We Keep Coding

How do i increase the value of an variable in a command line?

a = 1
for i in range(5):
browser.find_element_by_xpath("/html/body/div[6]/div/div/div[2]/div/div/div[1]/div[3]/button").click()
sleep(1)
I want to increase the 1 in div[1] by 1+ every loop, but how can i do that?
i thought i need to add a value, do "+a+" and last of all a "a = a + 1" to increase the value every time, but it didnt worked.
a = 1
for i in range(5):
browser.find_element_by_xpath("/html/body/div[6]/div/div/div[2]/div/div/div["+a+"]/div[3]/button").click()
a = a + 1
sleep(1)

for i in range(1,6):
browser.find_element_by_xpath("/html/body/div[6]/div/div/div[2]/div/div/div["+str(i)+"]/div[3]/button").click()
sleep(1)
you don't need 2 variables, just one variable i in the loop, convert it to string with str() and add it to where you need it, pretty simple. the value of i increases for every iteration of the loop going from 1 to 5 doing exactly what you need.

alternatively to Elyes' answer, you can use the 'global' keyword at the top of your function then a should increment 'correctly'.

You don't really need two variables for this unless you are going to use the second variable for something. However, look at the following code and it will show you that both i and a will give you the same result:
from time import sleep
a = 1
for i in range(1, 6):
path = "/html/body/div[6]/div/div/div[2]/div/div/div[{idx}]/div[3]/button".format(idx=i)
print(path, 'using i')
path = "/html/body/div[6]/div/div/div[2]/div/div/div[{idx}]/div[3]/button".format(idx=a)
a += 1
print(path, 'using a')
sleep(1)
Result:
/html/body/div[6]/div/div/div[2]/div/div/div[1]/div[3]/button using i
/html/body/div[6]/div/div/div[2]/div/div/div[1]/div[3]/button using a
/html/body/div[6]/div/div/div[2]/div/div/div[2]/div[3]/button using i
/html/body/div[6]/div/div/div[2]/div/div/div[2]/div[3]/button using a
/html/body/div[6]/div/div/div[2]/div/div/div[3]/div[3]/button using i
/html/body/div[6]/div/div/div[2]/div/div/div[3]/div[3]/button using a
/html/body/div[6]/div/div/div[2]/div/div/div[4]/div[3]/button using i
/html/body/div[6]/div/div/div[2]/div/div/div[4]/div[3]/button using a
/html/body/div[6]/div/div/div[2]/div/div/div[5]/div[3]/button using i
/html/body/div[6]/div/div/div[2]/div/div/div[5]/div[3]/button using a
You can read up on range here

How do I solve my program's counting problem?

