I have a Stata program that outputs a local scalar of space-separated variable names.
I have to run the program twice on two samples (same dta) and store the union (intersection - variable names appearing in both scalars) as a new space-separated local scalar (for input to another program).
I can't figure out how to split (per spaces) and or test the occurrences of variable names in each.
Stata has a bunch of extended macro functions to use on lists that you can find with help macrolists, where you can see that A & B returns the intersection of A and B. If A="a b c d" and B="b c f g", then A & B = "b c".
This allows you to do something like this:
clear
scalar l1="vara varb varc"
scalar l2="varc vard vare"
local l1 = scalar(l1)
local l2 = scalar(l2)
local inter: list l1 & l2
scalar inter="`inter'"
scalar list inter
You convert the scalars to locals, get their union, and convert that into a scalar. It is probably easier to just modify your code to use locals rather than scalars so you don't have to deal with conversions.
I am not sure I perfectly understand your question, if this is not the appropriate answer, please add an example for us to work with.
Here is the code that checks two space-separated macros and gets their intersection, even if it's not the most elegant, unless your macros are huge it should still be quite fast.
local list1 first list here
local list2 list two here
local intersection
foreach l1 in `list1' {
foreach l2 in `list2' {
// if they overlap, add to the intersection macro
if "`l1'" == "`l2'" {
local intersection `intersection' `l1'
}
}
}
mac list // show the macros stored currently in the do file
Related
I have a program which receives input from another program and use it for further operations. The input can be a list, set, tuple but for further operations a list is needed. So I am converting input to list.
The problem arises when input my program receives is a list/set/tuple with just one element like below. The
import itertools
def not_mine(c):
d = {'John':['mid', 'forward'],
'Lana':['mid'],
'Jacob':['defence', 'mid'],
'Ian':['goal', 'mid']}
n = itemgetter(*c)(d)
n = list(set(itertools.chain.from_iterable(n)))
return n
def mine(c):
name = not_mine(c)
name_1 = list(name)
print(name_1)
mine(['Jacob', 'Ian'])
['defence', 'goal', 'mid']
mine(['Lana'])
['i', 'm', 'd']
Is there any way to prevent the second case? It should be a list of one element ['mid'].
Iterators
The function set uses the first argument as an iterator to create a sequence of items. str is natively an iterator. In other words, you can loop over a str and you'll assign to the for variable each character in the string per iteration.
for whatami in "hi!":
print(whatami)
h
i
!
If you want to treat a single string input as a single item, explicitly pass an iterator argument to set (list works the same way, BTW) with a single item in it. Tuple is, also, an iterator. Let's try to use it to prove our theory
t1 = ('ourstring', )
print(f"t1 is of type {type(t1)}")
s1 = set(t1)
print(s1)
t1 is of type <class 'tuple'>
{'ourstring'}
It works!
What we've done with ('ourstring', ) is explicitly define a tuple with one item. There's a familiar delimiter, ,, used to say "this tuple is instantiated with only one item".
Input
To separate situations between ingesting a list of items and one string item, you can consider two approaches.
The most straight-forward way is to agree on a delimiter in the input such as comma separated values. firstvalue,secondvalue,etc. The down side of this is that you'll quickly run into limitations of what kind of data you can receive.
To ease your development, argparse is strongly recommended command line arguments. It is a built-in, battle-hardened package made for this type of task. The docs's first example even shows a multi-value field.
I started coding in Python 4 days ago, so I'm a complete newbie. I have a dataset that comprises an undefined number of dictionaries. Each dictionary is the x and y of a point in the coordinates.
I'm trying to compute the summatory of xy by nesting the loop that multiplies xy within the loop that sums the products.
However I haven't been able to figure out how to multiply the values for the two keys in each dictionary (so far I only got to multiply all the x*y)
So far I've got this:
If my data set were to be d= [{'x':0, 'y':0}, {'x':1, 'y':1}, {'x':2, 'y':3}]
I've got the code for the function that calculates the product of each pair of x and y:
def product_xy (product_x_per_y):
prod_xy =[]
n = 0
for i in range (len(d)):
result = d[n]['x']*d[n]['y']
prod_xy.append(result)
n+1
return prod_xy
I also have the function to add up the elements of a list (like prod_xy):
def total_xy_prod (sum_prod):
all = 0
for s in sum_prod:
all+= s
return all
I've been trying to find a way to nest this two functions so that I can iterate through the multiplication of each x*y and then add up all the products.
