1) I have a little problem while operating with hex-values. Why is sendValue(hex(12289)) leading to an error (aborting script), while sendValue(0x3001) works?
def sendValue(value):
for i in range(16):
if (value & 0x8000):
print "1" # later this bit will be sent to a LC
else:
print "0" # later this bit will be sent to a LC
value <<= 1 # corrected this
def main():
sendValue(0x3001)
sendValue(hex(12289))
if __name__ == '__main__':
main()
2) I expected the output
0
0
1
1
0
0
0
0
0
0
0
0
0
0
0
1
But I just get 0
Your sendValue() function just prints the most significant bit of the number 16 times. You need to scan through each of the bits. Eg,
#!/usr/bin/env python
def sendValue(value):
print hex(value),
for i in range(16):
if (value & 0x8000):
print "1",
else:
print "0",
#Right-shift value to put next bit into the MSB position
value <<= 1
print
def main():
sendValue(0x3001)
sendValue(12289)
sendValue(0x123f)
if __name__ == '__main__':
main()
**output**
0x3001 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1
0x3001 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1
0x123f 0 0 0 1 0 0 1 0 0 0 1 1 1 1 1 1
Note that Python has a built-in bin() function to convert integers to bit strings:
>>> bin(0x3001)
'0b11000000000001'
>>> bin(0xfed)
'0b111111101101'
But if you don't want to use bin() for some reason, there are other ways to do it that are more compact than my earlier code. Eg, shifting value to the left and masking it with & 1, like this:
def sendValue(value):
print hex(value),
print ' '.join([str(value >> i & 1) for i in range(16, -1, -1)])
You got this error because the hex function returns string. Hexadecimal, binary, decimal are only representations of the integer value. 12289 and 0x3001 is the same, So you can just do that:
def main():
sendValue(0x3001)
sendValue(12289)
# Or convert string to int if you need
sendValue(int(hex(12289), 16))
Related
I would like to create some number of for-loops equal to the length of a list, and iterate through the values in that list. For example, if I had the list:
[1,2,3,4]
I would like the code to function like:
for i in range(1):
for j in range(2):
for k in range(3):
for l in range(4):
myfunc(inputs)
I understand I would need to do this recursively, but I'm not quite sure how. Ideally, I would even be able to iterate through these list values by a variable step; perhaps I want to count by two's for one loop, by .8's for another, etc. In that case, I would probably deliver the information in a format like this:
[[value,step],[value,step] ... [value,step],[value,step]]
So, how could I do this?
Not quite sure what you want at the very end, but here's a way to recursively set-up your loops:
test = [1,2,3,4]
def recursive_loop(test):
if len(test) == 1:
for i in range(test[0]):
print('hi') # Do whatever you want here
elif len(test) > 1:
for i in range(test[0]):
recursive_loop(test[1:])
recursive_loop(test)
You can certainly do it with recursion, but there's already a library function for that:
itertools.product
from itertools import product
def nested_loops(myfunc, l):
for t in product(*(range(n) for n in l)):
myfunc(*t)
## OR EQUIVALENTLY
# def nested_loops(myfunc, l):
# for t in product(*map(range, l)):
# myfunc(l)
nested_loops(print, [1, 2, 3, 4])
# 0 0 0 0
# 0 0 0 1
# 0 0 0 2
# 0 0 0 3
# 0 0 1 0
# 0 0 1 1
# ...
# 0 1 2 1
# 0 1 2 2
# 0 1 2 3
You can of course include steps too. Library function zip can be useful.
