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evaluate a python string expression using dictionary values

How to merge two dictionaries in a single expression?Calling an external command in PythonWhat are metaclasses in Python?How can I safely create a nested directory?Does Python have a ternary conditional operator?How do I sort a dictionary by value?Add new keys to a dictionary?Check if a given key already exists in a dictionaryIterating over dictionaries using 'for' loopsDoes Python have a string 'contains' substring method?

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-3

I am parsing a text file which contain python "string" inside it. For e.g.:

'my_home1' in houses.split(',') and '2018' in iphone.split(',') and 14 < mask

for the example above, I wrote a possible dictionary:

dict = 'house' : 'my_home1,my_home2', 'iphone' : '2015,2018', 'mask' : '15' 

but dict is not know in advance and will be adapted during the running time of the python script.

I would like to replace all the appearance of keys inside the string above in a dynamic way and then to evaluate the expression.

python python-3.x

share|improve this question

edited Mar 26 at 10:28

Eagle

asked Mar 26 at 5:10

EagleEagle

1,30144 gold badges2424 silver badges3737 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  what is in the variable houses
  
  – Jeril
  Mar 26 at 5:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I hope now it is clearer
  
  – Eagle
  Mar 26 at 5:20
  
  

  
  
  
  

add a comment | 

-3

I am parsing a text file which contain python "string" inside it. For e.g.:

'my_home1' in houses.split(',') and '2018' in iphone.split(',') and 14 < mask

for the example above, I wrote a possible dictionary:

dict = 'house' : 'my_home1,my_home2', 'iphone' : '2015,2018', 'mask' : '15' 

but dict is not know in advance and will be adapted during the running time of the python script.

I would like to replace all the appearance of keys inside the string above in a dynamic way and then to evaluate the expression.

python python-3.x

share|improve this question

edited Mar 26 at 10:28

Eagle

asked Mar 26 at 5:10

EagleEagle

1,30144 gold badges2424 silver badges3737 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  what is in the variable houses
  
  – Jeril
  Mar 26 at 5:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I hope now it is clearer
  
  – Eagle
  Mar 26 at 5:20
  
  

  
  
  
  

add a comment | 

I am parsing a text file which contain python "string" inside it. For e.g.:

'my_home1' in houses.split(',') and '2018' in iphone.split(',') and 14 < mask

for the example above, I wrote a possible dictionary:

dict = 'house' : 'my_home1,my_home2', 'iphone' : '2015,2018', 'mask' : '15' 

but dict is not know in advance and will be adapted during the running time of the python script.

I would like to replace all the appearance of keys inside the string above in a dynamic way and then to evaluate the expression.

python python-3.x

share|improve this question

edited Mar 26 at 10:28

Eagle

asked Mar 26 at 5:10

EagleEagle

1,30144 gold badges2424 silver badges3737 bronze badges

'my_home1' in houses.split(',') and '2018' in iphone.split(',') and 14 < mask

dict = 'house' : 'my_home1,my_home2', 'iphone' : '2015,2018', 'mask' : '15' 

share|improve this question

edited Mar 26 at 10:28

Eagle

asked Mar 26 at 5:10

EagleEagle

1,30144 gold badges2424 silver badges3737 bronze badges

share|improve this question

edited Mar 26 at 10:28

Eagle

asked Mar 26 at 5:10

EagleEagle

1,30144 gold badges2424 silver badges3737 bronze badges

asked Mar 26 at 5:10

EagleEagle

1,30144 gold badges2424 silver badges3737 bronze badges

asked Mar 26 at 5:10

EagleEagle

1,30144 gold badges2424 silver badges3737 bronze badges

2 Answers
2

active

oldest

votes

0

At the end I found a solution to my problem:

text = "'my_home1' in houses.split(',') and '2018' in iphone.split(',') and 14 < mask"
dict = 'house' : 'my_home1,my_home2', 'iphone' : '2015,2018', 'mask' : '15' 

expression = 'false'

for key in dict.keys():
 if isinstance(dict.get(key), str):
 text = re.sub(key, ''''.format(dict.get(key)), text)
 else:
 text = re.sub(key, dict.get(key), text)

eval(text)

share|improve this answer

answered Mar 27 at 20:34

EagleEagle

1,30144 gold badges2424 silver badges3737 bronze badges

add a comment | 

-1

Is the following what you were expecting:

dict_ = 'house': 'my_home1,my_home2', 'iphone': '2015,2018'
print('my_home1' in dict_['house'].split(',') and '2018' in dict_[
 'iphone'].split(',') and '2019' not in dict_['iphone'].split(','))

share|improve this answer

edited Mar 26 at 7:26

answered Mar 26 at 5:45

JerilJeril

3,50422 gold badges2222 silver badges4242 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks for your answer, unfortunately this is not what i am searching. I updated my question.
  
  – Eagle
  Mar 26 at 7:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  i have updated my solution, check now
  
  – Jeril
  Mar 26 at 7:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  my difficulty with your solution is that I do not know in advance the keys inside dict.
  
  – Eagle
  Mar 26 at 7:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  i am not able to get you, can you come again
  
  – Jeril
  Mar 26 at 9:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I have adapted my example above. The difficulty is that you you do not know which textsstrings in the expressions are in the dictionary and need to be evaluate with 'dict_'.
  
