Im currently using micropython and it does not have the .zfill method.
What Im trying to get is to get the YYMMDDhhmmss of the UTC.
The time that it gives me for example is
t = (2019, 10, 11, 3, 40, 8, 686538, None)
I'm able to access the ones that I need by using t[:6]. Now the problem is with the single digit numbers, the 3 and 8. I was able to get it to show 1910113408, but I need to get 19101034008 I would need to get the zeroes before those 2. I used
t = "".join(map(str,t))
t = t[2:]
So my idea was to iterate over t and then check if the number is less than 10. If it is. I will add zeroes in front of it, replacing the number . And this is what I came up with.
t = (2019, 1, 1, 2, 40, 0)
t = list(t)
for i in t:
if t[i] < 10:
t[i] = 0+t[i]
t[i] = t[i]
print(t)
However, this gives me IndexError: list index out of range
Please help, I'm pretty new to coding/python.
When you use
for i in t:
i is not index, each item.
>>> for i in t:
... print(i)
...
2019
10
11
3
40
8
686538
None
If you want to use index, do like following:
>>> for i, v in enumerate(t):
... print("{} is {}".format(i,v))
...
0 is 2019
1 is 10
2 is 11
3 is 3
4 is 40
5 is 8
6 is 686538
7 is None
another way to create '191011034008'
>>> t = (2019, 10, 11, 3, 40, 8, 686538, None)
>>> "".join(map(lambda x: "%02d" % x, t[:6]))
'20191011034008'
>>> "".join(map(lambda x: "%02d" % x, t[:6]))[2:]
'191011034008'
note that:
%02d add leading zero when argument is lower than 10 otherwise (greater or equal 10) use itself. So year is still 4digit string.
This lambda does not expect that argument is None.
I tested this code at https://micropython.org/unicorn/
edited :
str.format method version:
"".join(map(lambda x: "{:02d}".format(x), t[:6]))[2:]
or
"".join(map(lambda x: "{0:02d}".format(x), t[:6]))[2:]
second example's 0 is parameter index.
You can use parameter index if you want to specify it (ex: position mismatch between format-string and params, want to write same parameter multiple times...and so on) .
>>> print("arg 0: {0}, arg 2: {2}, arg 1: {1}, arg 0 again: {0}".format(1, 11, 111))
arg 0: 1, arg 2: 111, arg 1: 11, arg 0 again: 1
I'd recommend you to use Python's string formatting syntax.
>> t = (2019, 10, 11, 3, 40, 8, 686538, None)
>> r = ("%d%02d%02d%02d%02d%02d" % t[:-2])[2:]
>> print(r)
191011034008
Let's see what's going on here:
%d means "display a number"
%2d means "display a number, at least 2 digits"
%02d means "display a number, at least 2 digits, pad with zeroes"
so we're feeding all the relevant numbers, padding them as needed, and cut the "20" out of "2019".
Related
I am trying to find the odd numbers and a multiple of 7 between a 1 to 100 and append them into an array. I have got this far:
var results: [Int] = []
for n in 1...100 {
if n / 2 != 0 && 7 / 100 == 0 {
results.append(n)
}
}
Your conditions are incorrect. You want to use "modular arithmetic"
Odd numbers are not divisible by 2. To check this use:
if n % 2 != 0
The % is the mod function and it returns the remainder of the division (e.g. 5 / 2 is 2.5 but integers don't have decimals, so the integer result is 2 with a remainder of 1 and 5 / 2 => 2 and 5 % 2 => 1)
To check if it's divisible by 7, use the same principle:
if n % 7 == 0
The remainder is 0 if the dividend is divisible by the divisor. The complete if condition is:
if n % 2 != 0 && n % 7 == 0
You can also use n % 2 == 1 because the remainder is always 1. The result of any mod function, a % b, is always between 0 and b - 1.
Or, using the new function isMultiple(of:, that final condition would be:
if !n.isMultiple(of: 2) && n.isMultiple(of: 7)
Swift 5:
Since Swift 5 has been released, you could use isMultiple(of:) method.
