If statement comparing numpy array raised ValueError

Question

I have two arrays. Both are 1D. However, I am getting the following Value Error. Below is what I tried.

R=np.arange(30,50,1)
T=np.arange(70,90,1)

H=[]

if (T > 8) and (R>10): 
   H.append(0.5 * (T + 61. + (T - 68.) * 1.2 + R * 0.094))
else:
   H.append(0 * 2)

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

Please include `T` and `R` so that your problem is reproducible — 7koFnMiP, Sep 01 '20 at 09:56
Have you seen this? [NumPy Error: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()](https://stackoverflow.com/questions/54191677/numpy-error-the-truth-value-of-an-array-with-more-than-one-element-is-ambiguous) — Carlo Zanocco, Sep 01 '20 at 10:04

Aviv Yaniv · Answer 1 · 2020-09-01T10:17:18.503

0

Python struggled to compare the values in the lists T to number.
The error is raised on T > 8.

For Lists

Use all() to check if all items in the list are above value:

all(x > value for x in list)

Use any() to check if any item in the list are above value:

any(x > value for x in list)

For NumPy Array

When having NumPy array, this can be shortened to:

T = np.array([1,2,3])
all(T > 8)

Fix

Fix to all:

H=[]

import numpy as np

T = np.array([30,34,56])
R = np.array([29,500,43])

if all(T > 8) and all(R > 10):
    H.append(0.5 * (T + 61. + (T - 68.) * 1.2 + R * 0.094))
else:
    H.append(0 * 2)

edited Sep 01 '20 at 10:17

answered Sep 01 '20 at 10:02

Aviv Yaniv

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3
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with numpy this can be shortened to just `(T > 8).all()`, etc. – Josef Sep 01 '20 at 10:07
I checked both T and R using `all(T > 80 for T in T)` and `all (T > 80 for T in T)`. I got `False` for both cases. – Chane Sep 01 '20 at 10:09
Then what you are looking for might by `.any()` – Josef Sep 01 '20 at 10:12
@Chane use: `all(T > 8) and all(R > 10)` to check all values or `any(T > 8) and any(R > 10)` to check if any of the values is above threshold – Aviv Yaniv Sep 01 '20 at 10:12

score 0 · Answer 2 · edited Sep 02 '20 at 04:42

0

Like the above answer states, you can use all or any. The fix using any is:

if any(t > value for t in T) and any(h > value for h in H):
   H.append(0.5 * (T + 61. + (T - 68.) * 1.2 + R * 0.094))
else:
   H.append(0 * 2)

The python interpreter needs 1 value at a time so that it can compare.

edited Sep 02 '20 at 04:42

David Duran

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answered Sep 01 '20 at 10:06

Vedant Matanhelia

117
1
10

score 0 · Answer 3 · answered Sep 01 '20 at 10:09

Simple as that.

import numpy as np
R=np.arange(30,50,1)
T=np.arange(70,90,1)

H=[]

if (T > 8).all() and (R>10).all():
   H.append(0.5 * (T + 61. + (T - 68.) * 1.2 + R * 0.094))
else:
   H.append(0 * 2)

print(H)

Here is an explanation.

If you really just want to know if atleast one element falls under the condition you should use any().

If statement comparing numpy array raised ValueError

3 Answers3