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'Truth value of an array is ambiguous' error with np.fromfunction

How do I sort a dictionary by value?ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()The truth value of an array with more than one element is ambiguous error? pythonPython Pandas add column for row-wise max value of selected columnsThe truth value of an array with more than one element is ambiguous. Use a.any() or a.all()?Error: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all() while using csr_matrixValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all() using while loopif array in list of arrays returns Value ErrorTruth value of an array is ambiguous

0

I'm trying to use the numpy.fromfunction to compute an array defined by a function but I got an error that I do not understand.

d_matrix is a distance matrix and the error message I get is "The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()". I put dtype = int in the np.fromfunction because I read that could the solution.

def v(r, i):
 return 1/N*np.sum(d_matrix[i,:]<r)

def rho_barre(r):
 return quad(rho, r, np.inf)[0]

def grad_F(i, j):
 return quad( lambda r : ( (v(r, i) + v(r, j))/2 - v_r) * rho_barre(max(r, d_matrix[i,j])), 0, np.inf)[0]

Grad_F = np.fromfunction(lambda i, j: grad_F(i,j), (N,N), dtype=int)

I'd like to know if someone may help me with this error and more generally if someone has an idea of what to do in order to compute an array defined by a function. I'm not sure I'm doing the fastest thing

python numpy

share|improve this question

asked Mar 25 at 15:47

MarcMMarcM

2844 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  fromfunction passes the whole np.indices(N,N) to your function, at once. It doesn't do it iteratively. Look at the code for fromfunction. I'd suggest using a double loop over the range(N)s instead. Posters often misunderstand the use of fromfunction.
  
  – hpaulj
  Mar 25 at 16:33
  
  

  
  
  
  

add a comment | 

0

I'm trying to use the numpy.fromfunction to compute an array defined by a function but I got an error that I do not understand.

d_matrix is a distance matrix and the error message I get is "The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()". I put dtype = int in the np.fromfunction because I read that could the solution.

def v(r, i):
 return 1/N*np.sum(d_matrix[i,:]<r)

def rho_barre(r):
 return quad(rho, r, np.inf)[0]

def grad_F(i, j):
 return quad( lambda r : ( (v(r, i) + v(r, j))/2 - v_r) * rho_barre(max(r, d_matrix[i,j])), 0, np.inf)[0]

Grad_F = np.fromfunction(lambda i, j: grad_F(i,j), (N,N), dtype=int)

I'd like to know if someone may help me with this error and more generally if someone has an idea of what to do in order to compute an array defined by a function. I'm not sure I'm doing the fastest thing

python numpy

share|improve this question

asked Mar 25 at 15:47

MarcMMarcM

2844 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  fromfunction passes the whole np.indices(N,N) to your function, at once. It doesn't do it iteratively. Look at the code for fromfunction. I'd suggest using a double loop over the range(N)s instead. Posters often misunderstand the use of fromfunction.
  
  – hpaulj
  Mar 25 at 16:33
  
  

  
  
  
  

add a comment | 

I'm trying to use the numpy.fromfunction to compute an array defined by a function but I got an error that I do not understand.

d_matrix is a distance matrix and the error message I get is "The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()". I put dtype = int in the np.fromfunction because I read that could the solution.

def v(r, i):
 return 1/N*np.sum(d_matrix[i,:]<r)

def rho_barre(r):
 return quad(rho, r, np.inf)[0]

def grad_F(i, j):
 return quad( lambda r : ( (v(r, i) + v(r, j))/2 - v_r) * rho_barre(max(r, d_matrix[i,j])), 0, np.inf)[0]

Grad_F = np.fromfunction(lambda i, j: grad_F(i,j), (N,N), dtype=int)

I'd like to know if someone may help me with this error and more generally if someone has an idea of what to do in order to compute an array defined by a function. I'm not sure I'm doing the fastest thing

python numpy

share|improve this question

asked Mar 25 at 15:47

MarcMMarcM

2844 bronze badges

def v(r, i):
 return 1/N*np.sum(d_matrix[i,:]<r)

def rho_barre(r):
 return quad(rho, r, np.inf)[0]

def grad_F(i, j):
 return quad( lambda r : ( (v(r, i) + v(r, j))/2 - v_r) * rho_barre(max(r, d_matrix[i,j])), 0, np.inf)[0]

Grad_F = np.fromfunction(lambda i, j: grad_F(i,j), (N,N), dtype=int)

share|improve this question

asked Mar 25 at 15:47

MarcMMarcM

2844 bronze badges

share|improve this question

asked Mar 25 at 15:47

MarcMMarcM

2844 bronze badges

asked Mar 25 at 15:47

MarcMMarcM

2844 bronze badges

asked Mar 25 at 15:47

MarcMMarcM

2844 bronze badges

1 Answer
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1

As pointed out in the comments, np.fromfunction provides arrays of indices, not individual index tuples. This is a common mistake, but working with index arrays is usually more efficient. If you really must produce a single value at a time, you could use a function like this instead:

import numpy as np

def fromfunction_iter(function, shape, dtype=None):
 # Iterator over all index tuples
 iter = np.ndindex(*shape)
 # First index
 idx = next(iter)
 # Produce first value
 value = function(*idx)
 # Make it into a NumPy value
 value = np.asarray(value, dtype=dtype)
 # Make output array of the right data type
 out = np.empty(shape, dtype=value.dtype)
 # Set first value
 out[idx] = value
 # Fill rest of values
 for idx in iter:
 out[idx] = function(*idx)
 return out

