Python: How to remove symbol in dict?












0















I get a xml response which is in bytes and parsed by xmltodict as shown below. Since I need to further use the response, I want to eliminate all the '@' in the key. I have tried two way to remove the sign but still not working. Are there any method to remove sign in dict?



XML response (resp.text): {'HOLDING_QUERY_RESPONSE': OrderedDict([('@LOGIN_ID', '011'), ('@CLIENT_ID', '011'), ('@CHANNEL', 'MB'), ('@SESSION_KEY', '111'), ('RESULT', OrderedDict([('@STATUS', 'OK'), ('BUYING_POWER', '98’),( 'USD_BUYING_POWER', '12'), ('AVAILABLE_MKT_VAL', '110'), ('CASH_HOLDING', OrderedDict([('LINEITEM', [OrderedDict([('@CURRENCY_CODE', 'USD'), ('@AVAILABLE_BAL', '-233'), ('@CASH_BALANCE', '0.00'), ('@CASH_ON_HOLD', '0.00'), ('@UNSETTLED_CASH', '0.00'), ('@ACCRUED_INTEREST', '0.00'), ('@BUY_SOLD_CONSIDERATION', '-233'), ('@EX_RATE', '1'), ('@AVAILABLE_VAL', '6'), ('@AMT', '0.00'), ('@DUE_AMT', '0.00')]),…….)])])]))]))])}


First tried method. The error is dict object has no attribute 'iteritems()'. If I remove the '.iteritem()', it will show Value Error: too many value to unpack.



info = {key.replace('@',''): item for key, item in xmltodict.parse(resp.text).iteritems()}
print(info)


Second tried method is changed it to string first and then translate it back to dict. But, still it seems the dict has not translate to string by json.dumps.



info = json.dumps(resp.text)
info.replace('@','')
to_dict = json.loads(info)
print(to_dict)









share|improve this question




















  • 2





    Try .items() instead of .iteritems(). You might be using Python 3.

    – Austin
    Nov 24 '18 at 3:53











  • Thanks for your reply, Austin. I tried this before, but still showing dict object has no item() attribute.

    – WILLIAM
    Nov 24 '18 at 5:48











  • it's items().

    – Austin
    Nov 24 '18 at 5:49











  • info.replace('@','') returns a new string, which you're ignoring, by the way

    – cricket_007
    Nov 25 '18 at 1:31
















0















I get a xml response which is in bytes and parsed by xmltodict as shown below. Since I need to further use the response, I want to eliminate all the '@' in the key. I have tried two way to remove the sign but still not working. Are there any method to remove sign in dict?



XML response (resp.text): {'HOLDING_QUERY_RESPONSE': OrderedDict([('@LOGIN_ID', '011'), ('@CLIENT_ID', '011'), ('@CHANNEL', 'MB'), ('@SESSION_KEY', '111'), ('RESULT', OrderedDict([('@STATUS', 'OK'), ('BUYING_POWER', '98’),( 'USD_BUYING_POWER', '12'), ('AVAILABLE_MKT_VAL', '110'), ('CASH_HOLDING', OrderedDict([('LINEITEM', [OrderedDict([('@CURRENCY_CODE', 'USD'), ('@AVAILABLE_BAL', '-233'), ('@CASH_BALANCE', '0.00'), ('@CASH_ON_HOLD', '0.00'), ('@UNSETTLED_CASH', '0.00'), ('@ACCRUED_INTEREST', '0.00'), ('@BUY_SOLD_CONSIDERATION', '-233'), ('@EX_RATE', '1'), ('@AVAILABLE_VAL', '6'), ('@AMT', '0.00'), ('@DUE_AMT', '0.00')]),…….)])])]))]))])}


First tried method. The error is dict object has no attribute 'iteritems()'. If I remove the '.iteritem()', it will show Value Error: too many value to unpack.



info = {key.replace('@',''): item for key, item in xmltodict.parse(resp.text).iteritems()}
print(info)


Second tried method is changed it to string first and then translate it back to dict. But, still it seems the dict has not translate to string by json.dumps.



info = json.dumps(resp.text)
info.replace('@','')
to_dict = json.loads(info)
print(to_dict)









share|improve this question




















  • 2





    Try .items() instead of .iteritems(). You might be using Python 3.