Make sure your code works as expected
First, your functions have a few mistakes. For example, in product_xy, you assign n=0, and later do n + 1; you probably meant to do n += 1 instead of n + 1. But n is also completely unnecessary; you can simply use the i from the range iteration to replace n like so: result = d[i]['x']*d[i]['y']
Nesting these two functions: part 1
To answer your question, it's fairly straightforward to get the sum of the products of the elements from your current code:
coord_sum = total_xy_prod(product_xy(d))
Nesting these two functions: part 2
However, there is a much shorter and more efficient way to tackle this problem. For one, Python provides the built-in function sum() to sum the elements of a list (and other iterables), so there's no need create total_xy_prod. Our code could at this point read as follows:
coord_sum = sum(product_xy(d))
But product_xy is also unnecessarily long and inefficient, and we could also replace it entirely with a shorter expression. In this case, the shortening comes from generator expressions, which are basically compact for-loops. The Python docs give some of the basic details of how the syntax works at list comprehensions, which are distinct, but closely related to generator expressions. For the purposes of answering this question, I will simply present the final, most simplified form of your desired result:
coord_sum = sum(e['x'] * e['y'] for e in d)
Here, the generator expression iterates through every element in d (using for e in d), multiplies the numbers stored in the dictionary keys 'x' and 'y' of each element (using e['x'] * e['y']), and then sums each of those products from the entire sequence.
There is also some documentation on generator expressions, but it's a bit technical, so it's probably not approachable for the Python beginner.
I have a simple issue with my code in Python 3.6.
I am reading a csv and storing int values in a list called Total.
I ask Python user to enter a number n=.. (in this example n=9).
I create n (9) empty list with :
for j in range (1,n+1):
command=""
command="list"+str(j)+"=[]
Now, I have list1, list2, ... list9
Then, I want to append these lists by reading Total starting at different elements and reading each n (9) elements.
For example:
list1=[Tot[0],Tot[8]...] list2=[Tot[1],Tot[9],...]
To do so, I want to something like
for k in range (0,n):
for a in range (0+k,len(Total),n):
listk.append(Total[a])
My problem is here, Python doesn't recognise the integer in listk such as:
list1, ... list9
Is there a certain way to do it ? Maybe by using a class ?
For your problem I suggest using a dictionary. A dictionary is a type of data structure in python that allows to store pairs of information - for example lists and their names. This way you can create n separate named lists, store them all inside one dictionary, and append them as necessary. To start at different indices, I make use of the modulo operator.
This should do the trick:
# shorthand syntax for creating a dictionary with n empty lists: list1,...,listn
lists = {"list" + str(i + 1): [] for i in range(n)}
for k in range(n):
list = lists.get("list" + str(k + 1))
for j in range(n):
list.append(Total[(j + k) % n])
A few sidenotes:
It seems you have a few misunderstandings concerning Python syntax. When you declare the variable "command" inside your loop, Python is saving a pointer to a specific place in stack memory. However you are inside a loop, so "command" is overwritten in every iteration of the loop, and the old pointer is lost.
The following line:
command="list"+str(j)+"=[]
is invalid in Python. The syntax to create lists using square brackets is:
variable = [item1, item2, item3] # or variable = [] for an empty list
You cannot mix between strings and list creation in the way you attempted to.
This might be a simple question. However, I wanted to get some clarifications of how the following code works.
a = np.arange(8)
a
array([1,2,3,4,5,6,7])
Example Function = a[0:-1]+a[1:]/2.0
In the Example Function, I want to draw your attention to the plus sign between the array a[0:-1]+a[1:]. How does that work? What does that look like?
For instance, is the plus sign (addition) adding the first index of each array? (e.g 1+2) or add everything together? (e.g 1+2+2+3+3+4+4+5+5+6+6+7)
Then, I assume /2.0 is just dividing it by 2...
