Get continent name from country using pycountry - python

How to convert continent name from country name using pycountry. I have a list of country like this
country = ['India', 'Australia', ....]
And I want to get continent name from it like.
continent = ['Asia', 'Australia', ....]


All the tools you need are provided in pycountry-convert.
Here is an example of how you can create your own function to make a direct conversion from country to continent name, using pycountry's tools:
import pycountry_convert as pc
def country_to_continent(country_name):
country_alpha2 = pc.country_name_to_country_alpha2(country_name)
country_continent_code = pc.country_alpha2_to_continent_code(country_alpha2)
country_continent_name = pc.convert_continent_code_to_continent_name(country_continent_code)
return country_continent_name
# Example
country_name = 'Germany'
print(country_to_continent(country_name))
Out[1]: Europe
Hope it helps!
Remember you can access function descriptions using '?? function_name'
?? pc.country_name_to_country_alpha2


Looking at the documentation, would something like this do the trick?:
country_alpha2_to_continent_code()
It converts country code(eg: NO, SE, ES) to continent name.
If first you need to acquire the country code you could use:
country_name_to_country_alpha2(cn_name, cn_name_format="default")
to get the country code from country name.
Full example:
import pycountry_convert as pc
country_code = pc.country_name_to_country_alpha2("China", cn_name_format="default")
print(country_code)
continent_name = pc.country_alpha2_to_continent_code(country_code)
print(continent_name)


from pycountry_convert import country_alpha2_to_continent_code, country_name_to_country_alpha2
continents = {
'NA': 'North America',
'SA': 'South America',
'AS': 'Asia',
'OC': 'Australia',
'AF': 'Africa',
'EU': 'Europe'
}
countries = ['India', 'Australia']
[continents[country_alpha2_to_continent_code(country_name_to_country_alpha2(country))] for country in countries]
Don't know what to do with Antarctica continent ¯_(ツ)_/¯

Related

Django: Filter by less verbose (database name) on TextChoices class

I have a model:
class RegulationModel(models.Model):
class Markets(models.TextChoices):
EUROPE = 'E', 'Europe'
US = 'US', 'US'
JAPAN = 'J', 'Japan'
KOREA = 'K', 'Korea'
BRAZIL = 'B', 'Brazil'
INDIA = 'I', 'India'
CHINA = 'C', 'China'
market = models.CharField(verbose_name='Market:', max_length=2, choices=Markets.choices)
[... more fields ..]
How do I filter on the more verbose choice in the Markets class?
For example say I want to filter with "China", i'd like to do something like:
regulation = RegulationModel.objects.get(market__verbose='China')
(I know this isn't possible but it's just to get an idea)
The reason why I am doing this is because I am getting the choice from AJAX therefore the format is not ["C","US"] .etc.
I was able to convert the verbose to the database value via:
>> market = 'China'
>> market_db = {y:x for x, y in RegulationModel.Markets._value2label_map_.items()}[market]
>> market_db
'C'
And then I could use this to query but it doesn't seem very pythonic.

I believe you can do following:
china_code = list(filter(lambda x: x[1] == 'China', RegulationModel.Market.choices)).pop()[1]
regulation = RegulationModel.objects.get(market=china_code)
But it is a bit hacky. I recommend using django-useful and its implementation of Choices.

How to avoid very long if-elif-elif-else statements in Python function

Is there a smart way to shorten very long if-elif-elif-elif... statements?
Let's say I have a function like this:
def very_long_func():
something = 'Audi'
car = ['VW', 'Audi', 'BMW']
drinks = ['Cola', 'Fanta', 'Pepsi']
countries = ['France', 'Germany', 'Italy']
if something in car:
return {'type':'car brand'}
elif something in drinks:
return {'type':'lemonade brand'}
elif something in countries:
return {'type':'country'}
else:
return {'type':'nothing found'}
very_long_func()
>>>> {'type': 'car brand'}
The actual function is much longer than the example. What would be the best way to write this function (not in terms of speed but in readability)
I was reading this, but I have trouble to apply it to my problem.

