I've got two generic lists in a vb.net program. I'd like to loop List_A and search List_A.ID in List_B, elements in common should be stored in a third list (LIST).
For Each n As BE_Busq In List_A
LIST = List_B.FindAll(Function(x As BE_Busq) x.ID = n.ID)
'' for each step, LIST should be incremented, not be replaced
Next
Method FindAll will return a generic list. How to increment LIST and not replaced it for each step in loop?
Try this:
LIST.addrange(List_B.FindAll(Function(x As BE_Busq) x.ID = n.ID))
You can use the AddRange method to add multiple items to a List.
For Each n As BE_Busq In List_A
LIST.AddRange(List_B.FindAll(Function(x As BE_Busq) x.ID = n.ID))
Next
Related
I created a form in Infopath with a rich text box field. The field is used to keep a list of usernames (first and last). I want to be able to keep a of count each entry and keep a tally. I then want to use that total # of entries to add or subtract from other fields. Is there any way to do that?
Is the rich text box field just a large string? If so you could just use python's built in split function, and either split by ("\r\n"), or (",").
Example:
u = "Bob, Michael, Jean"
x = u.split(",")
X will be a list of usernames. If you are using line breaks for each new username, then replace (",") with ("\r\n").
Now to count the items in a list you just need to iterate on the list you created with a for loop.
Example:
b = 0
u = "Bob, Michael, Jean"
x = u.split(",")
for i in x:
b += 1 // b will be the number of usernames
I have a dynamic list which is created based on value of n.
n = 3
drop_lst = ['a' + str(i) for i in range(n)]
df.drop(drop_lst)
But the above is not working.
Note:
My use case requires a dynamic list.
If I just do the below without list it works
df.drop('a0','a1','a2')
How do I make drop function work with list?
Spark 2.2 doesn't seem to have this capability. Is there a way to make it work without using select()?
You can use the * operator to pass the contents of your list as arguments to drop():
df.drop(*drop_lst)
You can give column name as comma separated list e.g.
df.drop("col1","col11","col21")
This is how drop specified number of consecutive columns in scala:
val ll = dfwide.schema.names.slice(1,5)
dfwide.drop(ll:_*).show
slice take two parameters star index and end index.
Use simple loop:
for c in drop_lst:
df = df.drop(c)
You can use drop(*cols) 2 ways .
df.drop('age').collect()
df.drop(df.age).collect()
Check the official documentation DataFrame.drop
I'm iterating through a dataframe (called hdf) and applying changes on a row by row basis. hdf is sorted by group_id and assigned a 1 through n rank on some criteria.
# Groupby function creates subset dataframes (a dataframe per distinct group_id).
grouped = hdf.groupby('group_id')
# Iterate through each subdataframe.
for name, group in grouped:
# This grabs the top index for each subdataframe
index1 = group[group['group_rank']==1].index
# If criteria1 == 0, flag all rows for removal
if(max(group['criteria1']) == 0):
for x in range(rank1, rank1 + max(group['group_rank'])):
hdf.loc[x,'remove_row'] = 1
I'm getting the following error:
TypeError: int() argument must be a string or a number, not 'Int64Index'
I get the same error when I try to cast rank1 explicitly I get the same error:
rank1 = int(group[group['auction_rank']==1].index)
Can someone explain what is happening and provide an alternative?
The answer to your specific question is that index1 is an Int64Index (basically a list), even if it has one element. To get that one element, you can use index1[0].
But there are better ways of accomplishing your goal. If you want to remove all of the rows in the "bad" groups, you can use filter:
hdf = hdf.groupby('group_id').filter(lambda group: group['criteria1'].max() != 0)
If you only want to remove certain rows within matching groups, you can write a function and then use apply:
def filter_group(group):
if group['criteria1'].max() != 0:
return group
else:
return group.loc[other criteria here]
hdf = hdf.groupby('group_id').apply(filter_group)
(If you really like your current way of doing things, you should know that loc will accept an index, not just an integer, so you could also do hdf.loc[group.index, 'remove_row'] = 1).
call tolist() on Int64Index object. Then the list can be iterated as int values.
simply add [0] to insure the getting the first value from the index
rank1 = int(group[group['auction_rank']==1].index[0])
How to sort somelist As List(of T) by the order set in another list sortorder As List(of Integer)? Both somelist and sortorder are of the same size and are indexed from 0 to n. Integers in the sortorder list determine the sort order: new index of item X in somelist = value of item X in sortorder.
Like this:
somelist = (itemA, itemB, itemC)
sortorder = (3, 1, 2)
somelist.sort()
somelist = (itemB, itemC, itemA)
I am trying to sort several equally sized lists using the predefined sort order.
You could use LINQ, although i hate the ugly method syntax in VB.NET:
somelist = somelist.
Select(Function(t, index) New With {.Obj = t, .Index = index}).
OrderBy(Function(x) sortorder(x.Index)).
Select(Function(x) x.Obj).
ToList()
This uses the overload of Enumerable.Select that projects the index of the item. The object and the index are stored in an anonymous type which is used for the ordering, finally i'm selecting the object and use ToList to build the ordered list.
Another way is to use Enumerable.Zip to merge both into an anonymous type:
Dim q = From x In somelist.Zip(sortorder, Function(t, sort) New With {.Obj = t, .Sort = sort})
Order By x.Sort Ascending
Select x.Obj
somelist = q.ToList()
If you want to order it descending, so the highest values first, use OrderByDescending in the method syntax and Order By x.Sort Descending in the query.
But why do you store such related informations in two different collections at all?
I have an array called #friend_comparisons which is populated with a number of user objects. I then sort the array using the following:
#friend_comparisons.sort! { |a,b| b.completions.where(:list_id => #list.id).first.counter <=> a.completions.where(:list_id => #list.id).first.counter }
This is sorting the array by a certain counter associated with each user (the specifics of which are not important to the question).
I want to find out how many user objects in the array have a counter that is greater than a certain number (let's say 5). How do I do this?
Here is how I am currently solving the problem:
#friends_rank = 1
for friend in #friend_comparisons do
if friend.completions.where(:list_id => #list.id).first.counter > #user_restaurants.count
#friends_rank = #friends_rank + 1
end
end
You can use Array#count directly.
#friend_comparisons.count {|friend| friend.counter >= 5 }
Docs: http://ruby-doc.org/core-2.2.0/Array.html#method-i-count
(same for ruby 1.9.3)
Array#select will get the job done.
Docs: http://www.ruby-doc.org/core-1.9.3/Array.html#method-i-select
You might do something like this:
number_of_users = #friend_comparisons.select{|friend| friend.counter >= 5 }.size