In your case, you should check if it is not multiple of ... :
if !n.isMultiple(of: 2)
Swift 5 is coming with isMultiple(of:) method for integers , so you can try
let res = Array(1...100).filter { !$0.isMultiple(of:2) && $0.isMultiple(of:7) }
Here is an efficient and concise way of getting the odd multiples of 7 less than or equal to 100 :
let results: [Int] = Array(stride(from: 7, through: 100, by: 14))
You can also use the built-in filter to do an operation on only qualified members of an array. Here is how that'd go in your case for example
var result = Array(1...100).filter { (number) -> Bool in
return (number % 2 != 0 && number % 7 == 0)
}
print(result) // will print [7, 21, 35, 49, 63, 77, 91]
You can read more about filter in the doc but here is the basics: it goes through each element and collects elements that return true on the condition. So it filters the array and returns what you want
I've got a list of some integers, e.g. [1, 2, 3, 4, 5, 10]
And I've another integer (N). For example, N = 19.
I want to check if my integer can be represented as a sum of any amount of numbers in my list:
19 = 10 + 5 + 4
or
19 = 10 + 4 + 3 + 2
Every number from the list can be used only once. N can raise up to 2 thousand or more. Size of the list can reach 200 integers.
Is there a good way to solve this problem?
4 years and a half later, this question is answered by Jonathan.
I want to post two implementations (bruteforce and Jonathan's) in Python and their performance comparison.
def check_sum_bruteforce(numbers, n):
# This bruteforce approach can be improved (for some cases) by
# returning True as soon as the needed sum is found;
sums = []
for number in numbers:
for sum_ in sums[:]:
sums.append(sum_ + number)
sums.append(number)
return n in sums
def check_sum_optimized(numbers, n):
sums1, sums2 = [], []
numbers1 = numbers[:len(numbers) // 2]
numbers2 = numbers[len(numbers) // 2:]
for sums, numbers_ in ((sums1, numbers1), (sums2, numbers2)):
for number in numbers_:
for sum_ in sums[:]:
sums.append(sum_ + number)
sums.append(number)
for sum_ in sums1:
if n - sum_ in sums2:
return True
return False
assert check_sum_bruteforce([1, 2, 3, 4, 5, 10], 19)
assert check_sum_optimized([1, 2, 3, 4, 5, 10], 19)
import timeit
print(
"Bruteforce approach (10000 times):",
timeit.timeit(
'check_sum_bruteforce([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 200)',
number=10000,
globals=globals()
)
)
print(
"Optimized approach by Jonathan (10000 times):",
timeit.timeit(
'check_sum_optimized([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 200)',
number=10000,
globals=globals()
)
)
Output (the float numbers are seconds):
Bruteforce approach (10000 times): 1.830944365834205
Optimized approach by Jonathan (10000 times): 0.34162875449254027
The brute force approach requires generating 2^(array_size)-1 subsets to be summed and compared against target N.
The run time can be dramatically improved by simply splitting the problem in two. Store, in sets, all of the possible sums for one half of the array and the other half separately. It can now be determined by checking for every number n in one set if the complementN-n exists in the other set.
This optimization brings the complexity down to approximately: 2^(array_size/2)-1+2^(array_size/2)-1=2^(array_size/2 + 1)-2
Half of the original.
Here is a c++ implementation using this idea.
#include <bits/stdc++.h>
using namespace std;
bool sum_search(vector<int> myarray, int N) {
//values for splitting the array in two
int right=myarray.size()-1,middle=(myarray.size()-1)/2;
set<int> all_possible_sums1,all_possible_sums2;
//iterate over the first half of the array
for(int i=0;i<middle;i++) {
//buffer set that will hold new possible sums
set<int> buffer_set;
//every value currently in the set is used to make new possible sums
for(set<int>::iterator set_iterator=all_possible_sums1.begin();set_iterator!=all_possible_sums1.end();set_iterator++)
buffer_set.insert(myarray[i]+*set_iterator);
all_possible_sums1.insert(myarray[i]);
//transfer buffer into the main set
for(set<int>::iterator set_iterator=buffer_set.begin();set_iterator!=buffer_set.end();set_iterator++)
all_possible_sums1.insert(*set_iterator);
}
//iterator over the second half of the array
for(int i=middle;i<right+1;i++) {
set<int> buffer_set;
for(set<int>::iterator set_iterator=all_possible_sums2.begin();set_iterator!=all_possible_sums2.end();set_iterator++)
buffer_set.insert(myarray[i]+*set_iterator);
all_possible_sums2.insert(myarray[i]);
for(set<int>::iterator set_iterator=buffer_set.begin();set_iterator!=buffer_set.end();set_iterator++)
all_possible_sums2.insert(*set_iterator);
}
//for every element in the first set, check if the the second set has the complemenent to make N
for(set<int>::iterator set_iterator=all_possible_sums1.begin();set_iterator!=all_possible_sums1.end();set_iterator++)
if(all_possible_sums2.find(N-*set_iterator)!=all_possible_sums2.end())
return true;
return false;
}
Ugly and brute force approach:
a = [1, 2, 3, 4, 5, 10]
b = []
a.size.times do |c|
b << a.combination(c).select{|d| d.reduce(&:+) == 19 }
end
puts b.flatten(1).inspect
ISSUE
I am trying to convert a 8 digit number into a date while in an array. Examples of the entries are 12282009 or 12202007. There are other malformed entries in the field including dates entered as strings. I want the 8 digit number to be formatted as 12/28/09 or 12/20/07 respectively. I keep getting a type mismatch error on the third to last line below. How do I do this??
