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7

0

I have a conceptual question about Python. This is the code

list1=['assistant manager', 'salesperson', 'doctor', 'production manager', 'sales manager', 'schoolteacher', 'mathematics teacher']

sub1 = "teacher"

sub2 = "sales"

ans=[]



for item in list1:

    if (sub1 and sub2) in item:

        ans.append(item)

Here, I expect the list to be empty as none of the items satisfy the condition if sub1 and sub2 in item: But when I print the list I get the output#1 as this

>>> ans

['salesperson', 'sales manager'] # I expected an empty list here

Also, when I use or instead of and as given below

for item in list1:

    if (sub1 or sub2) in item:

        ans.append(item)

the output#2 I get is

>>> ans

['schoolteacher', 'mathematics teacher'] # I expected a list of words containing sub1 or sub2 as their substrings

I saw a similar looking solution here, but it does not exactly solve my problem. Both the times I get a result which I do not expect while using and and or. Why is this happening during both these operations?

python logical-operators

share|improve this question

edited 28 mins ago

Peter Mortensen

13.7k1986112

asked 7 hours ago

satyasatya

15012

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  (sub1 and sub2) in item what will be the result of the expression in the brackets? That will be checked against item.
  
  – Klaus D.
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Change (sub1 and sub2) in item to sub1 in item and sub2 in item
  
  – Tom Karzes
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Possible duplicate of can Python 'and' return None?
  
  – user5173426
  7 hours ago
  

  
  
  
  

add a comment | 

7

0

I have a conceptual question about Python. This is the code

list1=['assistant manager', 'salesperson', 'doctor', 'production manager', 'sales manager', 'schoolteacher', 'mathematics teacher']

sub1 = "teacher"

sub2 = "sales"

ans=[]



for item in list1:

    if (sub1 and sub2) in item:

        ans.append(item)

Here, I expect the list to be empty as none of the items satisfy the condition if sub1 and sub2 in item: But when I print the list I get the output#1 as this

>>> ans

['salesperson', 'sales manager'] # I expected an empty list here

Also, when I use or instead of and as given below

for item in list1:

    if (sub1 or sub2) in item:

        ans.append(item)

the output#2 I get is

>>> ans

['schoolteacher', 'mathematics teacher'] # I expected a list of words containing sub1 or sub2 as their substrings

I saw a similar looking solution here, but it does not exactly solve my problem. Both the times I get a result which I do not expect while using and and or. Why is this happening during both these operations?

python logical-operators

share|improve this question

edited 28 mins ago

Peter Mortensen

13.7k1986112

asked 7 hours ago

satyasatya

15012

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  (sub1 and sub2) in item what will be the result of the expression in the brackets? That will be checked against item.
  
  – Klaus D.
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Change (sub1 and sub2) in item to sub1 in item and sub2 in item
  
  – Tom Karzes
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Possible duplicate of can Python 'and' return None?
  
  – user5173426
  7 hours ago
  

  
  
  
  

add a comment | 

I have a conceptual question about Python. This is the code

list1=['assistant manager', 'salesperson', 'doctor', 'production manager', 'sales manager', 'schoolteacher', 'mathematics teacher']

sub1 = "teacher"

sub2 = "sales"

ans=[]



for item in list1:

    if (sub1 and sub2) in item:

        ans.append(item)

Here, I expect the list to be empty as none of the items satisfy the condition if sub1 and sub2 in item: But when I print the list I get the output#1 as this

>>> ans

['salesperson', 'sales manager'] # I expected an empty list here

Also, when I use or instead of and as given below

for item in list1:

    if (sub1 or sub2) in item:

        ans.append(item)

the output#2 I get is

>>> ans

['schoolteacher', 'mathematics teacher'] # I expected a list of words containing sub1 or sub2 as their substrings

I saw a similar looking solution here, but it does not exactly solve my problem. Both the times I get a result which I do not expect while using and and or. Why is this happening during both these operations?

