TLDR: I have 2 arrays indices = numpy.arange(9) and another that contains some of the numbers in indices (maybe none at all, maybe it'll contain [2,4,7]). The output I'd like for this example is [0,1,3,5,6,8]. What method can be used to achieve this?
Edit: I found a method which works somewhat: casting both arrays to a set then taking the difference of the two does give the correct result, but as a set, even if I pass this result to a numpy.array(). I'll update this if I find a solution for that.
Edit2: Casting the result of the subtraction to a list, then casting passing that to a numpy.array() resolved my issue.
I guess I posted this question a little prematurely, given that I found the solution for it myself, but maybe this'll be useful to somebody in future!
You can make use of boolean masking:-
indices[~numpy.isin(indices,[2,4,7])]
Explanation:-
we are using numpy.isin() method to find out the values exists or not in incides array and then using ~ so that this gives opposite result and finally we are passing this boolean mask to indices
Related
I'm planning to overwrite a Field value using pandas but that does not seem to work. Am i missing anything as part of the code below?
`for row_no in range(df.shape[0]):
rowIndex = df.index[row_no]
if re.search('Fex|Process|PIP|VIP|Generic|Mobility', df.loc[rowIndex].VPC_Sub_Cat, re.I):
print(df.loc[rowIndex].Headline)
print(df.loc[rowIndex].VPC_Sub_Cat)
print(df.loc[rowIndex].Final_Result)
df.loc[rowIndex].Final_Result = 0
print(df.loc[rowIndex].Final_Result)
break`
The output that I get after running this piece of code is:
This is the description of the issue...
VPC-Generic
1
1
Also can i achieve the same thing using a function and applying that on a data frame? kindly let me know.
df.loc[rowIndex].Final_Result is equals to (in most situation though...)
df.loc[rowIndex]['Final_Result']
This will cause a chained assignment (see here #Warning)
Whether a copy or a reference is returned for a setting operation, may depend on the context. This is sometimes called chained assignment and should be avoided. See Returning a View versus Copy.
And then from Returning a view versus a copy
But it turns out that assigning to the product of chained indexing has inherently unpredictable results.
So using df.loc[rowIndex, 'Final_Result'] to make sure that the value you assigned is view, not copy.
I've been working on this for a few days now. Searched Stackoverflow and other sites for solutions but none of them appear to work. Most of the postings I've found are quite old (before 2013) so I'm thinking this is not the right way to do this.
I thought this would work:
[localMutableDictionary addEntriesFromDictionary:deviceDictionary];
localMutableDictionary remains null
I've worked around this using an array of integers instead of a mutable dictionary. But that doesn't give me the right result when I add the array to an NSDictionary for subsequent processing with NSJSONSerialisation. Values from my array don't get double quote marks around them. The json receiver / parser is expecting values in quotes (runs with json produced in VB code for a similar app). I can use an alternative parser to work around this, but I would rather get a clean solution.
This is probably a case of there being a simple syntax that I haven't managed to find, or that I'm just using an out-of-date style. Or I may just be adding my array to the NSDictionary "the wrong way". A solution for either method would work for me - thank you.
I have created a simple LabView program shown below that attempts to flatten an array [1,0,3] and then unflatten it and print out the contents.
However, I am unsuccessful in doing so. What am I doing wrong?
What am I doing wrong?
You're not going through tutorials or you're not reading the context help for the unflatten function (Ctrl+H) or you're not reading the full help for the function (right click>>Help) or you're not looking at the examples (from the help or Help>>Find Examples). Take your pick (preferably all four).
If you want an actual answer it is that LV is strictly typed, and therefore you need to tell the unflatten function which data type you want it to output (1D DBL array) and you're not doing that, but the real answer is what's in the previous paragraph - you should use those tools to learn how to find such an answer yourself.
The string returned by Flatten to String only contains the data, not the description of what data type was passed in, so in order to unflatten it again you need to tell Unflatten from String what type it was. You do this by wiring some data of the appropriate type (any data - if it's an array it can be an empty one) to the Type terminal.
I don't think this is immediately obvious from the LabVIEW 2012 help but I think it's fairly clear if you follow the link from the Unflatten from String help page to one of the examples. The Read Flattened Data.vi example has an array wired to the Type input.
Specifically, I have a model that has a field like this
pub_date = models.DateField("date published")
I want to be able to easily grab the object with the most recent pub_date. What is the easiest/best way to do this?
Would something like the following do what I want?
Edition.objects.order_by('pub_date')[:-1]
obj = Edition.objects.latest('pub_date')
You can also simplify things by putting get_latest_by in the model's Meta, then you'll be able to do
obj = Edition.objects.latest()
See the docs for more info. You'll probably also want to set the ordering Meta option.
Harley's answer is the way to go for the case where you want the latest according to some ordering criteria for particular Models, as you do, but the general solution is to reverse the ordering and retrieve the first item:
Edition.objects.order_by('-pub_date')[0]
Note:
Normal python lists accept negative indexes, which signify an offset from the end of the list, rather than the beginning like a positive number. However, QuerySet objects will raise AssertionError: Negative indexing is not supported. if you use a negative index, which is why you have to do what insin said: reverse the ordering and grab the 0th element.
Be careful of using
Edition.objects.order_by('-pub_date')[0]
as you might be indexing an empty QuerySet. I'm not sure what the correct Pythonic approach is, but the simplest would be to wrap it in an if/else or try/catch:
try:
last = Edition.objects.order_by('-pub_date')[0]
except IndexError:
# Didn't find anything...
But, as #Harley said, when you're ordering by date, latest() is the djangonic way to do it.
This has already been answered, but for more reference, this is what Django Book has to say about Slicing Data on QuerySets:
Note that negative slicing is not supported:
>>> Publisher.objects.order_by('name')[-1]
Traceback (most recent call last):
...
AssertionError: Negative indexing is not supported.
This is easy to get around, though. Just change the order_by()
statement, like this:
>>> Publisher.objects.order_by('-name')[0]
Refer the link for more such details. Hope that helps!
Here is the single line from one of my functions to test if any objects in my array have a given property with a matching value
Return ((From tag In DataCache.Tags Where (tag.FldTag = strtagname) Select tag).Count = 1)
WHERE....
DataCache.Tags is an array of custom objects
strtagname = "brazil"
and brazil is definitely a tag name stored within one of the custom objects in the array.
However the function continually returns false.
Can someone confirm to me that the above should or should not work.
and if it wont work can someone tell me the best way to test if any of the objects in the array contain a property with a specific value.
I suppose in summary I am looking for the equivalent of a SQL EXISTS statement.
Many thanks in hope.
Your code is currently checking whether the count is exactly one.
The equivalent of EXISTS in LINQ is Any. You want something like:
Return DataCache.Tags.Any(Function(tag) tag.FldTag = strtagname)
(Miraculously it looks like that syntax may be about right... it looks like the docs examples...)
Many Thanks for the response.
Your code did not work. Then I realised that I was comparing to an array value so it would be case sensitive.
However glad I asked the question, as I found a better way than mine.
Many thanks again !