Error strtrim s argument must be a string or cellstring

Удаление ведущего пробела из вектора символа

Создайте вектор символов с пробелами и символом табуляции как ведущий пробел.

chr = sprintf('  t   Remove    leading whitespace')

chr = 
'  	   Remove    leading whitespace'

Удалите начальные табуляцию и пробелы.

newChr = 
'Remove    leading whitespace'

strtrim удаляет ведущие пробельные символы, но не пробел между другими символами.

Удаление начальных и конечных пробелов из массива строк

Начиная в R2017a, можно создать строки с помощью двойных кавычек. Создайте массив строк и удалите начальный и конечный пробел с strtrim функция.

str = ["   Gemini    ","   Apollo    ";
       "   ISS       ","   Skylab    "]

str = 2x2 string
    "   Gemini    "    "   Apollo    "
    "   ISS       "    "   Skylab    "

newStr = 2x2 string
    "Gemini"    "Apollo"
    "ISS"       "Skylab"

Удаление ведущий и запаздывание пробела от массива ячеек

Удалите начальный и конечный пробел из всех векторов символов в массиве ячеек и отобразите их.

chr = {'     Trim leading whitespace';
       'Trim trailing whitespace     '}

chr = 2x1 cell
    {'     Trim leading whitespace' }
    {'Trim trailing whitespace     '}

newChr = 2x1 cell
    {'Trim leading whitespace' }
    {'Trim trailing whitespace'}

Хранение символа неразрывного пробела

Создайте вектор символов, который включает символ неразрывного пробела, char(160), как конечный пробельный символ.

chr = '     Keep nonbreaking space';
chr = [chr char(160) '     '];

Отобразите chr между | символы, чтобы показать начальный и конечный пробел.

ans = 
'|     Keep nonbreaking space |'

Удалите начальные и конечные пробельные символы.

Отобразите newChr между | символы. strtrim удаляет пробелы, но оставляет неразрывный пробел в конце newChr.

ans = 
'|Keep nonbreaking space |'

str — Введите текст
символьный массив | массив ячеек символьных массивов | массив строк

Введите текст в виде символьного массива или как массив ячеек символьных массивов или массив строк.

• Входной текст должен быть строковым скаляром или символьным массивом.

• Входные значения должны быть в области значений 0–127.

• Сгенерированный код возвращает пустой выходной параметр как символьный массив 1 на 0.

• str должен быть массив строк или массив ячеек из символьных векторов.

import autograd.numpy as np
from autograd import grad

def f(A):

    B = np.zeros((4,4))

    B[:2, :2] = A

    return B.sum()

A = np.random.randn(2,2)

df = grad(f)

print df(A)

# expected: [[ 1.,  1.],[ 1.,  1.]]

Traceback (most recent call last):
  File "/Users/tristanhearn/Dropbox/code/adcomponent/src/adcomponent/test.py", line 17, in <module>
    print df(A)
  File "/Users/tristanhearn/Documents/thearn_repos/autograd/autograd/core.py", line 20, in gradfun
    return backward_pass(*forward_pass(fun,args,kwargs,argnum))
  File "/Users/tristanhearn/Documents/thearn_repos/autograd/autograd/core.py", line 61, in forward_pass
    end_node = fun(*args, **kwargs)
  File "/Users/tristanhearn/Dropbox/code/adcomponent/src/adcomponent/test.py", line 9, in f
    B[:2, :2] = A
TypeError: float() argument must be a string or a number

Thanks for the encouragement, and for making a minimal working example!

This is indeed a known limitation of autograd — indexed assignment isn’t yet supported, unfortunately. You can usually get around this by building a list and calling np.concatenate().

We’ll add a try/catch block to give a more informative error message when this occurs.

Also, we’d love to hear about what you’re using autograd for, and if you have any other feature requests.

Also, we’d love to hear about what you’re using autograd for, and if you have any other feature requests.