However, this will generally be much slower, and in fact if you have to run some iterative algorithm like this with NumPy data you may need to look into something like Numba if you want to make it run really fast.

share|improve this answer

answered Mar 25 at 18:14

jdehesajdehesa

31.2k44 gold badges3939 silver badges6060 bronze badges

add a comment | 

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1

As pointed out in the comments, np.fromfunction provides arrays of indices, not individual index tuples. This is a common mistake, but working with index arrays is usually more efficient. If you really must produce a single value at a time, you could use a function like this instead:

import numpy as np

def fromfunction_iter(function, shape, dtype=None):
 # Iterator over all index tuples
 iter = np.ndindex(*shape)
 # First index
 idx = next(iter)
 # Produce first value
 value = function(*idx)
 # Make it into a NumPy value
 value = np.asarray(value, dtype=dtype)
 # Make output array of the right data type
 out = np.empty(shape, dtype=value.dtype)
 # Set first value
 out[idx] = value
 # Fill rest of values
 for idx in iter:
 out[idx] = function(*idx)
 return out

However, this will generally be much slower, and in fact if you have to run some iterative algorithm like this with NumPy data you may need to look into something like Numba if you want to make it run really fast.

share|improve this answer

answered Mar 25 at 18:14

jdehesajdehesa

31.2k44 gold badges3939 silver badges6060 bronze badges

add a comment | 

1

As pointed out in the comments, np.fromfunction provides arrays of indices, not individual index tuples. This is a common mistake, but working with index arrays is usually more efficient. If you really must produce a single value at a time, you could use a function like this instead:

import numpy as np

def fromfunction_iter(function, shape, dtype=None):
 # Iterator over all index tuples
 iter = np.ndindex(*shape)
 # First index
 idx = next(iter)
 # Produce first value
 value = function(*idx)
 # Make it into a NumPy value
 value = np.asarray(value, dtype=dtype)
 # Make output array of the right data type
 out = np.empty(shape, dtype=value.dtype)
 # Set first value
 out[idx] = value
 # Fill rest of values
 for idx in iter:
 out[idx] = function(*idx)
 return out

However, this will generally be much slower, and in fact if you have to run some iterative algorithm like this with NumPy data you may need to look into something like Numba if you want to make it run really fast.

share|improve this answer

answered Mar 25 at 18:14

jdehesajdehesa

31.2k44 gold badges3939 silver badges6060 bronze badges

add a comment | 

As pointed out in the comments, np.fromfunction provides arrays of indices, not individual index tuples. This is a common mistake, but working with index arrays is usually more efficient. If you really must produce a single value at a time, you could use a function like this instead:

import numpy as np

def fromfunction_iter(function, shape, dtype=None):
 # Iterator over all index tuples
 iter = np.ndindex(*shape)
 # First index
 idx = next(iter)
 # Produce first value
 value = function(*idx)
 # Make it into a NumPy value
 value = np.asarray(value, dtype=dtype)
 # Make output array of the right data type
 out = np.empty(shape, dtype=value.dtype)
 # Set first value
 out[idx] = value
 # Fill rest of values
 for idx in iter:
 out[idx] = function(*idx)
 return out

However, this will generally be much slower, and in fact if you have to run some iterative algorithm like this with NumPy data you may need to look into something like Numba if you want to make it run really fast.

share|improve this answer

answered Mar 25 at 18:14

jdehesajdehesa

31.2k44 gold badges3939 silver badges6060 bronze badges

import numpy as np

def fromfunction_iter(function, shape, dtype=None):
 # Iterator over all index tuples
 iter = np.ndindex(*shape)
 # First index
 idx = next(iter)
 # Produce first value
 value = function(*idx)
 # Make it into a NumPy value
 value = np.asarray(value, dtype=dtype)
 # Make output array of the right data type
 out = np.empty(shape, dtype=value.dtype)
 # Set first value
 out[idx] = value
 # Fill rest of values
 for idx in iter:
 out[idx] = function(*idx)
 return out

share|improve this answer

answered Mar 25 at 18:14

jdehesajdehesa

31.2k44 gold badges3939 silver badges6060 bronze badges

answered Mar 25 at 18:14

jdehesajdehesa

31.2k44 gold badges3939 silver badges6060 bronze badges

answered Mar 25 at 18:14

jdehesajdehesa

31.2k44 gold badges3939 silver badges6060 bronze badges