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Unpack a dictionary to format


Unpack a dictionary to format

By : Henrique Serraglia
Date : November 23 2020, 03:01 PM
I hope this helps you . Unpacking a dictionary to format gives named arguments, so you need to name the placeholders as well; you might need:
code :
mydict = {'yesterday':'2017-10-19','today':'2017-10-20','tomorrow':'2017-10-21'}
​
"yesterday:{yesterday}, today:{today}, tomorrow:{tomorrow}".format(**mydict)
# 'yesterday:2017-10-19, today:2017-10-20, tomorrow:2017-10-21'
"{}:{}, {}:{}, {}:{}".format(*(x for kv in mydict.items() for x in kv))
# 'today:2017-10-20, tomorrow:2017-10-21, yesterday:2017-10-19'


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Dictionary.py too many values to unpack

Dictionary.py too many values to unpack


By : Raybiz IlkTest
Date : March 29 2020, 07:55 AM
I wish this help you Whenever I try to reload my dictionary.DB file, it keeps telling me: , You can do:
code :
raw = word.split()
word, name, definition, wordtime = raw[0], raw[1], ' '.join(raw[2:-1]), raw[-1]
word = 'dog charles a cute cute animal 1387199870'
# definition -> 'a cute cute animal'
Unpack dictionary without knowledge of its key name

Unpack dictionary without knowledge of its key name


By : user2965849
Date : March 29 2020, 07:55 AM
This might help you If I have some dictionary like: , You can use iterable unpacking:
code :
>>> d = {'keyname': ['foo', 'bar']}
>>> [(k, v)] = d.items()
>>> k
'keyname'
>>> v
['foo', 'bar']
>>>
>>> from timeit import timeit
>>> d = {'keyname': ['foo', 'bar']}
>>> timeit('[(k, v)] = d.items()', 'from __main__ import d')
0.786108849029695
>>> timeit('[(k, v)] = d.iteritems()', 'from __main__ import d')
0.6730112346680928
>>>
Unpack Tuple within dictionary

Unpack Tuple within dictionary


By : Nguyễn Trường
Date : March 29 2020, 07:55 AM
hop of those help? Your value is not a tuple, it is a list with one element, which is a tuple; the nested tuple has 3 elements. Because the outer container has only one element, your unpacking assignment fails.
The following does work, by adding another layer, provided there is always exactly one element:
code :
for key, ((cut, fIn, fOUT),) in cutDict.iteritems():
for key, [(cut, fIn, fOUT)] in cutDict.iteritems():
>>> cutDict = {'scene1': [('scene3', 1001, 1125)]}
>>> for key, ((cut, fIn, fOUT),) in cutDict.iteritems():
...     print key, cut, fIn, fOut
...
scene1 scene3 1001 1125
>>> for key, [(cut, fIn, fOUT)] in cutDict.iteritems():
...     print key, cut, fIn, fOut
...
scene1 scene3 1001 1125
for key, cuts in cutDict.iteritems():
    for cut, fIn, fOUT in cuts:
        # ...
Passing variable containing format in struct.unpack format specifier

Passing variable containing format in struct.unpack format specifier


By : user7130857
Date : March 29 2020, 07:55 AM
This might help you I have length of a string a variable such as, , Found a way to do that.
code :
struct.unpack(('<{0}L'.format(variable/4)), another_variable)
Unpack Dictionary of Dataframes

Unpack Dictionary of Dataframes


By : byteo
Date : March 29 2020, 07:55 AM
hope this fix your issue I have used sqlalchemy to read in tables from a PostgreSQL database, and they are stored in a dictionary (keys are table names, values are the dataframes). I can access them one at a time, but would like to scale the solution so that I have all tables from the given database/schema as pandas dataframes. My code as-is to import is the following: , One thing I could think of is to use namedtuple from collections:
code :
from collections import namedtuple
import pandas as pd

supposedly_your_dict = {
    'table_1': pd.DataFrame(columns=['t_1_col_1', 't_1_col_2']),
    'table_2': pd.DataFrame(columns=['t_2_col_1', 't_2_col_2']),
    'table_3': pd.DataFrame(columns=['t_3_col_1', 't_3_col_2'])
}

DBSnapshot = namedtuple('DBSnapshot', supposedly_your_dict.keys())

db_tables = DBSnapshot(**supposedly_your_dict)
Empty DataFrame
Columns: [t_1_col_1, t_1_col_2]
Index: []
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