I am working on a problem in Python and don't understand the answer.
for number in range(1, 10):
if number % 2 == 0:
print(number)
The answer to this problem is 2,4,6,8
Can anyone explain this answer?
range is a function in python which generates a sequence of integers, for example:
r=range(3)
returns a iterable object range(0,3) which generates sequence of integers from 0 to 3-1(2),inorder for you to see the elements in it , you can loop through it:
for i in r:
print(i)
#prints number from 0 to 3-1
Or, wrap it in a list:
list(range(3)) //returns [0,1,2]
range can take 3 params as input start,end and optionally step.The parameters start and end are basically lower and upper bounds to the sequence.In the above example since we have given only one integer range considers start as 0 and end as 3. This function range(start,end,[step]) generates integers in the following manner: start,start+1....end-1 considering the above example 0,0+1...3-1
if you give both the start and the end params to the range, the function generates integers from start upto but not including end, Example:
for i in range(3,8):print(i) #prints numbers from 3 to 8-1
if you give the third parameter which is the step(which is usually 1 by default), then range adds that number to the sequence :
list(range(3,8)) or list(range(3,8,1)) # will return [3,4,5,6,7],sequence generation will be like:3,3+1,(3+1)+1...
list(range(3,8,2)) #returns [3,5,7];3,3+2,(3+2)+2....
So , coming to your question now :
for number in range(1, 10): if number % 2 == 0: print(number)
In the above code you are basically telling python to loop over the sequence of integeres between 1 to 9 and print the numbers which are divisible by 2,which prints 2,4,6,8.
Hope this helped you :)
Related
I have column in a df with a size range with different sizeformats.
artikelkleurnummer size
6725 0161810ZWA B080
6726 0161810ZWA B085
6727 0161810ZWA B090
6728 0161810ZWA B095
6729 0161810ZWA B100
in the sizerange are also these other size formats like XS - XXL, 36-50 , 36/38 - 52/54, ONE, XS/S - XL/XXL, 363-545
I have tried to get the prefix '0' out of all sizes with start with a letter in range (A:K). For exemple: Want to change B080 into B80. B100 stays B100.
steps:
1 look for items in column ['size'] with first letter of string in range (A:K),
2 if True change second position in string into ''
for range I use:
from string import ascii_letters
def range_alpha(start_letter, end_letter):
return ascii_letters[ascii_letters.index(start_letter):ascii_letters.index(end_letter) + 1]
then I've tried a for loop
for items in df['size']:
if df.loc[df['size'].str[0] in range_alpha('A','K'):
df.loc[df['size'].str[1] == ''
message
SyntaxError: unexpected EOF while parsing
what's wrong?
You can do it with regex and the pd.Series.str.replace -
df = pd.DataFrame([['0161810ZWA']*5, ['B080', 'B085', 'B090', 'B095', 'B100']]).T
df.columns = "artikelkleurnummer size".split()
replacement = lambda mpat: ''.join(g for g in mpat.groups() if mpat.groups().index(g) != 1)
df['size_cleaned'] = df['size'].str.replace(r'([a-kA-K])(0*)(\d+)', replacement)
Output
artikelkleurnummer size size_cleaned
0 0161810ZWA B080 B80
1 0161810ZWA B085 B85
2 0161810ZWA B090 B90
3 0161810ZWA B095 B95
4 0161810ZWA B100 B100
TL;DR
Find a pattern "LetterZeroDigits" and change it to "LetterDigits" using a regular expression.
Slightly longer explanation
Regexes are very handy but also hard. In the solution above, we are trying to find the pattern of interest and then replace it. In our case, the pattern of interest is made of 3 parts -
A letter in from A-K
Zero or more 0's
Some more digits
In regex terms - this can be written as r'([a-kA-K])(0*)(\d+)'. Note that the 3 brackets make up the 3 parts - they are called groups. It might make a little or no sense depending on how exposed you have been to regexes in the past - but you can get it from any introduction to regexes online.
Once we have the parts, what we want to do is retain everything else except part-2, which is the 0s.
The pd.Series.str.replace documentation has the details on the replacement portion. In essence replacement is a function that takes all the matching groups as the input and produces an output.
In the first part - where we identified three groups or parts. These groups are accessed with the mpat.groups() function - which returns a tuple containing the match for each group. We want to reconstruct a string with the middle part excluded, which is what the replacement function does
sizes = [{"size": "B080"},{"size": "B085"},{"size": "B090"},{"size": "B095"},{"size": "B100"}]
def range_char(start, stop):
return (chr(n) for n in range(ord(start), ord(stop) + 1))
for s in sizes:
if s['size'][0].upper() in range_char("A", "K"):
s['size'] = s['size'][0]+s['size'][1:].lstrip('0')
print(sizes)
Using a List/Dict here for example.
I am trying to get my head around the Pandas module and started learning about the Series data structure.