CODE
Dim del()
ReDim del(1 To importwsRowCount, 1 To 1)
del = Range("AH1:AH" & importwsRowCount).Value
Dim delChars As Long
Dim delType As String
For i = LBound(del, 1) To UBound(del, 1)
delChars = Len(del(i, 1)) 'Determine length of entry
If IsNumeric(del(i, 1)) = True Then 'Determine datatype of entry
delType = "Numeric"
del(i, 1) = Abs(del(i, 1))
Else
delType = "String"
del(i, 1) = UCase(del(i, 1))
End If
If delType = "Numeric" Then
If delChars = 8 Then
del(i, 1) = DateSerial((Right(del(i, 1), 4)), (Left(del(i, 1), 2)), (Mid(del(i, 1), 3, 2))) '<-- TYPE MISMATCH ERROR
End If
End If
ENTRY TEMPLATES
SEPT. 25, 20 (No year, no year! Delete.)
SEPT (No year, useless, delete.)
N/A (Rubbish! Deleted.)
LONG TIME AG (What moron thought this was a good idea, delete.)
JUNE 30, 200 (Apparently the field will only hold 12 characters, delete.)
CHARGED OFF (Useless, delete.)
94 DAYS (Take all characters preceding space and subtract from other field containing order date to obtain delinquent date.)
94 DPD (DPD in someones bright mind stands for Days Past Due I believe. Same as above.)
2008-7-15 12 (Not sure what additional number is, take all characters before space and transform.)
INVALID (Delete.)
BLANK (Do nothing.)
4/2/4/09 (Malformed date, delete.)
1/1/009 (Same as above.)
12282009 (Use nested LEFT and RIGHT and CONCATENATE with / in between.)
9202011 (Add leading zero, then same as above.)
92410 (Add leading zero, this will transform to 09/24/10)
41261 (Days since 31/12/1899, this will transform to 12/08/12)
1023 (Days since delinquent, subtract from ORDER DATE to get delinquent date.)
452 (Same as above.)
12 (Same as above.)
1432.84 (Monetary value, mistakenly entered by low IQ lackey. Delete.)
Right(Left(del(i, 1), 2), 6) is nonsensical.
The Left(del(i, 1), 2) part happens first and returns a 2-character string. If you then apply Right(..., 6) to that 2-character string you get an error.
