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Python convert String to Variable in the loop

I would like to do something like this:
x = 0
y = 3
TEST0 = 10
TEST1 = 20
TEST2 = 30
while x < y:
result = exec('TEST{}'.format(x))
print(result)
x += 1
And have the output:
10
20
30
Somehow convert TEST{variable} to the actual variable, or what is the way to do it?
Currently, I have result as:
None
None
None

Welcome to Python! What you need is a list:
x = 0
y = 3
TEST = [10, 20, 30]
while x < y:
result = TEST[x]
print(result)
x += 1
A list is created by putting the values between []. You access a particular element in the list by writing the name of the variable, followed by the index enclosed in []. Read more about lists here in the official tutorial.
Instead of the while loop with explicit indexing, it's nicer to use a for loop instead:
TEST = [10, 20, 30]
for element in TEST:
result = element
print(result)

I would recommend using a dictionary in most cases, however if you don't want to use a dictionary you can always use globals() or vars(). They work as follows:
global_var = 5
>> 5
print(globals()["global_var"])
>> 5
vars() works in the same way but at module scope.
In your case, do the following:
x = 0
y = 3
TEST0 = 10
TEST1 = 20
TEST2 = 30
while x < y:
result = globals()['TEST{}'.format(x)]
print(result)
x += 1

select region from sequence with Phobius output in Python

I need to use a certain program, to validate some of my results. I am relatively new in Python. The output is so different for each entry, see a snippit below:
SEQENCE ID TM SP PREDICTION
YOL154W_Q12512_Saccharomyces_cerevisiae 0 Y n8-15c20/21o
YDR481C_P11491_Saccharomyces_cerevisiae 1 0 i34-53o
YAL007C_P39704_Saccharomyces_cerevisiae 1 Y n5-20c25/26o181-207i
YAR028W_P39548_Saccharomyces_cerevisiae 2 0 i51-69o75-97i
YBL040C_P18414_Saccharomyces_cerevisiae 7 0 o6-26i38-56o62-80i101-119o125-143i155-174o186-206i
YBR106W_P38264_Saccharomyces_cerevisiae 1 0 o28-47i
YBR287W_P38355_Saccharomyces_cerevisiae 8 0 o12-32i44-63o69-90i258-275o295-315i327-351o363-385i397-421o
So, I need the last transmembrane region, in this case its always the last numbers between o and i or vise versa. if TM = 0, there is no transmembrane region, so I want the numbers if TM > 0
output I need:
34-53
181-207
75-97
186-206
28-47
397-421
preferably in seperate values, like:
first_number = 34
second_number = 53
Because I will be using a loop the values will be overwritten anyway. To summarize: I need the last region between the o and i or vise versa, with very variable strings (both in length and composition).
Trouble: If I just search (for example with regular expression) for the last region between o and i, I will sometimes pick the wrong region.

If the Phobius output is stored in a file, change 'Phobius_output' to the path, then the following code should give the expected result:
with open('Phobius_output') as file:
for line in file.readlines()[1:]:
if int(line.split()[1]) > 0:
prediction = line.split()[3]
i_idx, o_idx = prediction.rfind('i'), prediction.rfind('o')
last_region = prediction[i_idx + 1:o_idx] if i_idx < o_idx else prediction[o_idx + 1:i_idx]
first_number, second_number = map(int, last_region.split('-'))
print(last_region)

While Loop to produce Mathematical Sequences?

I've been asked to do the following:
Using a while loop, you will write a program which will produce the following mathematical sequence:
1 * 9 + 2 = 11(you will compute this number)
12 * 9 + 3 = 111
123 * 9 + 4 = 1111
Then your program should run as far as the results contain only "1"s. You can build your numbers as string, then convert to ints before calculation. Then you can convert the result back to a string to see if it contains all "1"s.
Sample Output:
1 * 9 + 2 = 11
12 * 9 + 3 = 111
123 * 9 + 4 = 1111
1234 * 9 + 5 = 11111
Here is my code:
def main():
Current = 1
Next = 2
Addition = 2
output = funcCalculation(Current, Addition)
while (verifyAllOnes(output) == True):
print(output)
#string concat to get new current number
Current = int(str(Current) + str(Next))
Addition += 1
Next += 1
output = funcCalculation(Current, Next)
def funcCalculation(a,b):
return (a * 9 + b)
def verifyAllOnes(val):
Num_str = str(val)
for ch in Num_str:
if(str(ch)!= "1"):
return False
return True
main()
The bug is that the formula isn't printing next to the series of ones on each line. What am I doing wrong?

Pseudo-code:
a = 1
b = 2
result = a * 9 + b
while string representation of result contains only 1s:
a = concat a with the old value of b, as a number
b = b + 1
result = a * 9 + b
This can be literally converted into Python code.