    – Austin
    Nov 24 '18 at 3:53











  • Thanks for your reply, Austin. I tried this before, but still showing dict object has no item() attribute.

    – WILLIAM
    Nov 24 '18 at 5:48











  • it's items().

    – Austin
    Nov 24 '18 at 5:49











  • info.replace('@','') returns a new string, which you're ignoring, by the way

    – cricket_007
    Nov 25 '18 at 1:31














0












0








0








I get a xml response which is in bytes and parsed by xmltodict as shown below. Since I need to further use the response, I want to eliminate all the '@' in the key. I have tried two way to remove the sign but still not working. Are there any method to remove sign in dict?



XML response (resp.text): {'HOLDING_QUERY_RESPONSE': OrderedDict([('@LOGIN_ID', '011'), ('@CLIENT_ID', '011'), ('@CHANNEL', 'MB'), ('@SESSION_KEY', '111'), ('RESULT', OrderedDict([('@STATUS', 'OK'), ('BUYING_POWER', '98’),( 'USD_BUYING_POWER', '12'), ('AVAILABLE_MKT_VAL', '110'), ('CASH_HOLDING', OrderedDict([('LINEITEM', [OrderedDict([('@CURRENCY_CODE', 'USD'), ('@AVAILABLE_BAL', '-233'), ('@CASH_BALANCE', '0.00'), ('@CASH_ON_HOLD', '0.00'), ('@UNSETTLED_CASH', '0.00'), ('@ACCRUED_INTEREST', '0.00'), ('@BUY_SOLD_CONSIDERATION', '-233'), ('@EX_RATE', '1'), ('@AVAILABLE_VAL', '6'), ('@AMT', '0.00'), ('@DUE_AMT', '0.00')]),…….)])])]))]))])}


First tried method. The error is dict object has no attribute 'iteritems()'. If I remove the '.iteritem()', it will show Value Error: too many value to unpack.



info = {key.replace('@',''): item for key, item in xmltodict.parse(resp.text).iteritems()}
print(info)


Second tried method is changed it to string first and then translate it back to dict. But, still it seems the dict has not translate to string by json.dumps.



info = json.dumps(resp.text)
info.replace('@','')
to_dict = json.loads(info)
print(to_dict)









share|improve this question
















I get a xml response which is in bytes and parsed by xmltodict as shown below. Since I need to further use the response, I want to eliminate all the '@' in the key. I have tried two way to remove the sign but still not working. Are there any method to remove sign in dict?



XML response (resp.text): {'HOLDING_QUERY_RESPONSE': OrderedDict([('@LOGIN_ID', '011'), ('@CLIENT_ID', '011'), ('@CHANNEL', 'MB'), ('@SESSION_KEY', '111'), ('RESULT', OrderedDict([('@STATUS', 'OK'), ('BUYING_POWER', '98’),( 'USD_BUYING_POWER', '12'), ('AVAILABLE_MKT_VAL', '110'), ('CASH_HOLDING', OrderedDict([('LINEITEM', [OrderedDict([('@CURRENCY_CODE', 'USD'), ('@AVAILABLE_BAL', '-233'), ('@CASH_BALANCE', '0.00'), ('@CASH_ON_HOLD', '0.00'), ('@UNSETTLED_CASH', '0.00'), ('@ACCRUED_INTEREST', '0.00'), ('@BUY_SOLD_CONSIDERATION', '-233'), ('@EX_RATE', '1'), ('@AVAILABLE_VAL', '6'), ('@AMT', '0.00'), ('@DUE_AMT', '0.00')]),…….)])])]))]))])}


First tried method. The error is dict object has no attribute 'iteritems()'. If I remove the '.iteritem()', it will show Value Error: too many value to unpack.



info = {key.replace('@',''): item for key, item in xmltodict.parse(resp.text).iteritems()}
print(info)


Second tried method is changed it to string first and then translate it back to dict. But, still it seems the dict has not translate to string by json.dumps.



info = json.dumps(resp.text)
info.replace('@','')
to_dict = json.loads(info)
print(to_dict)






python json xml response






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 25 '18 at 1:19







WILLIAM

















asked Nov 24 '18 at 3:51









WILLIAMWILLIAM

475




475








  • 2





    Try .items() instead of .iteritems(). You might be using Python 3.