The Mid function is needed here: Mid(del(i, 1), 3, 2)
Running the Abs function earlier changed the array entry from being a Variant with subtype String to being a Variant with subtype Double. This shouldn't necessarily affect the Left/Mid/Right functions but try:
del(i, 1) = CStr(del(i, 1))
del(i, 1) = DateSerial((Right(del(i, 1), 4)), (Left(del(i, 1), 2)), (Mid(del(i, 1), 3, 2)))
We need to identify what the actual value causing the error is so:
If delType = "Numeric" Then
If delChars = 8 Then
On Error Goto DateMismatchError
del(i, 1) = DateSerial((Right(del(i, 1), 4)), (Left(del(i, 1), 2)), (Mid(del(i, 1), 3, 2))) '<-- TYPE MISMATCH ERROR
On Error Goto 0
End If
End If
' at the end of your Sub or Function - I'm assuming Sub here
Exit Sub
DateMismatchError:
MsgBox "Date mismatch: error number " & Err.Number & ", " & Err.Description & _
" caused by data value: |" & del(i, 1) & "| at row " & i & ". Original data " & _
"value is |" & Range("AH" & i).Value2 & "|, displayed value is |" & _
Range("AH" & i).Text & "|, number format is |" & Range("AH" & i).NumberFormat & "|"
End Sub
You can use this shorter code to replace your array elements with formatted dates
It cuts down the amount of testing inside the loop to two IFs. If numeric test is run first - there is no point running a longer lenint test for strings that are not 8 characters
The string functions Left$, Mid$ etc are much quicker than their variant cousins Left, Mid etc
I have made a substituion for your importwsRowCount variable in the code below
Updated code to handle and dump results, now handles string tests and non-compliantnumbers as per barrowc comments
The code below puts the new dates into a second array, skipping the invalid dates
The second array is then dumped at `AI``
Sub ReCut2()
Dim del()
Dim X()
Dim lngCnt As Long
del = Range("AH1:Ah10").Value2
ReDim X(1 To UBound(del, 1), 1 To UBound(del, 2))
Dim delChars As Long
Dim delType As String
For lngCnt = LBound(del, 1) To UBound(del, 1)
If IsNumeric(del(lngCnt, 1)) Then
If Len(Int((del(lngCnt, 1)))) = 8 Then X(lngCnt, 1) = DateSerial(Right$(del(lngCnt, 1), 4), Left$(del(lngCnt, 1), 2), Mid$(del(lngCnt, 1), 3, 2))
End If
Next
[ai1].Resize(UBound(X, 1), UBound(X, 2)).Value2 = X
End Sub
I noticed that Verilog rounds my real number results into integer results. For example when I look at simulator, it shows the result of 17/2 as 9. What should I do? Is there anyway to define something like a: output real reg [11:0] output_value ? Or is it something that has to be done by simulator settings?
Simulation only (no synthesis). Example:
x defined as a signed input and output_value defined as output reg.
output_value = ((x >>> 1) + x) + 5;
If x=+1 then output value has to be: 13/2=6.5.
However when I simulate I see output_value = 6.
Code would help, but I suspect your not dividing reals at all. 17 and 2 are integers, and so a simple statement like that will do integer division.
17 / 2 = 8 (not 9, always rounds towards 0)
17.0 / 2.0 = 8.5
In your second case
output_value = ((x >>> 1) + x) + 5
If x is 1, x >>> 1 is 0, not 0.5 because you've just gone off the bottom of the word.
output_value = ((1 >>> 1) + 1) + 5 = 0 + 1 + 5 = 6
There's nothing special about verilog here. This is true for the majority of languages.
Didn't find any similar question about this.
This is a final round Facebook question:
You are given a ring of boxes. Each box has a non-negative number on it, can be duplicate.
Write a function/algorithm that will tell you the order at which you select the boxes, that will give you the max sum.
The catch is, if you select a box, it is taken off the ring, and so are the two boxes next to it (to the right and the left of the one you selected).
so if I have a ring of
{10 3 8 12}
If I select 12, 8 and 10 will be destroyed and you are left with 3.
The max will be selecting 8 first then 10, or 10 first then 8.
I tried re-assign the boxes their value by take its own value and then subtracts the two next to is as the cost.
So the old ring is {10 3 8 12}
the new ring is {-5, -15, -7, -6}, and I will pick the highest.
However, this definitely doesn't work if you have { 10, 19, 10, 0}, you should take the two 10s, but the algorithm will take the 19 and 0.
Help please?
It is most likely dynamic programming, but I don't know how.
The ring can be any size.
Here's some python that solves the problem:
def sublist(i,l):
if i == 0:
return l[2:-1]
elif i == len(l)-1:
return l[1:-2]
else:
return l[0:i-1] + l[i+2:]
def val(l):
if len(l) <= 3:
return max(l)
else:
return max([v+val(m) for v,m in [(l[u],sublist(u,l)) for u in range(len(l))]])
def print_indices(l):
print("Total:",val(l))
while l:
vals = [v+val(m) for v,m in [(l[u],sublist(u,l)) for u in range(len(l)) if sublist(u,l)]]
if vals:
i = vals.index(max(vals))
else:
i = l.index(max(l))
print('choice:',l[i],'index:',i,'new list:',sublist(i,l))
l = sublist(i,l)
print_indices([10,3,8,12])
print_indices([10,19,10,0])
Output:
Total: 18
choice: 10 index: 0 new list: [8]
choice: 8 index: 0 new list: []
Total: 20
choice: 10 index: 0 new list: [10]
choice: 10 index: 0 new list: []
No doubt it could be optimized a bit. The key bit is val(), which calculates the total value of a given ring. The rest is just bookkeeping.