如何检查列表（或数组）上的条件，然后在Python中根据该条件对元素进行排序

Question

I have a list of complex numbers with n elements, I want to check that how many of the elements are real (the imaginary part is zero) and then sort my list in a way that it starts with the real numbers.

For example for the list below:

a = [ 7 + 0j, -2 + 3j, -2 - 3j, 5 + 6j, 5 - 6j, -1+ 0j, -8 + 4j, -8 - 4j]

two elements are real ( first element and sixth element) I want to know that I have 2 real elements in my list and then I want to have a sort like below that starts with real numbers and the other elements remains unchanged:

b = [ 7 + 0j, -1+ 0j, -2 + 3j, -2 - 3j, 5 + 6j, 5 - 6j, -8 + 4j, -8 - 4j]

How can I do that? Thanks

Answer 1

Python has a sorted function that takes a key argument.

The value of the key parameter should be a function that takes a single argument and returns a key to use for sorting purposes.

You can sort based on a lambda function that checks the presence of non zero imaginary component.

a = [ 7 + 0j, -2 + 3j, -2 - 3j, 5 + 6j, 5 - 6j, -1+ 0j, -8 + 4j, -8 - 4j]
sorted(a, key = lambda x: x.imag != 0) 
#Output: [(7+0j), (-1+0j), (-2+3j), (-2-3j), (5+6j), (5-6j), (-8+4j), (-8-4j)]

Answer 2

You can sort the numbers by comparing their imaginary part against zero. Those with 0's will come first, remaining numbers will be unchanged in their order.

a = sorted(a, key=lambda x:x.imag!=0)
print(a)

Outputs

[(7+0j), (-1+0j), (-2+3j), (-2-3j), (5+6j), (5-6j), (-8+4j), (-8-4j)]

To get number of such entries use

sum(x.imag == 0 for x in a)