I have a list that I want to join it together as a string but I don't want it to be in the same order as the list. it gives me : "sdfg123" but I want it to provide me with the members in different orders each time that I run the program.
~
password=['s','d','f','g','1','2','3']
join_pass=''.join(password)
print(join_pass)
~
You could try to shuffle the list before joining:
import random
password=['s','d','f','g','1','2','3']
random.shuffle(password)
join_pass = ''.join(password)
print(join_pass)
Related
I am new to photon and I would like to know how I can create a list of elements in a list in order to print them on the terminal manually with a number in front of it.
lst = ["country", "inhabitants", "CO2-Emissions" ]
def make_header(lst):
first_column = []
for column_names in range(len(lst)):
first_column.append([])
for list in range(1):
column_names[first_column].append(list)
print(lst)
make_header(lst)
the output should be:
country
inhabitants
Co2 Emissions
Sadly I have no clue how to do that correctly. Could you give me a hint on that?
I want the columns displayed on the terminal as above.
I am not allowed to use maps, arrays, imports and other "fancy" and practical stuff so I approached for a nested loop.
Kind regards
Boris
i have two lists listA and listB of type object
ListA[name=abc, age=34, weight=0, height=0] data collected from excel sheet
ListB[name=null, age=0, weight=70, height=6] data collected from database
Now i want to combine both the lists into a single list
MergedList[name=abc, age=34, weight=70, height=6]
Note: my obj class has more than 15 properties so adding each property one by one using getProperty() will be time-consuming.is there a better way?
Convert them to a Map where the key is the name of the object ( you denoting the elements as name=abc suggests they are name/value pairs ).
Map<String,MyMysteriousObject> converted = list.stream().collect( Collectors.toMap(MyMysteriousObject::getName, Function.identity() ) );
( replace the getName with what ever function you use to get the name of your object )
And then just merge the maps. How to merge maps is described here for example.
While at it, consider replacing the List with Map in your entire code. Will surely save a lot of work elsewhere too.
But if you have to have a list again, just List<MyMysteriousObject> resultList = new ArrayList<>(resultMap);
I have the following code which I have to build upon (i.e. it can't be written a different way). I know there are other better ways of achieving the end result, but I want to use this code and then repeat it to make a list.
from random import choice
number_list = range(1,1001) # Creates a list from 1 to 1000
random_from_list = choice(number_list) # Chooses a random number from the list
I want to now repeat the choice function above 100 times, then print that list of 100 random numbers that I have chosen from my list of 1000 numbers. I have read up on "for" loops but I can't see how to apply it here.
If you don't need to build up your list you could just print them one at a time:
for _ in range(100):
print(choice(number_list))
If you want to build your list first you can use a "list comprehension":
choices = [choice(number_list) for _ in range(100)]
print(choices)
for i in range(100):
print(choice(number_list))
I'm using scrapy to iteratively scrape some data, and the data is being output as two lists through each iteration. I want to combine the two lists into one list at each iteration, so that in the end I will have one big list with many sublists(each sublist being the combination of the two lists created from each iteration)
That may be confusing so I will show my current output and code:
using Scrapy I"m iterating in the following way,
for i in response.css(''tr.insider....."):
i.css(a.tab-link:text).extract() #creating the first list
i.css('td::text').extract() #creating the second list
So the current output is something like this
[A,B,C] #first iteration
[1,2,3]
[D,E,F] #second iteration
[4,5,6]
[G,H,I] #third iteration
[7,8,9]
Desired output is
[[A,B,C,1,2,3], [D,E,F,4,5,6],[G,H,I,7,8,9]]
I tried the following code but I'm getting a list of None.
x =[]
for i in response.css(''tr.insider....."):
x.append(i.css(a.tablink::text).extract().extend(i.css('td::text').extract()))
But the return is just
None
None
None
None
None.....
Thanks!
extend function returns None, so you always append None to x.
For your purpose, I this is what you want:
for i in response.css(''tr.insider....."):
i.css('a.tab-link:text, td::text').extract()
You can simply add two lists together and append them to your results list.
results = []
for i in response.css("tr.insider....."):
first = i.css(a.tab-link:text).extract()
second = i.css('td::text').extract()
# combine both and append to results
results.append(first + second)
print(results)
# e.g.: [[A,B,C,1,2,3], [D,E,F,4,5,6],[G,H,I,7,8,9]]
I have a list of pk's and I would like to get the result in the same order that my list is defined... But the order of the elements is begging changed. How any one help me?
print list_ids
[31189, 31191, 31327, 31406, 31352, 31395, 31309, 30071, 31434, 31435]
obj_opor=Opor.objects.in_bulk(list_ids).values()
for o in obj_oportunidades:
print o
31395 31435 31434 30071 31309 31406 31189 31191 31352 31327
This object should be used in template to show some results to the user... But how you can see, the order is different from the original list_ids
Would have been nice to have this feature in SQL - sorting by a known list of values.
Instead, what you could do is:
obj_oportunidades=Opor.objects.in_bulk(list_ids).values()
all_opor = []
for o in obj_oportunidades:
print o
all_opor.append(o)
for i in list_ids:
if i in all_opor:
print all_opor.index(i)
Downside is that you have to get all the result rows first and store them before getting them in the order you want. (all_opor could be a dictionary above, with the table records stored in the values and the PKeys as dict keys.)
Other way, create a temp table with (Sort_Order, Pkey) and add that to the query:
Sort_Order PKey
1 31189
2 31191
...
So when you sort on Sort_Order and Opor.objects, you'll get Pkeys it in the order you specify.
I found a solution in: http://davedash.com/2010/02/11/retrieving-elements-in-a-specific-order-in-django-and-mysql/ it's suited me perfectly.
ids = [a_list, of, ordered, ids]
addons = Addon.objects.filter(id__in=ids).extra(
select={'manual': 'FIELD(id,%s)' % ','.join(map(str,ids))},
order_by=['manual'])
This code do something similiar to MySQL "ORDER BY FIELD".
This guy: http://blog.mathieu-leplatre.info/django-create-a-queryset-from-a-list-preserving-order.html
Solved the problem for both MySQL and PostgreSQL!
If you are using PostgreSQL go to that page.