python logical-operators

share|improve this question

edited 28 mins ago

Peter Mortensen

13.7k1986112

asked 7 hours ago

satyasatya

15012

list1=['assistant manager', 'salesperson', 'doctor', 'production manager', 'sales manager', 'schoolteacher', 'mathematics teacher']

sub1 = "teacher"

sub2 = "sales"

ans=[]



for item in list1:

    if (sub1 and sub2) in item:

        ans.append(item)

>>> ans

['salesperson', 'sales manager'] # I expected an empty list here

for item in list1:

    if (sub1 or sub2) in item:

        ans.append(item)

>>> ans

['schoolteacher', 'mathematics teacher'] # I expected a list of words containing sub1 or sub2 as their substrings

share|improve this question

edited 28 mins ago

Peter Mortensen

13.7k1986112

asked 7 hours ago

satyasatya

15012

share|improve this question

edited 28 mins ago

Peter Mortensen

13.7k1986112

asked 7 hours ago

satyasatya

15012

edited 28 mins ago

Peter Mortensen

13.7k1986112

edited 28 mins ago

Peter Mortensen

13.7k1986112

asked 7 hours ago

satyasatya

15012

asked 7 hours ago

satyasatya

15012

2 Answers
2

active

oldest

votes

14

("teacher" and "sales") in "salesmanager" do not mean the same in Python and in English.

In English, it is synonynous to ("teacher" in "salesmanager") and ("sales" in "salesmanager") (which Python would understand as you thought it should, and evaluate to False).

Python on the other hand will first evaluate "teacher" and "sales", because it is in parentheses, and thus has higher priority. and will return the first argument if falsy, otherwise the second argument. "teacher" is not falsy, so "teacher" and "sales" evaluates as "sales". Then, Python continues to evaluate "sales" in "salesmanager", and returns True.

share|improve this answer

answered 7 hours ago

AmadanAmadan

131k13143196

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So far as I know, all languages give parentheses first priority, not just Python. But you are correct in explaining what Python returns for ( X and Y )
  
  – Carl Witthoft
  29 mins ago
  

  
  
  
  

add a comment | 

5

The and and or operators don't do what you think they do. Try breaking up your expressions:

if sub1 in item or sub2 in item:



if sub1 in item and sub2 in item:

The and operator evaluates its left-hand operand and, if the result is truthy, returns the right-hand operand, otherwise the left-hand operand.

The or operator evaluates its left-hand operand and, if the result is falsy, returns the right-hand operand, otherwise the left-hand operand.

So, in your first expression evaluates as follows:

(sub1 and sub2) in item

("teacher" and "sales") in item

("sales") in item

which is not what you expected.

Similarly for your second expression:

(sub1 or sub2) in item

("teacher" or "sales") in item

("teacher") in item

share|improve this answer

answered 7 hours ago

RobᵩRobᵩ

116k13138222

add a comment | 

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14

("teacher" and "sales") in "salesmanager" do not mean the same in Python and in English.

In English, it is synonynous to ("teacher" in "salesmanager") and ("sales" in "salesmanager") (which Python would understand as you thought it should, and evaluate to False).

Python on the other hand will first evaluate "teacher" and "sales", because it is in parentheses, and thus has higher priority. and will return the first argument if falsy, otherwise the second argument. "teacher" is not falsy, so "teacher" and "sales" evaluates as "sales". Then, Python continues to evaluate "sales" in "salesmanager", and returns True.

share|improve this answer

answered 7 hours ago

AmadanAmadan

131k13143196

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So far as I know, all languages give parentheses first priority, not just Python. But you are correct in explaining what Python returns for ( X and Y )
  
  – Carl Witthoft
  29 mins ago
  

  
  
  
  

add a comment | 

14

("teacher" and "sales") in "salesmanager" do not mean the same in Python and in English.

In English, it is synonynous to ("teacher" in "salesmanager") and ("sales" in "salesmanager") (which Python would understand as you thought it should, and evaluate to False).