  – Eagle
  Mar 26 at 10:35
  

  
  
  
  

 | 
show 1 more comment

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0

At the end I found a solution to my problem:

text = "'my_home1' in houses.split(',') and '2018' in iphone.split(',') and 14 < mask"
dict = 'house' : 'my_home1,my_home2', 'iphone' : '2015,2018', 'mask' : '15' 

expression = 'false'

for key in dict.keys():
 if isinstance(dict.get(key), str):
 text = re.sub(key, ''''.format(dict.get(key)), text)
 else:
 text = re.sub(key, dict.get(key), text)

eval(text)

share|improve this answer

answered Mar 27 at 20:34

EagleEagle

1,30144 gold badges2424 silver badges3737 bronze badges

add a comment | 

0

At the end I found a solution to my problem:

text = "'my_home1' in houses.split(',') and '2018' in iphone.split(',') and 14 < mask"
dict = 'house' : 'my_home1,my_home2', 'iphone' : '2015,2018', 'mask' : '15' 

expression = 'false'

for key in dict.keys():
 if isinstance(dict.get(key), str):
 text = re.sub(key, ''''.format(dict.get(key)), text)
 else:
 text = re.sub(key, dict.get(key), text)

eval(text)

share|improve this answer

answered Mar 27 at 20:34

EagleEagle

1,30144 gold badges2424 silver badges3737 bronze badges

add a comment | 

At the end I found a solution to my problem:

text = "'my_home1' in houses.split(',') and '2018' in iphone.split(',') and 14 < mask"
dict = 'house' : 'my_home1,my_home2', 'iphone' : '2015,2018', 'mask' : '15' 

expression = 'false'

for key in dict.keys():
 if isinstance(dict.get(key), str):
 text = re.sub(key, ''''.format(dict.get(key)), text)
 else:
 text = re.sub(key, dict.get(key), text)

eval(text)

share|improve this answer

answered Mar 27 at 20:34

EagleEagle

1,30144 gold badges2424 silver badges3737 bronze badges

text = "'my_home1' in houses.split(',') and '2018' in iphone.split(',') and 14 < mask"
dict = 'house' : 'my_home1,my_home2', 'iphone' : '2015,2018', 'mask' : '15' 

expression = 'false'

for key in dict.keys():
 if isinstance(dict.get(key), str):
 text = re.sub(key, ''''.format(dict.get(key)), text)
 else:
 text = re.sub(key, dict.get(key), text)

eval(text)

share|improve this answer

answered Mar 27 at 20:34

EagleEagle

1,30144 gold badges2424 silver badges3737 bronze badges

answered Mar 27 at 20:34

EagleEagle

1,30144 gold badges2424 silver badges3737 bronze badges

answered Mar 27 at 20:34

EagleEagle

1,30144 gold badges2424 silver badges3737 bronze badges

-1

Is the following what you were expecting:

dict_ = 'house': 'my_home1,my_home2', 'iphone': '2015,2018'
print('my_home1' in dict_['house'].split(',') and '2018' in dict_[
 'iphone'].split(',') and '2019' not in dict_['iphone'].split(','))

share|improve this answer

edited Mar 26 at 7:26

answered Mar 26 at 5:45

JerilJeril

3,50422 gold badges2222 silver badges4242 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks for your answer, unfortunately this is not what i am searching. I updated my question.
  
  – Eagle
  Mar 26 at 7:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  i have updated my solution, check now
  
  – Jeril
  Mar 26 at 7:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  my difficulty with your solution is that I do not know in advance the keys inside dict.
  
  – Eagle
  Mar 26 at 7:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  i am not able to get you, can you come again
  
  – Jeril
  Mar 26 at 9:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I have adapted my example above. The difficulty is that you you do not know which textsstrings in the expressions are in the dictionary and need to be evaluate with 'dict_'.
  
  – Eagle
  Mar 26 at 10:35
  

  
  
  
  

 | 
show 1 more comment

-1

Is the following what you were expecting:

dict_ = 'house': 'my_home1,my_home2', 'iphone': '2015,2018'
print('my_home1' in dict_['house'].split(',') and '2018' in dict_[
 'iphone'].split(',') and '2019' not in dict_['iphone'].split(','))

share|improve this answer

edited Mar 26 at 7:26

answered Mar 26 at 5:45

JerilJeril

3,50422 gold badges2222 silver badges4242 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks for your answer, unfortunately this is not what i am searching. I updated my question.
  
  – Eagle
  Mar 26 at 7:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  i have updated my solution, check now
  
  – Jeril
  Mar 26 at 7:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  my difficulty with your solution is that I do not know in advance the keys inside dict.
  
  – Eagle
  Mar 26 at 7:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  i am not able to get you, can you come again
  
  – Jeril
  Mar 26 at 9:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I have adapted my example above. The difficulty is that you you do not know which textsstrings in the expressions are in the dictionary and need to be evaluate with 'dict_'.
  
  – Eagle
  Mar 26 at 10:35
  

  
  
  
  

 | 
show 1 more comment

Is the following what you were expecting:

dict_ = 'house': 'my_home1,my_home2', 'iphone': '2015,2018'
print('my_home1' in dict_['house'].split(',') and '2018' in dict_[
 'iphone'].split(',') and '2019' not in dict_['iphone'].split(','))

share|improve this answer

edited Mar 26 at 7:26

answered Mar 26 at 5:45

JerilJeril

3,50422 gold badges2222 silver badges4242 bronze badges

dict_ = 'house': 'my_home1,my_home2', 'iphone': '2015,2018'
print('my_home1' in dict_['house'].split(',') and '2018' in dict_[
 'iphone'].split(',') and '2019' not in dict_['iphone'].split(','))

share|improve this answer

edited Mar 26 at 7:26

answered Mar 26 at 5:45

JerilJeril

3,50422 gold badges2222 silver badges4242 bronze badges

answered Mar 26 at 5:45

JerilJeril

3,50422 gold badges2222 silver badges4242 bronze badges

answered Mar 26 at 5:45

JerilJeril

3,50422 gold badges2222 silver badges4242 bronze badges