
3 4.0 4 91.481930
4 5.0 5 69.767368
5 6.0 6 63.030212
6 7.0 70 59.591631
7 8.0 8 56.422723
8 9.0 9 54.673108
9 10.0 10 55.946732
Traceback (most recent call last):
File "/Users/willhutchins/Desktop/minitest.py", line 17, in <module>
df['y']=np.where(df['Z']>=(14.6413819224756*(df['X']**0.5)+64.4092780704338*(np.log(df['X'])**-2)
File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/pandas/core/ops/__init__.py", line 1229, in wrapper
res = na_op(values, other)
File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/pandas/core/ops/__init__.py", line 1115, in na_op
result = method(y)
ValueError: operands could not be broadcast together with shapes (10,) (3,)
为什么方程式可以作为独立项工作，而在np.where中使用时却不能工作?
Why does the equation work as a stand-alone item but not work when used in np.where?
推荐答案
让我们
expr = 14.6413819224756*(df['X']**0.5)+64.4092780704338*(np.log(df['X'])**-2)+1675.7498523727*(np.exp(-df['X']))+3.07221083927051*np.cos(df['X'])
然后您会发现您的代码是:
then you'll find out that your code is:
df['y']=np.where(df['Z']>=(expr,8,9))
df['Z']的shape是(10,)，这意味着它是具有10行的一维pandas.Series对象.但是，(expr,8,9)是一个简单的tuple，其中包含3个项目(但是expr的确是10行pandas.Series).
The shape of df['Z'] is (10,), which means it is a one-dimensional pandas.Series object which has 10 rows. However, (expr,8,9) is a simple tuple which has 3 items (expr is indeed a 10-row pandas.Series however).
这就是提示为operands could not be broadcast together with shapes (10,) (3,) 的原因，因为numpy不知道如何将 10 行pandas.Series与 3 项目tuple.
That's why the hint is operands could not be broadcast together with shapes (10,) (3,) , since numpy doesn't know how to compare a 10-row pandas.Series with a 3-item tuple.
再次检查您的方程式，并对其进行修改以满足您的需求.
Check your equation again and get it modified to meet your needs.
更新:
根据注释，8和9是np.where(condition,x,y)的两个参数，分别是x和y.但是您错误地将它们放在df['Z']>=之后的expr中，这使得>=运算符将pandas.Series与tuple而不是两个pandas.Series进行比较.
According to the comment, the 8 and 9 are two arguments to np.where(condition,x,y) as the x and y. But you put them in the expr after df['Z']>= by mistake, which makes the >= operator compares a pandas.Series's with a tuple, but not two pandas.Series.
只需移动最后一个括号即可，代码将运行良好:
Just move the last parentheses and the code will work well:
df['y']=np.where(df['Z']>=(14.6413819224756*(df['X']**0.5)+64.4092780704338*(np.log(df['X'])**-2)
+1675.7498523727*(np.exp(-df['X']))+3.07221083927051*np.cos(df['X'])),8,9)
结果应为:
X Z y
0 1.4 1 9
1 2.0 2000 8
2 3.0 3 9
3 4.0 4 9
4 5.0 5 9
5 6.0 6 9
6 7.0 70 8
7 8.0 8 9
8 9.0 9 9
9 10.0 10 9
更新2:
要在满足两个条件的情况下执行np.where，或者说要执行and，只需使用np.where((condition1) & (conditions),x,y).例如:
To do np.where while two conditions are met, or to say, an and operation, just use np.where((condition1) & (conditions),x,y). For example:
df['foo']=np.where((df['Z']>3) & (df['Z']<100),True,False)
请注意，在&之前和之后的括号是必需的.您将通过数据获得此信息:
Note, the parentheses here before and after & is necessary. You'll get this with your data:
X Z y foo
0 1.4 1 999.999293 False
1 2.0 2000 380.275104 False
2 3.0 3 159.114194 False
3 4.0 4 91.481930 True
4 5.0 5 69.767368 True
5 6.0 6 63.030212 True
6 7.0 70 59.591631 True
7 8.0 8 56.422723 True
8 9.0 9 54.673108 True
9 10.0 10 55.946732 True
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