def nested_loops_with_steps_v1(myfunc, upperbounds, steps):
for t in product(*(range(0, n, s) for n,s in zip(upperbounds, steps))):
myfunc(*t)
nested_loops_with_steps_v1(print, [1,2,8,10], [1,1,4,5])
# 0 0 0 0
# 0 0 0 5
# 0 0 4 0
# 0 0 4 5
# 0 1 0 0
# 0 1 0 5
# 0 1 4 0
# 0 1 4 5
Or if your steps and upperbounds are already zipped together:
def nested_loops_with_steps_v2(myfunc, l):
for t in product(*(range(0, n, s) for n,s in l)):
myfunc(*t)
nested_loops_with_steps_v2(print, [(1,1),(2,1),(8,4),(10,5)])
# 0 0 0 0
# 0 0 0 5
# 0 0 4 0
# 0 0 4 5
# 0 1 0 0
# 0 1 0 5
# 0 1 4 0
# 0 1 4 5
I have the following:
df['PositionLong'] = 0
df['PositionLong'] = np.where(df['Alpha'] == 1, 1, (np.where(np.logical_and(df['PositionLong'].shift(1) == 1, df['Bravo'] == 1), 1, 0)))
This lines basically only take in df['Alpha'] but not the df['PositionLong'].shift(1).. It cannot recognize it but I dont understand why?
It produces this:
df['Alpha'] df['Bravo'] df['PositionLong']
0 0 0
1 1 1
0 1 0
1 1 1
1 1 1
However what I wanted the code to do is this:
df['Alpha'] df['Bravo'] df['PositionLong']
0 0 0
1 1 1
0 1 1
1 1 1
1 1 1
I believe the solution is to loop each row, but this will take very long.
Can you help me please?
You are looking for a recursive function, since a previous PositionLong value depends on Alpha, which itself is used to determine PositionLong.
But numpy.where is a regular function, so df['PositionLong'].shift(1) is evaluated as a series of 0 values, since you initialise the series with 0.
A manual loop need not be expensive. You can use numba to efficiently implement your recursive algorithm:
from numba import njit
#njit
def rec_algo(alpha, bravo):
res = np.empty(alpha.shape)
res[0] = 1 if alpha[0] == 1 else 0
for i in range(1, len(res)):
if (alpha[i] == 1) or ((res[i-1] == 1) and bravo[i] == 1):
res[i] = 1
else:
res[i] = 0
return res
df['PositionLong'] = rec_algo(df['Alpha'].values, df['Bravo'].values).astype(int)
Result:
print(df)
Alpha Bravo PositionLong
0 0 0 0
1 1 1 1
2 0 1 1
3 1 1 1
4 1 1 1
I am interested in generating all binary combination of N variables without having to implement a manual loop of iterating N times over N and each time looping over N/2 and so on.
Do we have such functionality in python?
E.g:
I have N binary variables:
pool=['A','B','C',...,'I','J']
len(pool)=10
I would like to generate 2^10=1024 space out of these such as:
[A B C ... I J]
iter0 = 0 0 0 ... 0 0
iter1 = 0 0 0 ... 0 1
iter2 = 0 0 0 ... 1 1
...
iter1022 = 1 1 1 ... 1 0
iter1023 = 1 1 1 ... 1 1
You see that I don't have repetitions here, each variable is enabled once per each of these iter's sequences. How can I do that using Python's itertools?
itertools.product with the repeat parameter is the simplest answer:
for A, B, C, D, E, F, G, H, I, J in itertools.product((0, 1), repeat=10):
The values of each variable will cycle fastest on the right, and slowest on the left, so you'll get:
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 1 1
0 0 0 0 0 0 0 1 0 0
etc. This may be recognizable to you: It's just the binary representation of an incrementing 10 bit number. Depending on your needs, you may actually want to just do:
for i in range(1 << 10):
then mask i with 1 << 9 to get the value of A, 1 << 8 for B, and so on down to 1 << 0 (that is, 1) for J. If the goal is just to print them, you can even get more clever, by binary stringifying and then using join to insert the separator:
for i in range(1 << 10):
print(' '.join('{:010b}'.format(i)))
# Or letting print insert the separator:
print(*'{:010b}'.format(i)) # If separator isn't space, pass sep='sepstring'
Working on a project for CS1 that prints out a grid made of 0s and adds shapes of certain numbered sizes to it. Before it adds a shape it needs to check if A) it will fit on the grid and B) if something else is already there. The issue I am having is that when run, the function that checks to make sure placement for the shapes is valid will always do the first and second shapes correctly, but any shape added after that will only "see" the first shape added when looking for a collision. I checked to see if it wasnt taking in the right list after the first time but that doesnt seem to be it. Example of the issue....