Testing all ones
Well, for starters, here is one easy way to check that the value is all ones:
def only_ones(n):
n_str = str(n)
return set(n_str) == set(['1'])
You could do something more "mathy", but I'm not sure that it would be any faster. It would much more easily
generalize to other bases (than 10) if that's something you were interested in though
def only_ones(n):
return (n % 10 == 1) and (n == 1 or only_ones2(n / 10))
Uncertainty about how to generate the specific recurrence relation...
As for actually solving the problem though, it's actually not clear what the sequence should be.
What comes next?
123456
1234567
12345678
123456789
?
Is it 1234567890? Or 12345678910? Or 1234567900?
Without answering this, it's not possible to solve the problem in any general way (unless in fact the 111..s
terminate before you get to this issue).
I'm going to go with the most mathematically appealing assumption, which is that the value in question is the
sum of all the 11111... values before it (note that 12 = 11 + 1, 123 = 111 + 11 + 1, 1234 = 1111 + 111 + 11 + 1, etc...).
A solution
In this case, you could do something along these lines:
def sequence_gen():
a = 1
b = 1
i = 2
while only_ones(b):
yield b
b = a*9 + i
a += b
i += 1
Notice that I've put this in a generator in order to make it easier to only grab as many results from this
sequence as you actually want. It's entirely possible that this is an infinite sequence, so actually running
the while code by itself might take a while ;-)
s = sequence_gen()
s.next() #=> 1
s.next() #=> 11
A generator gives you a lot of flexibility for things like this. For instance, you could grab the first 10 values of the sequence using the itertools.islice
function:
import itertools as it
s = sequence_gen()
xs = [x for x in it.islice(s, 10)]
print xs

How to toggle between two values?

I want to toggle between two values in Python, that is, between 0 and 1.
For example, when I run a function the first time, it yields the number 0. Next time, it yields 1. Third time it's back to zero, and so on.
Sorry if this doesn't make sense, but does anyone know a way to do this?

Use itertools.cycle():
from itertools import cycle
myIterator = cycle(range(2))
myIterator.next() # or next(myIterator) which works in Python 3.x. Yields 0
myIterator.next() # or next(myIterator) which works in Python 3.x. Yields 1
# etc.
Note that if you need a more complicated cycle than [0, 1], this solution becomes much more attractive than the other ones posted here...
from itertools import cycle
mySmallSquareIterator = cycle(i*i for i in range(10))
# Will yield 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 0, 1, 4, ...

You can accomplish that with a generator like this:
>>> def alternate():
... while True:
... yield 0
... yield 1
...
>>>
>>> alternator = alternate()
>>>
>>> alternator.next()
0
>>> alternator.next()
1
>>> alternator.next()
0

You can use the mod (%) operator.
count = 0 # initialize count once
then
count = (count + 1) % 2
will toggle the value of count between 0 and 1 each time this statement is executed. The advantage of this approach is that you can cycle through a sequence of values (if needed) from 0 - (n-1) where n is the value you use with your % operator. And this technique does not depend on any Python specific features/libraries.
e.g.
count = 0
for i in range(5):
count = (count + 1) % 2
print(count)
gives:
1
0
1
0
1

You may find it useful to create a function alias like so:
import itertools
myfunc = itertools.cycle([0,1]).next
then
myfunc() # -> returns 0
myfunc() # -> returns 1
myfunc() # -> returns 0
myfunc() # -> returns 1

In python, True and False are integers (1 and 0 respectively). You could use a boolean (True or False) and the not operator:
var = not var
Of course, if you want to iterate between other numbers than 0 and 1, this trick becomes a little more difficult.
To pack this into an admittedly ugly function:
def alternate():
alternate.x=not alternate.x
return alternate.x
alternate.x=True #The first call to alternate will return False (0)
mylist=[5,3]
print(mylist[alternate()]) #5
print(mylist[alternate()]) #3
print(mylist[alternate()]) #5

from itertools import cycle
alternator = cycle((0,1))
next(alternator) # yields 0
next(alternator) # yields 1
next(alternator) # yields 0
next(alternator) # yields 1
#... forever

var = 1
var = 1 - var
That's the official tricky way of doing it ;)

Using xor works, and is a good visual way to toggle between two values.
count = 1
count = count ^ 1 # count is now 0
count = count ^ 1 # count is now 1

To toggle variable x between two arbitrary (integer) values,
e.g. a and b, use:
# start with either x == a or x == b
x = (a + b) - x
# case x == a:
# x = (a + b) - a ==> x becomes b
# case x == b:
# x = (a + b) - b ==> x becomes a
Example:
Toggle between 3 and 5
x = 3
x = 8 - x (now x == 5)
x = 8 - x (now x == 3)
x = 8 - x (now x == 5)
This works even with strings (sort of).
YesNo = 'YesNo'
answer = 'Yes'
answer = YesNo.replace(answer,'') (now answer == 'No')
answer = YesNo.replace(answer,'') (now answer == 'Yes')
answer = YesNo.replace(answer,'') (now answer == 'No')

Using the tuple subscript trick:
value = (1, 0)[value]

Using tuple subscripts is one good way to toggle between two values:
toggle_val = 1
toggle_val = (1,0)[toggle_val]
If you wrapped a function around this, you would have a nice alternating switch.

