Python：使用循环语句将图像分配给变量

Question

I need to assign a image to three variables ie dsp1, dsp2, dsp3 by looping. On execution, I get a Syntax error.

SyntaxError: can't assign to operator.

for i in range(0,3):    
    dsp+str(i)=Image.open("L1.jpg")

What is the problem with 'str(i)' ?

Can any one explain with simple example ?

Answer 1

You can not assign to an operator.

Look at your code line:

 dsp + str(i)  =  Image.open("L1.jpg")

You have dsp + str(i) on the left side, an expression containing the sum operator + . Even if that would get evaluated properly, the result would be a string like "dsp1" for example. You can't assign any value to a string.

And because such uses make no sense, Python does not support operators on the left side of an assignment.

You want to dynamically create a variable name instead of hard-coding it. Although this is possible using exec() , that is strongly discouraged, as it easily leads to bugs in your code, is hard to read and even harder to debug. It may even be a security risk (code injection) if anything getting evaluated this way is untrusted data like user input.

What you should use instead is a simple list :

dsp = []
for i in range(0,3):    
    dsp[i] = Image.open("L1.jpg")  # for whatever reason you open the same file 3 times...

You create a list simply using the square brackets. If you want to initialize it with some values, simply write them inside, separated by commas:

my_list = ["zero", 1, 2, "three", 4.5, True]

You access elements of a list by specifying an index value, starting with 0 for the first element:

print(my_list[3])  # output: three

You can also easily loop over all elements of a list:

for item in my_list:
    print(item)
# output:
# zero
# 1
# 2
# three
# 4.5
# True

Answer 2

Instead of generating dynamic variables, place these images in a list:

images = []
for i in range(3):
    images[i] = Image.open("L1.jpg")

Using this method, the L1.jpg is assigned to the following:

images[0]
images[1]
images[2]

---

Alternatively, you can use a dictionary to get closer to the variable name format you are using:

images = {}
for i in range(3):
    images['dsp' + str(i)] = Image.open("L1.jpg")

This produces a dictionary that has the following layout:

{
'dsp2': <image object>, 
'dsp1': <image object>, 
'dsp0': <image object>
}

You can access any of these images by using the key (ie. image['dsp1'] )

In both of these cases, you don't need to worry about dynamic variables. Instead, everything you will be using sits in either a single list or dictionary.

Answer 3

You can't just string text together to create a variable name like that unfortunately. The problem is with dsp+str(i)= , not just str(i) .

If you absolutely must do it this way, you can do it using globals like so:

for i in range(0,3):    
    globals()["dsp" + str(i)] = Image.open("L1.jpg")

print(dsp1) # This will contain something

This will allow you to access those variables as if you had created them the 'normal' way first.

Ideally though, you should probably assign your results to a list instead, rather than discrete variable names.

Answer 4

The problem is that you're trying to generate variable names on the fly. If I were you I would try to use a dictionary instead of generation dynamic variable names. If you REALLY need dynamic variable names I would go for something like this:

exec("dsp%d = Image.open(\\"L1.jpg\\")" % (i));

but I really do not recommend it. Just use a dictonary!