You can't hash lists as dictionary values. So go other way round. Create a mapping of type -> list. And initialize your output with the default type. This allows you to keep on adding new types to your mapping without changing any code.
def very_long_func():
something = 'Audi'
car = ['VW', 'Audi', 'BMW']
drinks = ['Cola', 'Fanta', 'Pepsi']
countries = ['France', 'Germany', 'Italy']
out = {'type': 'nothing found'} # If nothing matches
mapping = {
'car brand': car,
'lemonade brand': drinks,
'country': countries
}
for k,v in mapping.items() :
if something in v:
out['type'] = k # update if match found
break
return out # returns matched or default value

you can create dictionary like this and then use map_dict.
from functools import reduce
car = ['VW', 'Audi', 'BMW']
drinks = ['Cola', 'Fanta', 'Pepsi']
countries = ['France', 'Germany', 'Italy']
li = [car, drinks, countries]
types = ['car brand', 'lemonade brand', 'country', 'nothing found']
dl = [dict(zip(l, [types[idx]]*len(l))) for idx, l in enumerate(li)]
map_dict = reduce(lambda a, b: dict(a, **b), dl)

Try this:
def create_dct(lst, flag):
return {k:flag for k in lst}
car = ['VW', 'Audi', 'BMW']
drinks = ['Cola', 'Fanta', 'Pepsi']
countries = ['France', 'Germany', 'Italy']
merge_dcts = {}
merge_dcts.update(create_dct(car, 'car brand'))
merge_dcts.update(create_dct(drinks, 'lemonade brand'))
merge_dcts.update(create_dct(countries, 'country'))
something = 'Audi'
try:
print("type: ", merge_dcts[something])
except:
print("type: nothing found")

You can simulate a switch statement with a helper function like this:
def switch(v): yield lambda *c: v in c
The your code could be written like this:
something = 'Audi'
for case in switch(something):
if case('VW', 'Audi', 'BMW'): name = 'car brand' ; break
if case('Cola', 'Fanta', 'Pepsi'): name = 'lemonade brand' ; break
if case('France', 'Germany', 'Italy'): name = 'country' ; break
else: name = 'nothing found'
return {'type':name}
If you don't have specific code to do for each value, then a simple mapping dictionary would probably suffice. For ease of maintenance, you can start with a category-list:type-name mapping and expand it before use:
mapping = { ('VW', 'Audi', 'BMW'):'car brand',
('Cola', 'Fanta', 'Pepsi'):'lemonade brand',
('France', 'Germany', 'Italy'):'country' }
mapping = { categ:name for categs,name in mapping.items() for categ in categs }
Then your code will look like this:
something = 'Audi'
return {'type':mapping.get(something,'nothing found')}
using a defaultdict would make this even simpler to use by providing the 'nothing found' value automatically so you could write: return {'type':mapping[something]}

Extracting text after string

I want to extract the string after "name=" from the following text. I have written the following regular expression but it isn't really working. The desired output is [Taal, Muntinlupa city]
text = [ "id='00e5885868b4d7ed', url='https://api.twitter.com/1.1/geo/id/00e5885868b4d7ed.json', place_type='city', name='Taal', full_name='Taal, Calabarzon', country_code='PH', country='Republic of the Philippines'",
"id='00c699d656122ebe', url='https://api.twitter.com/1.1/geo/id/00c699d656122ebe.json', place_type='city', name='Muntinlupa City', full_name='Muntinlupa City, National Capital Region', country_code='PH', country='Republic of the Philippines']
matched_vals = [re.findall(r'(?<=name\=).*(?=\s)',tweet) for tweet in text]