A numpy array uses vector algebra in that you can only add two arrays if they have the same dimensions as you are adding element by element
a = [1,2,3,4,5]
b = [1,1,1]
a+b # will throw an error
whilst
a = [1,2,3,4,5]
b = [1,1,1,1,1]
a+b # is ok
The division is also element by element.
Now to your question about the indexing
a = [1,2,3,4,5]
a[0:-1]= [1,2,3,4]
a[1:] = [2,3,4,5]
or more generally a[index_start: index_end] is inclusive at the start_index but exclusive at the end_index - unless you are given a a[start_index:]where it includes everything up to and including the last element.
My final tip is just to try and play around with the structures - there is no harm in trying different things, the computer will not explode with a wrong value here or there. Unless you trying to do so of course.
If arrays have identical shapes, they can be added:
new_array = first_array.__add__(second_array)
This simple operation adds each value from first_array to each value in second_array and puts result into new_array.
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.
Closed 10 years ago.
Ive been working on a code that reads lines in a file document and then the code organizes them. However, i got stuck at one point and my friend told me what i could use. the code works but it seems that i dont know what he is doing at line 7 and 8 FROM THE BOTTOM. I used #### so you guys know which lines it is.
So, essentially how can you re-write those 2 lines of codes and why do they work? I seem to not understand dictionaries
from sys import argv
filename = input("Please enter the name of a file: ")
file_in=(open(filename, "r"))
print("Number of times each animal visited each station:")
print("Animal Id Station 1 Station 2")
animaldictionary = dict()
for line in file_in:
if '\n' == line[-1]:
line = line[:-1]
(a, b, c) = line.split(':')
ac = (a,c)
if ac not in animaldictionary:
animaldictionary[ac] = 0
animaldictionary[ac] += 1
alla = []
for key, value in animaldictionary:
if key not in alla:
alla.append(key)
print ("alla:",alla)
allc = []
for key, value in animaldictionary:
if value not in allc:
allc.append(value)
print("allc", allc)
for a in sorted(alla):
print('%9s'%a,end=' '*13)
for c in sorted(allc):
ac = (a,c)
valc = 0
if ac in animaldictionary:
valc = animaldictionary[ac]
print('%4d'%valc,end=' '*19)
print()
print("="*60)
print("Animals that visited both stations at least 3 times: ")
for a in sorted(alla):
x = 'false'
for c in sorted(allc):
ac = (a,c)
count = 0
if ac in animaldictionary:
count = animaldictionary[ac]
if count >= 3:
x = 'true'
if x is 'true':
print('%6s'%a, end=' ')
print("")
print("="*60)
print("Average of the number visits in each month for each station:")
#(alla, allc) =
#for s in zip(*animaldictionary.keys()):
# (alla,allc).append(s)
#print(alla, allc)
(alla,allc,) = (set(s) for s in zip(*animaldictionary.keys())) ##### how else can you write this
##### how else can you rewrite the next code
print('\n'.join(['\t'.join((c,str(sum(animaldictionary.get(ac,0) for a in alla for ac in ((a,c,),))//12)))for c in sorted(allc)]))
print("="*60)
print("Month with the maximum number of visits for each station:")
print("Station Month Number")
print("1")
print("2")
The two lines you indicated are indeed rather confusing. I'll try to explain them as best I can, and suggest alternative implementations.
The first one computes values for alla and allc:
(alla,allc,) = (set(s) for s in zip(*animaldictionary.keys()))
This is nearly equivalent to the loops you've already done above to build your alla and allc lists. You can skip it completely if you want. However, lets unpack what it's doing, so you can actually understand it.
The innermost part is animaldictionary.keys(). This returns an iterable object that contains all the keys of your dictionary. Since the keys in animaldictionary are two-valued tuples, that's what you'll get from the iterable. It's actually not necessary to call keys when dealing with a dictionary in most cases, since operations on the keys view are usually identical to doing the same operation on the dictionary directly.