    – Austin
    Nov 24 '18 at 3:53











  • Thanks for your reply, Austin. I tried this before, but still showing dict object has no item() attribute.

    – WILLIAM
    Nov 24 '18 at 5:48











  • it's items().

    – Austin
    Nov 24 '18 at 5:49











  • info.replace('@','') returns a new string, which you're ignoring, by the way

    – cricket_007
    Nov 25 '18 at 1:31














  • 2





    Try .items() instead of .iteritems(). You might be using Python 3.

    – Austin
    Nov 24 '18 at 3:53











  • Thanks for your reply, Austin. I tried this before, but still showing dict object has no item() attribute.

    – WILLIAM
    Nov 24 '18 at 5:48











  • it's items().

    – Austin
    Nov 24 '18 at 5:49











  • info.replace('@','') returns a new string, which you're ignoring, by the way

    – cricket_007
    Nov 25 '18 at 1:31








2




2





Try .items() instead of .iteritems(). You might be using Python 3.

– Austin
Nov 24 '18 at 3:53





Try .items() instead of .iteritems(). You might be using Python 3.

– Austin
Nov 24 '18 at 3:53













Thanks for your reply, Austin. I tried this before, but still showing dict object has no item() attribute.

– WILLIAM
Nov 24 '18 at 5:48





Thanks for your reply, Austin. I tried this before, but still showing dict object has no item() attribute.

– WILLIAM
Nov 24 '18 at 5:48













it's items().

– Austin
Nov 24 '18 at 5:49





it's items().

– Austin
Nov 24 '18 at 5:49













info.replace('@','') returns a new string, which you're ignoring, by the way

– cricket_007
Nov 25 '18 at 1:31





info.replace('@','') returns a new string, which you're ignoring, by the way

– cricket_007
Nov 25 '18 at 1:31












1 Answer
1






active

oldest

votes


















1














Try this:



info = {key.replace('@',''): item for key, item in xmltodict.parse(resp.text).items()}
print(info)


As mentioned in the comments, xmltodict.parse returns a dictionary, and in your case, you might be working with Python 3, so use the .items() attribute to access the dictionary.



As an example:



d = ('@CASH', '0.00'), ('@INTEREST', '0.00'), ('@BUY', '-233'), ('@RATE', '1.000000'), ('@AVAILABLE', '6,228'), ('@AMT', '0.00')
info = {key.replace('@',''): item for key, item in d}

print(info)
>>
{'CASH': '0.00', 'INTEREST': '0.00', 'BUY': '-233', 'RATE': '1.000000', 'AVAILABLE': '6,228', 'AMT': '0.00'}


Updated question soution



Similarly, if it is a nested dictionary, you have to access the key you want to replace @ in its values, from your updated example, you should be able to access the nested dictionary with xmltodict.parse(resp.text)['HOLDING_QUERY_RESPONSE'].



from collections import OrderedDict

#based on your updated data structure
d = OrderedDict([('@CASH', '0.00'), ('@INTEREST', '0.00'), ('@BUY', '-233'), ('@RATE', '1.000000'), ('@AVAILABLE', '6,228'), ('@AMT', '0.00')])
xml_response = {}
xml_response['HOLDING_QUERY_RESPONSE'] = d

xml_response
{'HOLDING_QUERY_RESPONSE': OrderedDict([('@CASH', '0.00'),
('@INTEREST', '0.00'),
('@BUY', '-233'),
('@RATE', '1.000000'),
('@AVAILABLE', '6,228'),
('@AMT', '0.00')])}

info = {key.replace('@',''):value for key, value in xml_response['HOLDING_QUERY_RESPONSE'].items()}
print(info)





share|improve this answer


























  • Thanks, now the error is gone, but I found that the @ sign is still there.

