I am trying to create a function that takes up a string of numbers and outputs the maximum and minimum values. Here is my code
def high_and_low(numbers):
numbers = map(int, numbers.split())
max_n = max(numbers)
print(max_n)
min_n = min(numbers)
return(max_n, min_n)
But I get the following error: ValueError: min() arg is an empty sequence. So I assume that it does not read the negative values, but I dont know why..
I assume you're using Python 3.x, where map() was changed to return a generator rather than an explicit list of results as in Python 2.x. Calling max() on this generator exhausted it, leaving no elements for min() to iterate over.
One solution would be to convert this generator to a list, perhaps numbers = list(numbers) as the second line of your function. As a list, you can iterate it as many times as you need to.
Related
I am trying to write a code which calculates the HCF of two numbers but I am either getting a error or an empty list as my answer
I was expecting the HCF, My idea was to get the factors of the 2 given numbers and then find the common amongst them then take the max out of that
For future reference, do not attach screenshots. Instead, copy your code and put it into a code block because stack overflow supports code blocks. To start a code block, write three tildes like ``` and to end it write three more tildes to close. If you add a language name like python, or javascript after the first three tildes, syntax highlighting will be enabled. I would also create a more descriptive title that more accurately describes the problem at hand. It would look like so:
Title: How to print from 1-99 in python?
for i in range(1,100):
print(i)
To answer your question, it seems that your HCF list is empty, and the python max function expects the argument to the function to not to be empty (the 'arg' is the HCF list). From inspection of your code, this is because the two if conditions that need to be satisfied before anything is added to HCF is never satisfied.
So it could be that hcf2[x] is not in hcf and hcf[x] is not in hcf[x] 2.
What I would do is extract the logic for the finding of the factors of each number to a function, then use built in python functions to find the common elements between the lists. Like so:
num1 = int(input("Num 1:")) # inputs
num2 = int(input("Num 2:")) # inputs
numberOneFactors = []
numberTwoFactors = []
commonFactors = []
# defining a function that finds the factors and returns it as a list
def findFactors(number):
temp = []
for i in range(1, number+1):
if number%i==0:
temp.append(i)
return temp
numberOneFactors = findFactors(num1) # populating factors 1 list
numberTwoFactors = findFactors(num2) # populating factors 2 list
# to find common factors we can use the inbuilt python set functions.
commonFactors = list(set(numberOneFactors).intersection(numberTwoFactors))
# the intersection method finds the common elements in a set.
I am using this line of code which has the replace method to combine low frequency values in the column
psdf['method_name'] = psdf['method_name'].replace(small_categoris, 'Other')
The error I am getting is:
'to_replace' should be one of str, list, tuple, dict, int, float
So I tried to run this line of code before the replace method
psdf['layer'] = psdf['layer'].astype("string")
Now the column is of type string but the same error still appears. For the context, I am working on pandas api on spark. Also, is there a more efficient way than replace? especially if we want to do the same for more than one column.
I want to count the number of occurences in a dataframe, and I need to do it using the following function:
for x in homicides_prec.reset_index().DATE.drop_duplicates():
count= homicides_prec.loc[x]['VICTIM_AGE'].count()
print(count)
However, this only works for when the specific Date is repeated. It does not work when dates only appear once, and I don't understand why. I get this error:
TypeError: count() takes at least 1 argument (0 given)
That said, it really doesn't make sense to me, because I get that error for this specific value (which only appears once on the dataframe):
for x in homicides_prec.reset_index().DATE[49:50].drop_duplicates():
count= homicides_prec.loc[x]['VICTIM_AGE'].count()
print(count)
However, I don't get the error if I run this:
homicides_prec.loc[homicides_prec.reset_index().DATE[49:50].drop_duplicates()]['VICTIM_AGE'].count()
Why does that happen??? I can't use the second option because I need to use the for loop.
More info, in case it helps: The problem seems to be that, when I run this (without counting), the output is just a number:
for x in homicides_prec.reset_index().DATE[49:50].drop_duplicates(): count= homicides_prec.loc[x]['VICTIM_AGE']
print(count)
Output: 33
So, when I add the .count it will not accept that input. How can I fix this?
There are a few issues with the code you shared, but the shortest answer is that when x appears only once you are not doing a slice, rather you are accessing some value.
if x == '2019-01-01' and that value appears twice then
homicides_prec.loc[x]
will be a pd.DataFrame with two rows, and
homicides_prec.loc[x]['VICTIM_AGE']
will be a pd.Series object with two rows, and it will happily take a .count() method.
However, if x == '2019-01-02' and that date is unique, then
homicides_prec.loc[x]
will be a pd.Series representing row where the index is x
From that we see that
homicides_prec.loc[x]['VICTIM_AGE']
is a single value, so .count() does not make sense.