Shape Sizes = 4, 3, 2, 1
Python Outputs:
4 4 4 4 1 2 3 0
4 4 4 4 2 2 3 0
4 4 4 4 3 3 3 0
4 4 4 4 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
It Should Output:
4 4 4 4 3 3 3 1
4 4 4 4 3 3 3 0
4 4 4 4 3 3 3 0
4 4 4 4 2 2 0 0
0 0 0 0 2 2 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
What's going on here? Full Code is below...
def binCreate(size):
binlist = [[0 for col in range(size)] for row in range(size)]
return binlist
def binPrint(lst):
for row in range(len(lst)):
for col in range(len(lst[row])):
print(lst[row][col], end = " ")
print()
def itemCreate(fileName):
lst = []
for i in open(fileName):
i = i.split()
lst = i
lst = [int(i) for i in lst]
return lst
def main():
size = int(input("Bin Size: "))
fileName = str(input("Item Size File: "))
binList = binCreate(size)
blockList = itemCreate(fileName)
blockList.sort(reverse = True)
binList = checker(binList, len(binList), blockList)
binPrint(binList)
def isSpaceFree(binList, r, c, size):
if r + size > len(binList[0]):
return False
elif c + size > len(binList[0]):
return False
for row in range(r, r + size):
for col in range(c, c + size):
if binList[r][c] != 0:
return False
elif binList[r][c] == size:
return False
return True
def checker(binList, gSize, blockList):
for i in blockList:
r = 0
c = 0
comp = False
while comp != True:
check = isSpaceFree(binList, r, c, i)
if check == True:
for x in range(c, c+ i):
for y in range(r, r+ i):
binList[x][y] = i
comp = True
else:
print(c)
print(r)
r += 1
if r > gSize:
r = 0
c += 1
if c > gSize:
print("Imcompadible")
comp = True
print(i)
binPrint(binList)
input()
return binList
Your code to test for open spaces looks in binList[r][c] (where r is a row value and c is a column value). However, the code that sets the values once an open space has been found sets binList[x][y] (where x is a column value and y is a row value).
The latter is wrong. You want to set binList[y][x] instead (indexing by row, then column).
That will get you a working solution, but it will still not be exactly what you say you expect (you'll get a reflection across the diagonal). This is because your code updates r first, then c only when r has exceeded the bin size. If you want to place items to the right first, then below, you need to swap them.
I'd suggest using two for loops for r and c, rather than a while too, but to make it work in an elegant way you'd probably need to factor out the "find one item's place" code so you could return from the inner loop (rather than needing some complicated code to let you break out of both of the nested loops).
I put in this code:
def game():
down, right = 0, 0
while 1:
for i in range(down):
print "00000000000000000000000000000000000000000000000000000000000000000000000000000000"
for j in range(right):
print "0",
print "S",
for j in range(80 - (right + 1)):
print "0",
for i in range(38 - (down + 1)):
print "00000000000000000000000000000000000000000000000000000000000000000000000000000000"
direction = raw_input("Direction?")
if(direction.upper() == "W"):
down -= 1
elif(direction.upper() == "S"):
down += 1
elif(direction.upper() == "A"):
right -= 1
elif(direction.upper() == "D"):
right += 1
game()
The 'print "S",' is supposed to have a comma at the end, to stop it printing on the next line the next time I call the print function.
The actual thing came up with something like this: (I've chopped some lines containing "000000")
S 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000
Why do the first zeroes have spaces in-between? There are no spaces in my print function.
print "0",
Means print "0", and print a space.
The comma basically causes print to output a space instead of a newline. If you want to circumvent this behavior, use sys.stdout.write() instead.
You can set softspace to False after every ,
print "S",
sys.stdout.softspace = False
but then you might as well use sys.stdout.write()