Python on the other hand will first evaluate "teacher" and "sales", because it is in parentheses, and thus has higher priority. and will return the first argument if falsy, otherwise the second argument. "teacher" is not falsy, so "teacher" and "sales" evaluates as "sales". Then, Python continues to evaluate "sales" in "salesmanager", and returns True.

share|improve this answer

answered 7 hours ago

AmadanAmadan

131k13143196

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So far as I know, all languages give parentheses first priority, not just Python. But you are correct in explaining what Python returns for ( X and Y )
  
  – Carl Witthoft
  29 mins ago
  

  
  
  
  

add a comment | 

("teacher" and "sales") in "salesmanager" do not mean the same in Python and in English.

In English, it is synonynous to ("teacher" in "salesmanager") and ("sales" in "salesmanager") (which Python would understand as you thought it should, and evaluate to False).

Python on the other hand will first evaluate "teacher" and "sales", because it is in parentheses, and thus has higher priority. and will return the first argument if falsy, otherwise the second argument. "teacher" is not falsy, so "teacher" and "sales" evaluates as "sales". Then, Python continues to evaluate "sales" in "salesmanager", and returns True.

share|improve this answer

answered 7 hours ago

AmadanAmadan

131k13143196

share|improve this answer

answered 7 hours ago

AmadanAmadan

131k13143196

answered 7 hours ago

AmadanAmadan

131k13143196

answered 7 hours ago

AmadanAmadan

131k13143196

5

The and and or operators don't do what you think they do. Try breaking up your expressions:

if sub1 in item or sub2 in item:



if sub1 in item and sub2 in item:

The and operator evaluates its left-hand operand and, if the result is truthy, returns the right-hand operand, otherwise the left-hand operand.

The or operator evaluates its left-hand operand and, if the result is falsy, returns the right-hand operand, otherwise the left-hand operand.

So, in your first expression evaluates as follows:

(sub1 and sub2) in item

("teacher" and "sales") in item

("sales") in item

which is not what you expected.

Similarly for your second expression:

(sub1 or sub2) in item

("teacher" or "sales") in item

("teacher") in item

share|improve this answer

answered 7 hours ago

RobᵩRobᵩ

116k13138222

add a comment | 

5

The and and or operators don't do what you think they do. Try breaking up your expressions:

if sub1 in item or sub2 in item:



if sub1 in item and sub2 in item:

The and operator evaluates its left-hand operand and, if the result is truthy, returns the right-hand operand, otherwise the left-hand operand.

The or operator evaluates its left-hand operand and, if the result is falsy, returns the right-hand operand, otherwise the left-hand operand.

So, in your first expression evaluates as follows:

(sub1 and sub2) in item

("teacher" and "sales") in item

("sales") in item

which is not what you expected.

Similarly for your second expression:

(sub1 or sub2) in item

("teacher" or "sales") in item

("teacher") in item

share|improve this answer

answered 7 hours ago

RobᵩRobᵩ

116k13138222

add a comment | 

The and and or operators don't do what you think they do. Try breaking up your expressions:

if sub1 in item or sub2 in item:



if sub1 in item and sub2 in item:

The and operator evaluates its left-hand operand and, if the result is truthy, returns the right-hand operand, otherwise the left-hand operand.

The or operator evaluates its left-hand operand and, if the result is falsy, returns the right-hand operand, otherwise the left-hand operand.

So, in your first expression evaluates as follows:

(sub1 and sub2) in item

("teacher" and "sales") in item

("sales") in item

which is not what you expected.

Similarly for your second expression:

(sub1 or sub2) in item

("teacher" or "sales") in item

("teacher") in item

share|improve this answer

answered 7 hours ago

RobᵩRobᵩ

116k13138222

if sub1 in item or sub2 in item:



if sub1 in item and sub2 in item:

(sub1 and sub2) in item

("teacher" and "sales") in item

("sales") in item

(sub1 or sub2) in item

("teacher" or "sales") in item

("teacher") in item

share|improve this answer

answered 7 hours ago

RobᵩRobᵩ

116k13138222

answered 7 hours ago

RobᵩRobᵩ

116k13138222

answered 7 hours ago

RobᵩRobᵩ

116k13138222