Use pattern r"name='(.+?)'"
Ex:
import re
text = [ "id='00e5885868b4d7ed', url='https://api.twitter.com/1.1/geo/id/00e5885868b4d7ed.json', place_type='city', name='Taal', full_name='Taal, Calabarzon', country_code='PH', country='Republic of the Philippines'",
"id='00c699d656122ebe', url='https://api.twitter.com/1.1/geo/id/00c699d656122ebe.json', place_type='city', name='Muntinlupa City', full_name='Muntinlupa City, National Capital Region', country_code='PH', country='Republic of the Philippines'"
]
for i in text:
print(re.search(r"name='(.+?)'", i).group(1))
Output:
Taal
Muntinlupa City

Create a dictionary out of the string, and that take the value of the key 'name':
dicts = []
for dic in text:
dicts.append(ast.literal_eval(dic))
and then you can you these name (and other data very efficient):
for d in dicts:
print(d['name'])

How to format strings properly in python?

I have some data scraped from a website in a string as shown below:
myDatastr =
'United States3.43M+57,9421M138K+282Brazil1.88M+20,2861.21M72,833+733India907K+28,701571K23,727+500Russia734K+6,537504K11,439+104Peru330K+3,797221K12,054+184Chile318K287K7,024Mexico304K+4,685189K35,491+485United Kingdom290K+65044,830+21South Africa288K138K4,172Iran260K+2,349223K13,032+203Spain256K+2,045150K28,406+3Pakistan254K+2,753171K5,320+69Italy243K+169195K34,967+13Saudi Arabia235K+2,852170K2,704+20Turkey214K+1,008196K5,382+19Germany200K+159185K9,138+1Bangladesh187K+3,09998,3172,391+39France172K78,59730,029Colombia154K+3,83265,8095,455+148Canada108K71,8418,790Qatar104K101K149Argentina103K+3,0991,903+58Mainland China85,568+3Egypt83,00124,9753,935Iraq79,73546,9883,250Indonesia76,981+1,28236,6893,656+50Sweden75,826+315,536+0Ecuador68,459+5895,9005,063+16Belarus65,114+18255,492468+4Belgium62,707Kazakhstan59,899+1,64634,190375+0Oman58,179+1,31837,257259+9Philippines57,006+74720,3711,599+65Kuwait55,50845,356393United Arab Emirates55,19845,513334Ukraine54,13326,5031,398Netherlands51,308+1016,156+0Bolivia49,25015,2941,866Panama47,17323,919932Portugal46,81831,0651,662Singapore46,283+32242,54126+0Dominican Republic45,50622,441903Israel40,632+1,33619,395365+1Poland38,190+29926,0481,576+5Afghanistan34,45521,2541,012Nigeria33,513+59513,671744+4Bahrain33,47628,425104Romania32,94821,6921,901Switzerland32,94629,6001,686Armenia32,151+18219,865573+8Guatemala29,7424,3211,244Honduras28,579+489789+15Ireland25,638+1023,3491,746+0Ghana24,98821,067139Azerbaijan24,570+52015,640313+8Japan21,868Algeria19,689+49414,0191,018+7Moldova19,439+17412,793649+2Austria18,94817,000708Serbia18,360Nepal16,945+1443,65238+0Morocco15,93612,934255Cameroon15,173+25711,928359+0Uzbekistan13,591+4878,03063+3South Korea13,512+3312,282289+0Czechia13,2388,373353Denmark13,147609Côte d'Ivoire12,766Kyrgyzstan11,538+488149+15Kenya10,2942,946197Sudan10,250+0+0Australia9,980+1837,769108+0El Salvador9,9785,755267Venezuela9,707+2422,67193+4Norway8,984+08,138253+0Malaysia8,725+78,520122Senegal8,198+1215,514150+3North