List comprehension with conditional

Question

My goal is to count how many elements of a given list correspond to a condition. To do so, I made the following lines of code:

cont = 0
seq = []
max = 10
for x in seq if x == max:
    cont = cont+1

However, I get syntax error in if x == max. How can I fix this?

Answer 1

The if cannot be on the same line as for.

cont = 0
seq = []
max = 10
for x in seq:
    if x == max:
        cont = cont + 1
print(cont)

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

Answer 2

To count the elements of a list that satisfy a certain condition, taking into account the question title ("list understanding") you can do:

seq = [1,3,4,21,10,30,10,24]
max = 10
cont = len([x for x in seq if x == max]) # 2

Note that for this specific purpose (since the condition is equality, ==) enough and advise to do:

cont = seq.count(max)

DEMONSTRATION

PS: This is because you have the if in the same line as for, is not correct syntax for a list comprehension

Answer 3

Hello. The syntax used by you in your code is incorrect based on the patterns of the language, and Python, a block is always delimited by two points ':' followed by a:

line break with 4-space tab already predefined by the IDE (which may be modified in accordance with the user preference),

the most accessible syntax based on your question would be as follows:.

cont = 0
seq = [0 , 5, 6, 7, 8, 9, 6, 10, 10, 25]
max = 10
for x in seq:
    if x == max:
        cont = cont+1

op = int(input("DESEJA VERIFICAR O RESULTADO? DIGITE 1 PARA VISUALIZAR OU OUTRO VALOR PARA SAIR " ))
if op == 1:
    print("FORAM ENCONTRADOS %i valores" %(cont))
else:
    print()

Note that I modified the original program and applied the preview option, so the user could choose between viewing how many values are between the conformities or leaving the program.

Answer 4

From what I understand you want to implement a list comprehension in which you can count the number of times the value "10" I repeated in that list.

Well, to resolve this issue we can use the following code:

print(len([x for x in list(map(int, input('Digite os valores: ').split())) if x == 10]))

Note that when we execute the code we receive the following message: Digite os valores: . Right now we must type all the values, in the same line, separated for a single space and press enter.

From that moment the values captured by input() will be stored in a list. Next the block for shall cover all the elements of that list and with the help of the if check whether the value of x is equal to 10. If yes, the value will be stored in the list. Then the function len() calculate the size of this list and display the amount of values 10.

Answer 5

As colleagues have already said, the two points are missing : after for x in seq and if x == max and python does not allow certain constructs to be positioned on the same line:

cont = 0
seq = [1, 2, 3, 6, 10, 77, 92, 10, 10]
max = 10
for x in seq:
  if x == max:
    cont = cont+1

print(cont)  

I would like to suggest another approach, functional programming using the module functools that offers functions that act on or return other functions.

In case the solution uses the function reduce() which bears the following signature:

functools.reduce( função , iterável [ , inicializador ] )

Applies to função of two arguments cumulatively to the items of the eternal left to right, so as to reduce the iterable to a single value.
For example, reduce(lambda x, y: x+y, [1, 2, 3, 4, 5]) calculates ((((1+2)+3)+4)+5).
The left argument, x, is the accumulator and the right argument, y, is a value from the eternal.
If the optional initializer is present, it is placed before the iterable items in the calculation and serves as the default when the iterable is empty.
If the initializer is not provided and the iterable contains only one item, the first item will be returned.

In the example was used a lambda function comparing an element e of seq with max. If both are equal sum 1 to the accumulator a or just give it back a:

from functools import reduce

seq = [1, 2, 3, 6, 10, 77, 92, 10, 10]
max = 10

cont = reduce(lambda a, v : a + (1 if v==max else 0), seq, 0)

print(cont)   #3

^{Test the code on Repl.it}