If a variable is previously defined and you want it to toggle between two values, you may use the
a if b else c form:
variable = 'value1'
variable = 'value2' if variable=='value1' else 'value1'
In addition, it works on Python 2.5+ and 3.x
See Expressions in the Python 3 documentation.

Simple and general solution without using any built-in. Just keep the track of current element and print/return the other one then change the current element status.
a, b = map(int, raw_input("Enter both number: ").split())
flag = input("Enter the first value: ")
length = input("Enter Number of iterations: ")
for i in range(length):
print flag
if flag == a:
flag = b;
else:
flag = a
Input:
3 835Output:38383
Means numbers to be toggled are 3 and 8
Second input, is the first value by which you want to start the sequence
And last input indicates the number of times you want to generate

One cool way you can do in any language:
variable = 0
variable = abs(variable - 1) // 1
variable = abs(variable - 1) // 0

Printing in a loop

I have the following file I'm trying to manipulate.
1 2 -3 5 10 8.2
5 8 5 4 0 6
4 3 2 3 -2 15
-3 4 0 2 4 2.33
2 1 1 1 2.5 0
0 2 6 0 8 5
The file just contains numbers.
I'm trying to write a program to subtract the rows from each other and print the results to a file. My program is below and, dtest.txt is the name of the input file. The name of the program is make_distance.py.
from math import *
posnfile = open("dtest.txt","r")
posn = posnfile.readlines()
posnfile.close()
for i in range (len(posn)-1):
for j in range (0,1):
if (j == 0):
Xp = float(posn[i].split()[0])
Yp = float(posn[i].split()[1])
Zp = float(posn[i].split()[2])
Xc = float(posn[i+1].split()[0])
Yc = float(posn[i+1].split()[1])
Zc = float(posn[i+1].split()[2])
else:
Xp = float(posn[i].split()[3*j+1])
Yp = float(posn[i].split()[3*j+2])
Zp = float(posn[i].split()[3*j+3])
Xc = float(posn[i+1].split()[3*j+1])
Yc = float(posn[i+1].split()[3*j+2])
Zc = float(posn[i+1].split()[3*j+3])
Px = fabs(Xc-Xp)
Py = fabs(Yc-Yp)
Pz = fabs(Zc-Zp)
print Px,Py,Pz
The program is calculating the values correctly but, when I try to call the program to write the output file,
mpipython make_distance.py > distance.dat
The output file (distance.dat) only contains 3 columns when it should contain 6. How do I tell the program to shift what columns to print to for each step j=0,1,....
For j = 0, the program should output to the first 3 columns, for j = 1 the program should output to the second 3 columns (3,4,5) and so on and so forth.
Finally the len function gives the number of rows in the input file but, what function gives the number of columns in the file?
Thanks.

Append a , to the end of your print statement and it will not print a newline, and then when you exit the for loop add an additional print to move to the next row:
for j in range (0,1):
...
print Px,Py,Pz,
print
Assuming all rows have the same number of columns, you can get the number of columns by using len(row.split()).
Also, you can definitely shorten your code quite a bit, I'm not sure what the purpose of j is, but the following should be equivalent to what you're doing now:
for j in range (0,1):
Xp, Yp, Zp = map(float, posn[i].split()[3*j:3*j+3])
Xc, Yc, Zc = map(float, posn[i+1].split()[3*j:3*j+3])
...

You don't need to:
use numpy
read the whole file in at once
know how many columns
use awkward comma at end of print statement
use list subscripting
use math.fabs()
explicitly close your file
Try this (untested):
with open("dtest.txt", "r") as posnfile:
previous = None
for line in posnfile:
current = [float(x) for x in line.split()]
if previous:
delta = [abs(c - p) for c, p in zip(current, previous)]
print ' '.join(str(d) for d in delta)
previous = current

just in case your dtest.txt grows larger and you don't want to redirect your output but rather write to distance.dat, especially, if you want to use numpy. Thank #John for pointing out my mistake in the old code ;-)
import numpy as np
pos = np.genfromtxt("dtest.txt")
dis = np.array([np.abs(pos[j+1] - pos[j]) for j in xrange(len(pos)-1)])
np.savetxt("distance.dat",dis)