    – WILLIAM
    Nov 24 '18 at 13:39











  • I added an example, it should work and its pretty straightforward. Are you accessing the right variables?

    – BernardL
    Nov 24 '18 at 16:13











  • I also edited the resp.text. Actually the dict is like this and have many layer. Does it affect the result of simply using the above code?

    – WILLIAM
    Nov 25 '18 at 1:21











  • @WILL this only works on the top level keys of the dictionary

    – cricket_007
    Nov 25 '18 at 1:30













  • I edited my answer based on the update, if any of the answers here solved your problem, kindly accept the best one as your answer.

    – BernardL
    Nov 25 '18 at 2:07











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














Try this:



info = {key.replace('@',''): item for key, item in xmltodict.parse(resp.text).items()}
print(info)


As mentioned in the comments, xmltodict.parse returns a dictionary, and in your case, you might be working with Python 3, so use the .items() attribute to access the dictionary.



As an example:



d = ('@CASH', '0.00'), ('@INTEREST', '0.00'), ('@BUY', '-233'), ('@RATE', '1.000000'), ('@AVAILABLE', '6,228'), ('@AMT', '0.00')
info = {key.replace('@',''): item for key, item in d}

print(info)
>>
{'CASH': '0.00', 'INTEREST': '0.00', 'BUY': '-233', 'RATE': '1.000000', 'AVAILABLE': '6,228', 'AMT': '0.00'}


Updated question soution



Similarly, if it is a nested dictionary, you have to access the key you want to replace @ in its values, from your updated example, you should be able to access the nested dictionary with xmltodict.parse(resp.text)['HOLDING_QUERY_RESPONSE'].



from collections import OrderedDict

#based on your updated data structure
d = OrderedDict([('@CASH', '0.00'), ('@INTEREST', '0.00'), ('@BUY', '-233'), ('@RATE', '1.000000'), ('@AVAILABLE', '6,228'), ('@AMT', '0.00')])
xml_response = {}
xml_response['HOLDING_QUERY_RESPONSE'] = d

xml_response
{'HOLDING_QUERY_RESPONSE': OrderedDict([('@CASH', '0.00'),
('@INTEREST', '0.00'),
('@BUY', '-233'),
('@RATE', '1.000000'),
('@AVAILABLE', '6,228'),
('@AMT', '0.00')])}

info = {key.replace('@',''):value for key, value in xml_response['HOLDING_QUERY_RESPONSE'].items()}
print(info)





share|improve this answer


























  • Thanks, now the error is gone, but I found that the @ sign is still there.

    – WILLIAM
    Nov 24 '18 at 13:39











  • I added an example, it should work and its pretty straightforward. Are you accessing the right variables?

    – BernardL
    Nov 24 '18 at 16:13











  • I also edited the resp.text. Actually the dict is like this and have many layer. Does it affect the result of simply using the above code?

    – WILLIAM
    Nov 25 '18 at 1:21











  • @WILL this only works on the top level keys of the dictionary

    – cricket_007
    Nov 25 '18 at 1:30













  • I edited my answer based on the update, if any of the answers here solved your problem, kindly accept the best one as your answer.

    – BernardL
    Nov 25 '18 at 2:07
















1














Try this:



info = {key.replace('@',''): item for key, item in xmltodict.parse(resp.text).items()}
print(info)


As mentioned in the comments, xmltodict.parse returns a dictionary, and in your case, you might be working with Python 3, so use the .items() attribute to access the dictionary.



As an example:



d = ('@CASH', '0.00'), ('@INTEREST', '0.00'), ('@BUY', '-233'), ('@RATE', '1.000000'), ('@AVAILABLE', '6,228'), ('@AMT', '0.00')
info = {key.replace('@',''): item for key, item in d}

print(info)
>>
{'CASH': '0.00', 'INTEREST': '0.00', 'BUY': '-233', 'RATE': '1.000000', 'AVAILABLE': '6,228', 'AMT': '0.00'}


Updated question soution



Similarly, if it is a nested dictionary, you have to access the key you want to replace @ in its values, from your updated example, you should be able to access the nested dictionary with xmltodict.parse(resp.text)['HOLDING_QUERY_RESPONSE'].