I’m writing an auto-differentiating component for NASA’s OpenMDAO framework. This will make it much easier to quickly & automatically provide numerical derivates across coupled engineering codes (without the potential instability of finite-difference approximations), which is very important for numerical optimization or sensitivity analysis. There are other alternatives as well (complex-step approximation, or closed-form specification of derivates of course), but AD has been on our radar for awhile.

Unfortunately, some of our disciplinary analysis codes (engine cycle analysis in particular) do use indexed assignment in a few places, so my AD component isn’t quite as general as I had hoped at the moment. But not all of our components require this.

Also, looking through the source, it looks like you guys have implemented a jacobian function (which I did as well). I wish I had noticed that before!

Again, awesome library 👍

Wow, sounds very cool! I hope autograd continues to help. Also, the more reasons we have to make indexed assignment work, the more impetus we’ll have to figure out an implementation.

By the way, there are currently two jacobian functions, one in core.py and another in convenience_wrappers.py. The one in core.py is better in at least three ways: it’s faster (only one forward pass is done), it avoids repeated prints or side-effects (also because of the single forward pass), and it’s more general (it can take jacobians of jacobians, which I think the wrapper version can’t do). I’ll probably delete the version in convenience_wrappers.py sometime today (unless someone protests).

Another by the way: on the current master branch, the error message produced by the code in your original test case should be somewhat clearer!

[... snip traceback ...]

/Users/mattjj/packages/autograd/issue58.py in f(x)
      6 def f(x):
      7     A = np.zeros((4,4))
----> 8     A[:2,:2] = x
      9     return A.sum()
     10

AutogradException: autograd doesn't support assigning into arrays
Sub-exception:
TypeError: float() argument must be a string or a number

@duvenaud what does it mean by «building a list and calling np.concatenate()»?

In the case of the OP’s code, I think he means rewriting lines like this:

B = np.zeros((4,4))
B[:2, :2] = A

as something like this:

row1 = np.concatenate([A, np.zeros((2,2))], axis=1)
row2 = np.zeros((2, 4))
B = np.concatenate([row1, row2], axis=0)

In general, instead of building arrays by allocating zeros and then assigning into blocks, you can instead build the blocks as separate arrays and then concatenate them. The latter method works with autograd, but the former method doesn’t.

@mattjj Can I give B a correct data type such that it can hold the autograd nodes? (Or is data type the cause of issue?)

No, autograd doesn’t support assignment into arrays. We could overload indexed assignment, but it would require a substantial amount of internal bookkeeping, and it might even obfuscate what happens when the program runs. In particular, to do reverse-mode autodiff, the intermediate values computed during the evaluation of the function need to be stored because they usually need to be read during the backward pass computation. Since assignment into arrays might clobber intermediate values, whenever an assignment happens we’d need to copy that data somewhere for use in the backward pass. Instead, by not supporting assignment, we basically force the user to be explicit about when this kind of copying happens.

If you need to use assignment, one thing you can do is mark your function as a primitive and then define its vjp manually. Within a primitive you’re just using naked numpy, so assignment works. In the case of the OP’s code, which might be too simplified to be interesting, that might look something like

import autograd.numpy as np
from autograd import grad
from autograd.core import primitive

@primitive
def f(A):
    B = np.zeros((4, 4))
    B[:2, :2] = A
    return B.sum()

def f_vjp(g, ans, vs, gvs, A):
    return g * np.ones((2, 2))

f.defvjp(f_vjp)

Dear Sir ,
Thank you for your efforts,
Please this is my part of code , when I try to run these lines , this error appears please can you kindly help me to solve it please ?

def interpolation(noisy , SNR , Number_of_pilot , interp):

noisy_image = np.zeros((40000,72,14,2))



noisy_image[:,:,:,0] = np.real(noisy)

noisy_image[:,:,:,1] = np.imag(noisy)

TypeError: float() argument must be a string or a number, not ‘dict’