Macedonia8,1974,326385Democratic Republic of the Congo8,075+623,620190+0Costa Rica8,0362,30431Ethiopia7,7662,430128Bulgaria7,525Finland7,2956,800329Palestine7,037Bosnia and Herzegovina6,981+4823,179226+5Haiti6,727+732,924139+4Tajikistan6,551+46+0French Guiana6,170+22129+3Guinea6,141+974,86237+0Gabon5,942+03,00446+0Mauritania5,275+149+3Kosovo5,118+1872,370108+6Djibouti4,972+4+0Luxembourg4,956+314,183111+0Madagascar4,867+289+1Central African Republic4,288+0+0Hungary4,247+133,073595+0Greece3,826+311,374193+0Croatia3,775+532,514119+0Albania3,5712,01495Thailand3,220+33,09058+0Equatorial Guinea3,071+084251+0Somalia3,059+81,30693+1Paraguay2,9801,29325Nicaragua2,846+01,75091+0Maldives2,762+672,29013+0Mayotte2,711+0+0Sri Lanka2,646+1061,98111+0Malawi2,43074739Cuba2,428+62,26887+0Mali2,411Lebanon2,334+166+0South Sudan2,148+9+0Republic of the Congo2,103+0+0Estonia2,0141,89569Slovakia1,90228Iceland1,90010Zambia1,895+01,34842+0Lithuania1,869Guinea-Bissau1,842+077326+0Slovenia1,841+14+0Cape Verde1,698+75+0Sierra Leone1,642+71,17563+0New Zealand1,545+022+0Hong Kong1,522+521,2178+1Yemen1,498+33424+7Libya1,433Benin1,378+0+0Eswatini1,351Rwanda1,337+38+1Tunisia1,263Montenegro1,221Jordan1,179+3+0Latvia1,173+01,01930+0Mozambique1,157+22+0Niger1,099+097868+0Burkina Faso1,033+13Uganda1,025Cyprus1,021+7+0Liberia1,010+12+4Georgia995+9Uruguay98731Zimbabwe985+3+0Chad880+6+1Namibia861+72281+0Andorra85580352Suriname780+3952618Jamaica759+110+0São Tomé and Príncipe727+228414+0Togo720San Marino716+3+0Malta674+06589+0Réunion593+16+0Tanzania509+0+0Angola506Taiwan4514387Syria417+2319+3Botswana399+85381+0Vietnam372+2Mauritius342+033010+0Isle of Man336+031224+0Myanmar (Burma)331+1+0Jersey329+431+0Comoros317+32967+0Guyana30015517Burundi269+82071+0Martinique255+0+0Guernsey25223813Guernsey252+0+0Lesotho245+49333+1Eritrea232+0Mongolia230+3Cayman Islands201+01941+0Guadeloupe190+0+0Faroe Islands188+01880Gibraltar180+0Cambodia1651330Bermuda150+01379+0Brunei141+01383+0Trinidad and Tobago133+01178+0Northern Cyprus1131044The Bahamas111+3+0Monaco1094Aruba105+0993Barbados103+5+0Seychelles100+0110Turkmenistan10000Liechtenstein85+0+0Bhutan84760Sint Maarten78+06315+0Antigua and Barbuda74+0573+0Turks and Caicos Islands72122The Gambia64+0343+0French Polynesia62+0600Macao46450Saint Martin44+0+0Belize37+0202+0Saint Vincent and the Grenadines35+0296Fiji26+0180Curaçao25+0241+0Timor-Leste24+0240Grenada23+0230Saint Lucia22+0190New Caledonia21+0210Laos19+0190Åland Islands19Dominica18+0180Saint Kitts and Nevis17150Falkland Islands (Islas Malvinas)13+0130Greenland13+0130Montserrat12101Vatican City12+0120Papua New Guinea1180British Virgin Islands8+071+0Caribbean Netherlands7+0Saint Barthélemy6+0Anguilla3+030Saint Pierre and Miquelon2+010Western Sahara'
I want to get the data like:
[United States,3.43M,+57,942,1M,138K,+282]
[Brazil,1.88M,+20,286,1.21M,72,833,+733]
Tried different things but did not work.

