if condition not met but still prints "if" in nested if statements with for loop

Question

Why does this code never goes in to "else" and print accordingly when the condition in "if" is not satisfied?

j=0
for i in data:
  if j<10:

    if i['product']['id'] == p_id:
        if (i['stop_price']!='None'):
            print("Order Type:" +  str(i['stop_order_type']))
            print("Stop Price: " + str(i['stop_price']))

        else:
            print("Order Type: " + str(i['order_type']))



        print("Limit Price: " + str(i['limit_price']))
        print("Side: " + str(i['side']))
        print("Size: " + str(i['size']))
        print("Unfilled Size: " + str(i['unfilled_size']))

        print("\n\n")

    j+=1

It prints the below output:

Order Type:stop_loss_order
Stop Price: 405.0
Limit Price: 400.0
Side: buy
Size: 1
Unfilled Size: 1



Order Type:None
Stop Price: None
Limit Price: 280.0
Side: sell
Size: 1
Unfilled Size: 0



Order Type:None
Stop Price: None
Limit Price: 300.0
Side: sell
Size: 1
Unfilled Size: 1

But the correct Output should be:

Order Type:stop_loss_order
Stop Price: 405.0
Limit Price: 400.0
Side: buy
Size: 1
Unfilled Size: 1



Order Type:Limit
Limit Price: 280.0
Side: sell
Size: 1
Unfilled Size: 0



Order Type:Limit
Limit Price: 300.0
Side: sell
Size: 1
Unfilled Size: 1

Tom Karzes · Accepted Answer · 2020-05-01T07:09:21.437

1

Change 'None' to a bare None. You're comparing it to the string 'None', which is generally going to be False unless it actually is the string 'None'.

It should look like:

    if (i['stop_price'] != None):

Note that in the case of a comparison with None, it's slightly more efficient to do:

    if (i['stop_price'] is not None):

Thierry Lathuille points out that the latter is recommended by the PEP 8 Programming Recommendations. Both versions should behave the same.

edited May 01 '20 at 07:09

answered May 01 '20 at 06:09

Tom Karzes

22,815
2
22
41

1

You should rather do `if (i['stop_price'] is not None):`, see https://stackoverflow.com/questions/14247373/python-none-comparison-should-i-use-is-or – Thierry Lathuille May 01 '20 at 06:41
@ThierryLathuille They're equivalent in the case of `None`, although `is not` is probably a little faster. I was addressing the specific problem that prevented OP's code from running, and I didn't want to confuse the issue by changing the comparison operator. – Tom Karzes May 01 '20 at 07:00
I understand why you left it the way it was - maybe you could add the 'canonical' way to do it to your answer. – Thierry Lathuille May 01 '20 at 07:02
@ThierryLathuille Ok, I added a note per your suggestion. – Tom Karzes May 01 '20 at 07:10

Achuth Varghese · Answer 2 · 2020-05-01T06:20:01.040

0

Check the value and type of i['stop_price'] in line if (i['stop_price']!='None') and modify the condition accordingly.

Here you are trying to check the value of i['stop_price'] not to be equal to 'None' of type String which can always be True unless the actual value in i['stop_price'] is equal to 'None' of type String.

edited May 01 '20 at 06:20

answered May 01 '20 at 06:13

Achuth Varghese

2,356
1
4
18

if condition not met but still prints "if" in nested if statements with for loop

2 Answers2