from collections import OrderedDict

#based on your updated data structure
d = OrderedDict([('@CASH', '0.00'), ('@INTEREST', '0.00'), ('@BUY', '-233'), ('@RATE', '1.000000'), ('@AVAILABLE', '6,228'), ('@AMT', '0.00')])
xml_response = {}
xml_response['HOLDING_QUERY_RESPONSE'] = d

xml_response
{'HOLDING_QUERY_RESPONSE': OrderedDict([('@CASH', '0.00'),
('@INTEREST', '0.00'),
('@BUY', '-233'),
('@RATE', '1.000000'),
('@AVAILABLE', '6,228'),
('@AMT', '0.00')])}

info = {key.replace('@',''):value for key, value in xml_response['HOLDING_QUERY_RESPONSE'].items()}
print(info)





share|improve this answer


























  • Thanks, now the error is gone, but I found that the @ sign is still there.

    – WILLIAM
    Nov 24 '18 at 13:39











  • I added an example, it should work and its pretty straightforward. Are you accessing the right variables?

    – BernardL
    Nov 24 '18 at 16:13











  • I also edited the resp.text. Actually the dict is like this and have many layer. Does it affect the result of simply using the above code?

    – WILLIAM
    Nov 25 '18 at 1:21











  • @WILL this only works on the top level keys of the dictionary

    – cricket_007
    Nov 25 '18 at 1:30













  • I edited my answer based on the update, if any of the answers here solved your problem, kindly accept the best one as your answer.

    – BernardL
    Nov 25 '18 at 2:07














1












1








1







Try this:



info = {key.replace('@',''): item for key, item in xmltodict.parse(resp.text).items()}
print(info)


As mentioned in the comments, xmltodict.parse returns a dictionary, and in your case, you might be working with Python 3, so use the .items() attribute to access the dictionary.



As an example:



d = ('@CASH', '0.00'), ('@INTEREST', '0.00'), ('@BUY', '-233'), ('@RATE', '1.000000'), ('@AVAILABLE', '6,228'), ('@AMT', '0.00')
info = {key.replace('@',''): item for key, item in d}

print(info)
>>
{'CASH': '0.00', 'INTEREST': '0.00', 'BUY': '-233', 'RATE': '1.000000', 'AVAILABLE': '6,228', 'AMT': '0.00'}


Updated question soution



Similarly, if it is a nested dictionary, you have to access the key you want to replace @ in its values, from your updated example, you should be able to access the nested dictionary with xmltodict.parse(resp.text)['HOLDING_QUERY_RESPONSE'].



from collections import OrderedDict

#based on your updated data structure
d = OrderedDict([('@CASH', '0.00'), ('@INTEREST', '0.00'), ('@BUY', '-233'), ('@RATE', '1.000000'), ('@AVAILABLE', '6,228'), ('@AMT', '0.00')])
xml_response = {}
xml_response['HOLDING_QUERY_RESPONSE'] = d

xml_response
{'HOLDING_QUERY_RESPONSE': OrderedDict([('@CASH', '0.00'),
('@INTEREST', '0.00'),
('@BUY', '-233'),
('@RATE', '1.000000'),
('@AVAILABLE', '6,228'),
('@AMT', '0.00')])}

info = {key.replace('@',''):value for key, value in xml_response['HOLDING_QUERY_RESPONSE'].items()}
print(info)





share|improve this answer















Try this:



info = {key.replace('@',''): item for key, item in xmltodict.parse(resp.text).items()}
print(info)


As mentioned in the comments, xmltodict.parse returns a dictionary, and in your case, you might be working with Python 3, so use the .items() attribute to access the dictionary.