Taking a subset of your string, splitting it into a multi-line string for readability.
First capture the countries to a list, and use this list to obtain all values between countries.
Finally output to a list of lists as desired:
import re
text = """
United States3.43M+57,9421M138K+282Brazil1.88M+20,2861.21M72,833+733
India907K+28,701571K23,727+500Russia734K+6,537504K11,439+104
Peru330K+3,797221K12,054+184Chile318K287K7,024Mexico304K+4,685189K35,491+485
United Kingdom290K+65044,830+21South Africa288K138K4,172
Iran260K+2,349223K13,032+203Spain256K+2,045150K28,406+3
Pakistan254K+2,753171K5,320+69Italy243K+169195K34,967+13
Saudi Arabia235K+2,852170K2,704+20Turkey214K+1,008196K5,382+19
Germany200K+159185K9,138+1
"""
pattern = r"([a-z]{3,}(?: [a-z]{2,})?)"
regex = re.compile(pattern, re.I)
countries = [''.join([i for i in r if not i.isdigit()]).rstrip('K') for r in re.findall(regex, text)]
out = []
for idx, country in enumerate(countries):
if idx < len(countries) -1:
pattern = fr'{country}(.*){countries[idx+1]}'
regex = re.compile(pattern, re.I | re.DOTALL)
result = re.search(regex, text).group(1).strip()
result = result.replace('+', ',+').split(',')
tmp = [country]
for i in result:
tmp.append(i)
out.append(tmp)
else:
result = text.split(countries[-1])[-1].strip()
result = result.replace('+', ',+').split(',')
tmp = [countries[-1]]
for i in result:
tmp.append(i)
out.append(tmp)
for country in out:
print(country)
Returns:
['United States', '3.43M', '+57', '9421M138K', '+282']
['Brazil', '1.88M', '+20', '2861.21M72', '833', '+733']
['India', '907K', '+28', '701571K23', '727', '+500']
['Russia', '734K', '+6', '537504K11', '439', '+104']
['Peru', '330K', '+3', '797221K12', '054', '+184']
['Chile', '318K287K7', '024']
['Mexico', '304K', '+4', '685189K35', '491', '+485']
['United Kingdom', '290K', '+65044', '830', '+21']
['South Africa', '288K138K4', '172']
['Iran', '260K', '+2', '349223K13', '032', '+203']
['Spain', '256K', '+2', '045150K28', '406', '+3']
['Pakistan', '254K', '+2', '753171K5', '320', '+69']
['Italy', '243K', '+169195K34', '967', '+13']
['Saudi Arabia', '235K', '+2', '852170K2', '704', '+20']
['Turkey', '214K', '+1', '008196K5', '382', '+19']
['Germany', '200K', '+159185K9', '138', '+1']

How to select only label from a dict for graph title (Plotly Dash)

I have a dictionary with labels and key as shown below.
country = pd.read_csv('assets/countries.csv')
country = country.set_index('Code')
country_options = [{'label': f'({cod}) {country.loc[cod, "Country"]} ',
'value': f'{cod}'} for cod in country.index]
[{'label': '(AFG) Afghanistan ', 'value': 'AFG'}, {'label': '(ALB) Albania ', 'value': 'ALB'}, {'label': '(DZA) Algeria '.....
I also have a callback to update a graph and would like to add the title depending on the option selected from a dropdown list.
def update_graph(indi=indicator_options,coun=country_options):
df = indicator_select(coun, indi)
graph_content = dict(data=line_plot(df),
layout=go.Layout(
title=coun,
yaxis=dict(hoverformat='.1f'),
)
)
return graph_content
However, the title=coun will return me the 3 letter symbol (value) of the dict, but I need the label with the symbol and the name. Is it possible?

The value prop of the dropdown will only ever return the value part of the options dictionary, and not the label part. What you could do is create a mapping for the values you need, like this:
country_labels = [{f'{cod}': f'({cod}) {country.loc[cod, "Country"]}'} for cod in country.index]
And then just do title=country_labels[coun],

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