I have created the following Series in Spyder :-
songs = pd.Series(data = [145,142,38,13], name = "Count")
I can obtain information about the Series index using the code:-
songs.index
The output of the above code is as follows:-
My question is where it states Start = 0 and Stop = 4, what are these referring to?
I have interpreted start = 0 as the first element in the Series is in row 0.
But i am not sure what Stop value refers to as there are no elements in row 4 of the Series?
Can some one explain?
Thank you.
This concept as already explained adequately in the comments (indexing is at minus one the count of items) is prevalent in many places.
For instance, take the list data structure-
z = songs.to_list()
[145, 142, 38, 13]
len(z)
4 # length is four
# however indexing stops at i-1 position 'i' being the length/count of items in the list.
z[4] # this will raise an IndexError
# you will have to start at index 0 going till only index 3 (i.e. 4 items)
z[0], z[1], z[2], z[-1] # notice how -1 can be used to directly access the last element
Python beginner here.
I already have the solution to the question but I'm not understanding why the "add" variable in the solution plays a role of creating exceptions to remove numbers between 6 and 9. I already tried Python Tutor but still not understanding. Many thanks in advance!
QUESTION: Return the sum of the numbers in the array, except ignore sections of numbers starting with a 6 and extending to the next 9 (every 6 will be followed by at least one 9). Return 0 for no numbers.
Sample Solution code
def summer_69(arr):
total = 0
add = True
for num in arr:
while add:
if num != 6:
total += num
break
else:
add = False
while not add:
if num != 9:
break
else:
add = True
break
return total
Sample answers:
summer_69([1, 3, 5]) --> 9
summer_69([4, 5, 6, 7, 8, 9]) --> 9
summer_69([2, 1, 6, 9, 11]) --> 14
You can think of the variable "add" as a flag. I think that might be a better name for this variable in this instance.
It is only being used to tell if you have run into a 6 within the sequence of numbers in the array, then once it has been set it goes through an arbitrary amount of numbers in the array until it gets a 9 and then it resets the flag.
It may help to rename the variable "add" as "flag". Have your new variable "flag" default to False and then if you run into a 6 set "flag" to true. Once the flag is on do not add any trailing numbers in the sequence until you run into the number 9 then reset to false.
Perhaps that will help the readability. Naming variables is the hardest part of programming.
I am trying to iterate over rows in a Pandas Dataframe using the itertuples()-method, which works quite fine for my case. Now i want to check if a specific value ('x') is in a specific tuple. I used the count() method for that, as i need to use the number of occurences of x later.
The weird part is, for some Tuples that works just fine (i.e. in my case (namedtuple[7].count('x')) + (namedtuple[8].count('x')) ), but for some (i.e. namedtuple[9].count('x')) i get an AttributeError: 'int' object has no attribute 'count'
Would appreciate your help very much!
Apparently, some columns of your DataFrame are of object type (actually a string)
and some of them are of int type (more generally - numbers).
To count occurrences of x in each row, you should:
Apply a function to each row which:
checks whether the type of the current element is str,
if it is, return count('x'),
if not, return 0 (don't attempt to look for x in a number).
So far this function returns a Series, with a number of x in each column
(separately), so to compute the total for the whole row, this Series should
be summed.
Example of working code:
Test DataFrame:
C1 C2 C3
0 axxv bxy 10
1 vx cy 20
2 vv vx 30
Code:
for ind, row in df.iterrows():
print(ind, row.apply(lambda it:
it.count('x') if type(it).__name__ == 'str' else 0).sum())
(in my opinion, iterrows is more convenient here).
The result is:
0 3
1 1
2 1
So as you can see, it is possible to count occurrences of x,
even when some columns are not strings.
if we write 12wkd3, how to choose/filter 123 as integer in octave?
example in octave:
A = input("A?\n")
A?
12wkd3
A = 123
while 12wkd3 is user keyboard input and A = 123 is the expected answer.
assuming that the general form you're looking for is taking an arbitrary string from the user input, removing anything non-numeric, and storing the result it as an integer:
A = input("A? /n",'s');
A = int32(str2num(A(isdigit(A))));
example output:
A?
324bhtk.p89u34
A = 3248934
to clarify what's written above:
in the input statement, the 's' argument causes the answer to get stored as a string, otherwise it's evaluated by Octave first. most inputs would produce errors, others may be interpreted as functions or variables.
isdigit(A) produces a logical array of values for A with a 1 for any character that is a 0-9 number, and 0 otherwise.
isdigit('a1 3 b.') = [0 1 0 1 0 0 0]
A(isdigit(A)) will produce a substring from A using only those values corresponding to a 1 in the logical array above.
A(isdigit(A)) = 13
that still returns a string, so you need to convert it into a number using str2num(). that, however, outputs a double precision number. so finally to get it to be an integer you can use int32()