Moving on, the keys gets wrapped up by a call to the zip function using zip(*keys). There's two things happening here. First, the * syntax unpacks the iterable from above into separate arguments. So if animaldictionary's keys were ("a1", "c1), ("a2", "c2"), ("a3", "c3") this would call zip with those three tuples as separate arguments. Now, what zip does is turn several iterable arguments into a single iterable, yielding a tuple with the first value from each, then a tuple with the second value from each, and so on. So zip(("a1", "c1"), ("a2", "c2"), ("a3", "c3")) would return a generator yielding ("a1", "a2", "a3") followed by ("c1", "c2", "c3").
The next part is a generator expression that passes each value from the zip expression into the set constructor. This serves to eliminate any duplicates. set instances can also be useful in other ways (e.g. finding intersections) but that's not needed here.
Finally, the two sets of a and c values get assigned to variables alla and allc. They replace the lists you already had with those names (and the same contents!).
You've already got an alternative to this, where you calculate alla and allc as lists. Using sets may be slightly more efficient, but it probably doesn't matter too much for small amounts of data. Another, more clear, way to do it would be:
alla = set()
allc = set()
for key in animaldict: # note, iterating over a dict yields the keys!
a, c = key # unpack the tuple key
alla.add(a)
allc.add(c)
The second line you were asking about does some averaging and combines the results into a giant string which it prints out. It is really bad programming style to cram so much into one line. And in fact, it does some needless stuff which makes it even more confusing. Here it is, with a couple of line breaks added to make it all fit on the screen at once.
print('\n'.join(['\t'.join((c,str(sum(animaldictionary.get(ac,0)
for a in alla for ac in ((a,c,),))//12)
)) for c in sorted(allc)]))
The innermost piece of this is for ac in ((a,c,),). This is silly, since it's a loop over a 1-element tuple. It's a way of renaming the tuple (a,c) to ac, but it is very confusing and unnecessary.
If we replace the one use of ac with the tuple explicitly written out, the new innermost piece is animaldictionary.get((a,c),0). This is a special way of writing animaldictionary[(a, c)] but without running the risk of causing a KeyError to be raised if (a, c) is not in the dictionary. Instead, the default value of 0 (passed in to get) will be returned for non-existant keys.
That get call is wrapped up in this: (getcall for a in alla). This is a generator expression that gets all the values from the dictionary with a given c value in the key
(with a default of zero if the value is not present).
The next step is taking the average of the values in the previous generator expression: sum(genexp)//12. This is pretty straightforward, though you should note that using // for division always rounds down to the next integer. If you want a more precise floating point value, use just /.
The next part is a call to '\t'.join, with an argument that is a single (c, avg) tuple. This is an awkward construction that could be more clearly written as c+"\t"+str(avg) or "{}\t{}".format(c, avg). All of these result in a string containing the c value, a tab character and the string form of the average calcualted above.
The next step is a list comprehension, [joinedstr for c in sorted(allc)] (where joinedstr is the join call in the previous step). Using a list comprehension here is a bit odd, since there's no need for a list (a generator expression would do just as well).
Finally, the list comprehension is joined with newlines and printed: print("\n".join(listcomp)). This is straightforward.
Anyway, this whole mess can be rewritten in a much clearer way, by using a few variables and printing each line separately in a loop:
for c in sorted(allc):
total_values = sum(animaldictionary.get((a,c),0) for a in alla)
average = total_values // 12
print("{}\t{}".format(c, average))
To finish, I have some general suggestions.
First, your data structure may not be optimal for the uses you are making of you data. Rather than having animaldict be a dictionary with (a,c) keys, it might make more sense to have a nested structure, where you index each level separately. That is, animaldict[a][c]. It might even make sense to have a second dictionaries containing the same values indexed in the reverse order (e.g. one is indexed [a][c] while another is indexed [c][a]). With this approach you might not need the alla and allc lists for iterating (you'd just loop over the contents of the main dictionary directly).
My second suggestion is about code style. Many of your variables are named poorly, either because their names don't have any meaning (e.g. c) or where the names imply a meaning that is incorrect. The most glaring issue is your key and value variables, which in fact unpack two pieces of the key (AKA a and c). In other situations you can get keys and values together, but only when you are iterating over a dictionary's items() view rather than on the dictionary directly.