As an example:



d = ('@CASH', '0.00'), ('@INTEREST', '0.00'), ('@BUY', '-233'), ('@RATE', '1.000000'), ('@AVAILABLE', '6,228'), ('@AMT', '0.00')
info = {key.replace('@',''): item for key, item in d}

print(info)
>>
{'CASH': '0.00', 'INTEREST': '0.00', 'BUY': '-233', 'RATE': '1.000000', 'AVAILABLE': '6,228', 'AMT': '0.00'}


Updated question soution



Similarly, if it is a nested dictionary, you have to access the key you want to replace @ in its values, from your updated example, you should be able to access the nested dictionary with xmltodict.parse(resp.text)['HOLDING_QUERY_RESPONSE'].



from collections import OrderedDict

#based on your updated data structure
d = OrderedDict([('@CASH', '0.00'), ('@INTEREST', '0.00'), ('@BUY', '-233'), ('@RATE', '1.000000'), ('@AVAILABLE', '6,228'), ('@AMT', '0.00')])
xml_response = {}
xml_response['HOLDING_QUERY_RESPONSE'] = d

xml_response
{'HOLDING_QUERY_RESPONSE': OrderedDict([('@CASH', '0.00'),
('@INTEREST', '0.00'),
('@BUY', '-233'),
('@RATE', '1.000000'),
('@AVAILABLE', '6,228'),
('@AMT', '0.00')])}

info = {key.replace('@',''):value for key, value in xml_response['HOLDING_QUERY_RESPONSE'].items()}
print(info)






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 25 '18 at 2:06

























answered Nov 24 '18 at 6:09









BernardLBernardL

2,37311130




2,37311130













  • Thanks, now the error is gone, but I found that the @ sign is still there.

    – WILLIAM
    Nov 24 '18 at 13:39











  • I added an example, it should work and its pretty straightforward. Are you accessing the right variables?

    – BernardL
    Nov 24 '18 at 16:13











  • I also edited the resp.text. Actually the dict is like this and have many layer. Does it affect the result of simply using the above code?

    – WILLIAM
    Nov 25 '18 at 1:21











  • @WILL this only works on the top level keys of the dictionary

    – cricket_007
    Nov 25 '18 at 1:30













  • I edited my answer based on the update, if any of the answers here solved your problem, kindly accept the best one as your answer.

    – BernardL
    Nov 25 '18 at 2:07



















  • Thanks, now the error is gone, but I found that the @ sign is still there.

    – WILLIAM
    Nov 24 '18 at 13:39











  • I added an example, it should work and its pretty straightforward. Are you accessing the right variables?

    – BernardL
    Nov 24 '18 at 16:13











  • I also edited the resp.text. Actually the dict is like this and have many layer. Does it affect the result of simply using the above code?

    – WILLIAM
    Nov 25 '18 at 1:21











  • @WILL this only works on the top level keys of the dictionary

    – cricket_007
    Nov 25 '18 at 1:30













  • I edited my answer based on the update, if any of the answers here solved your problem, kindly accept the best one as your answer.

    – BernardL
    Nov 25 '18 at 2:07

















Thanks, now the error is gone, but I found that the @ sign is still there.

– WILLIAM
Nov 24 '18 at 13:39





Thanks, now the error is gone, but I found that the @ sign is still there.

– WILLIAM
Nov 24 '18 at 13:39













I added an example, it should work and its pretty straightforward. Are you accessing the right variables?

– BernardL
Nov 24 '18 at 16:13





I added an example, it should work and its pretty straightforward. Are you accessing the right variables?

– BernardL
Nov 24 '18 at 16:13













I also edited the resp.text. Actually the dict is like this and have many layer. Does it affect the result of simply using the above code?

– WILLIAM
Nov 25 '18 at 1:21





I also edited the resp.text. Actually the dict is like this and have many layer. Does it affect the result of simply using the above code?

– WILLIAM
Nov 25 '18 at 1:21













@WILL this only works on the top level keys of the dictionary

– cricket_007
Nov 25 '18 at 1:30







@WILL this only works on the top level keys of the dictionary

– cricket_007
Nov 25 '18 at 1:30















I edited my answer based on the update, if any of the answers here solved your problem, kindly accept the best one as your answer.

– BernardL
Nov 25 '18 at 2:07





I edited my answer based on the update, if any of the answers here solved your problem, kindly accept the best one as your answer.

– BernardL
Nov 25 '18 at 2:07




















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