Working with Julia 1.1:
The following minimal code works and does what I want:
function test()
df = DataFrame(NbAlternative = Int[], NbMonteCarlo = Int[], Similarity = Float64[])
append!(df.NbAlternative, ones(Int, 5))
df
end
Appending a vector to one column of df. Note: in my whole code, I add a more complicated Vector{Int} than ones' return.
However, #code_warntype test() does return:
%8 = invoke DataFrames.getindex(%7::DataFrame, :NbAlternative::Symbol)::AbstractArray{T,1} where T
Which means I suppose, thisn't efficient. I can't manage to get what this #code_warntype error means. More generally, how can I understand errors returned by #code_warntype and fix them, this is a recurrent unclear issue for me.
EDIT: #BogumiłKamiński's answer
Then how one would do the following code ?
for na in arr_nb_alternative
#show na
for mt in arr_nb_montecarlo
println("...$mt")
append!(df.NbAlternative, ones(Int, nb_simulations)*na)
append!(df.NbMonteCarlo, ones(Int, nb_simulations)*mt)
append!(df.Similarity, compare_smaa(na, nb_criteria, nb_simulations, mt))
end
end
compare_smaa returns a nb_simulations length vector.
You should never do such things as it will cause many functions from DataFrames.jl to stop working properly. Actually such code will soon throw an error, see https://github.com/JuliaData/DataFrames.jl/issues/1844 that is exactly trying to patch this hole in DataFrames.jl design.
What you should do is appending a data frame-like object to a DataFrame using append! function (this guarantees that the result has consistent column lengths) or using push! to add a single row to a DataFrame.
Now the reason you have type instability is that DataFrame can hold vector of any type (technically columns are held in a Vector{AbstractVector}) so it is not possible to determine in compile time what will be the type of vector under a given name.
EDIT
What you ask for is a typical scenario that DataFrames.jl supports well and I do it almost every day (as I do a lot of simulations). As I have indicated - you can use either push! or append!. Use push! to add a single run of a simulation (this is not your case, but I add it as it is also very common):
for na in arr_nb_alternative
#show na
for mt in arr_nb_montecarlo
println("...$mt")
for i in 1:nb_simulations
# here you have to make sure that compare_smaa returns a scalar
# if it is passed 1 in nb_simulations
push!(df, (na, mt, compare_smaa(na, nb_criteria, 1, mt)))
end
end
end
And this is how you can use append!:
for na in arr_nb_alternative
#show na
for mt in arr_nb_montecarlo
println("...$mt")
# here you have to make sure that compare_smaa returns a vector
append!(df, (NbAlternative=ones(Int, nb_simulations)*na,
NbMonteCarlo=ones(Int, nb_simulations)*mt,
Similarity=compare_smaa(na, nb_criteria, nb_simulations, mt)))
end
end
Note that I append here a NamedTuple. As I have written earlier you can append a DataFrame or any data frame-like object this way. What "data frame-like object" means is a broad class of things - in general anything that you can pass to DataFrame constructor (so e.g. it can also be a Vector of NamedTuples).
Note that append! adds columns to a DataFrame using name matching so column names must be consistent between the target and appended object.
This is different in push! which also allows to push a row that does not specify column names (in my example above I show that a Tuple can be pushed).
I have a custom keyword in the robot framework which counts the items of a list. This works already in my underlying python file and prints the number 5 when five elements exists in a list.
Then I want to bring this value to the robot framework. But instead of a number I get:
${N_groups} is <built-in method count of list object at 0x03B01D78>
The code of the robot file:
*** Test Cases ***
Count Groups
${N_groups} Setup Groups Count Groups
log to console ${N_groups}
How to get item-count of the list as an integer value?
Here is a part of my python file:
#keyword(name="Count Groups")
def count_groups(self):
N = self.cur_page.count_groups()
return N
And a more low level python file:
def count_groups(self):
ele_tc = self._wait_for_treecontainer_loaded(self._ef.get_setup_groups_treecontainer())
children_text = self._get_sublist_filter(ele_tc, lambda ele: ele.find_element_by_tag_name('a').text,
True)
return children_text.count
Your function count_groups is returning children_text.count. children_text is a list, and you're returning the count method of that object, which explains the error that you're seeing. It's no different than if you did something like return [1,2,3].count.
Perhaps you intend to actually call the count function and return the results? Or, perhaps you are intending to return the length of the list? It's hard to see what the intent of the code is.
In either case, robot is reporting exactly what you're doing: you're returning a reference to a function, not an integer. My guess is that what you really want to do is return the number of items in the list, in which case you should change the return statement to:
return len(children_text)