(Apologies this is gonna be a long question)
I just have a bug in my code that I have not been able to resolve for a very long time. I would really appreciate if someone could help me find out what the problem is.
Context:
I have a long string of letters - lets call this subject - containing the letters A, G, T and C (like DNA) and the whole point of my algorithms is to correctly count how many of each of the following STRs are found within subject. The STRs are:
AGATC
TTTTTTCT
AATG
TCTAG
GATA
TATC
GAAA
TCTG
I must count how many of each are within subject. Counting works by going sequentially letter by letter until the start of one of above STRs are found. If the rest of the STR follows, the program should update the counter of the respective STR and then boost the searching index to account of the length of the STR and then keep going. It should stop when it reaches the end of subject.
(Hope it makes sense).
My Code:
STRs = ['AGATC','TTTTTTCT','AATG','TCTAG','GATA','TATC','GAAA','TCTG']
subject = "GCTAAATTTGTTCAGCCAGATGTAGGCTTACAAATCAAGCTGTCCGCTCGGCACGGCCTACACACGTCGTGTAACTACAACAGCTAGTTAATCTGGATATCACCATGACCGAATCATAGATTTCGCCTTAAGGAGCTTTACCATGGCTTGGGATCCAATACTAAGGGCTCGACCTAGGCGAATGAGTTTCAGGTTGGCAATCAGCAACGCTCGCCATCCGGACGACGGCTTACAGTTAGTAGCATAGTACGCGATTTTCGGGAAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCCCGTCAACTCATTCACACCGCATCCTTTCCTGCCACTGTAACTAGTCGACTGGGGAACCTCATCATCCATACTCTCCCACATTATGCCTCCCAACCTTGTTAAGCGTGGCATGCTTGGGATTGCATTGATGCTTCTTGGAGAGGACGCTTTCGTTTTGGAGATTACAGGGATCCAATTTTATCATCGGTTCGACTCCCGTAACGACTTAGCAGTAAGGGTGCTAGTTCCTGGTTAGAATCTTAATAAATCACGTCGCTTGGAGCAAGACAAAGATCGTCGTAATGCCAAGTGCACGACCACCTTCAGACTTGCAGGACCCGTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTCGATAGCTATGCGGTTCAATACAATCTTAACGCAATGCAGCGATGTGGTTTCGTACACTTAGCATAAAACCCCCCACATTAAATCGATGTACCCGCCCTCTTAGACGCCAATTTCAATGCCGAACCTCCGGCGGGTATCTCTGCACTAGGAGAAGTAGCACGTCGCTGTAGCGAACTCCTATCGTGAGATAATTTGTAGAGCTGCTCTTATAATACAATAGCTCAGATGGATTATTCCATGGACATCCCCGTGCGTTGTTTCGAGGATGGTAGGTGGAAATTTTGCCAGACCTCTAGTCTTAAACATGGTTGACGTTATAGGCGCTATCTCTTGCGTCTGGAAGTGTTAATCCGTGAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAACACGCAACTCTGGAGGAGGGCACTGCACTGCAAACTTGCGTAATATCCTTCACCCACACTTGCCTGGCCTCCTTGCTTAAAGCTCTGGCGATGCGATTTTTCGGCCCAGTAGCTGAATAGGTCATGAAATGGGCACCGAACTGGAAAGACCCATATATTCGATACTCACAACTTAATGATAGCGCGATTAAGAGCGACACCAAAAACCAAATTACGTTCACGAACCTTTGAGAGTCAAGGAGACTTAGACCGAATTGAATGATCACTGATGCGCCCGCTGATACTGAGCCTCACCATTAATCGCCGACCAATACGGCGTGTACCGGGCGCGGCCTTGCCGCATAACGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATATCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTACACAGCCCCGTCCTCATTGCTAAGTGCACTGGCAACTGGACCTAAAGATTTTTCGAGTATGGCCCTCGAATCAAGCGCCCACCCAGAAACCTACGAGCCAGTAACCCCAGTAAACAAGCATTAGTGCTATATGCTTGCTGCCCACTAGGACCCTTATGGTTCATACCAGGGTGACGTGTCTTGCGGGCCAAGGATGAACCAGAAGCAAGATCCTTAGATGGACGACTGTCTCATTGCTTAAACTCCACATACCAAAGGGCGCGGTAAACGATAGTTTTAGGTAATGTTAGTCGGATGGTTGTCTGCAGCTACCAATACAGCCTGGCACCCAGGGTCTGAACAATAACGCGTGAGAGCAGCTCTCCCGCGTGTGGTGGATTTGCCGTCTATGAAATTGAGGCTCTTGCAACTATTCGCACTCGGAATGCCCTCATATCTGGTGCCTAGCGGCCTTTGCCCCGTGCCGGTAGGACTAAACTCTACGGATCGTTGACGGATCTCGATGTGGAAGATGGTTATGAAAGATAACAACGCGTGTGCTAATTGATTTAGACAAGTATTGCGGCAGTAAAAGATAATCGGCTGCAGAGTTACGAAAGACTTCCATGCATGGATTCCATTCCTTCTAGTATAGGACCCACTCTGAATACACGTCTTGCGGGCCGATCATCTCCACCGCTGCGGAAGAAAGCAATTAAGAATCTATGCTCATTAAGAGTGCGACTATAATGCGGATCTTACAGTGCTAATGATCAGGACGTCGTCCAAGCAGGCTGCATGCCGAATTTAGCTTACGTCAGGATCAGGCGTTATAGCCTGGGAATCGGACTATGAGGACGCCACGACCTCTGGGAGAAAGCTATATACATTGAGGATCGCGCCATCTTTATGAGACTCAAATGAATCTAGATAGGTAGCATTGCGGACTTGAGTTAGCACATCGGTATTGGAAGGTGAGGGTCCTGCCGCTCGTTCTATGTTCGGTTTATAGTATACAAATAGGTCATCCCGAACGTTGAAGTTAAACTCATGACACGTTGTCGTAATGAAACGGGCCTGTTATTAGGGATACAGACAAAAGGCACAAGCTGGCTTGCACATTAAGGCGCACTAGAGATCCTCACAACCGTTGCCCGCACGGAGGTCGTGTCTAACAGACAGTGAACCAGCCGTATTGGGGTGGATGACCTGAGCTTCTTGGGGCCTGTTGTACACCGCGTGTGGTTCAACTGGTACACATACTACGAATATTCGAAATCATTGTACTGTGCTCTTCGGTGCTACTGACTGTGAGCGAATGCATCCCAATCCCAAACAATGCTTGTGGTAGGAGAATTGAAACTCTCGAAGCCTGGCCCAATGTCATCTACTTTTAACATGTCGGGCCAGGAGTTACGGGCATTGCTTACTTACTTTGCCCCCTTACACCACAGCAGCGCGATTCTTGTTGTAGTAGATTTTATACGACTCGCGAATTAAATGGAACTTGTCTGTCCCATATCGATCGTGTCCATCGTAAGATGAGATTGTAGGAGCATTCGGAAGTCTATGCGGCCCAGGGACTACTACGTTAAATCTGGTCAGACGTGGTTTACAAGGCGTCCCGATCTTCTCAGAACATATGGGAAAGCACTACCGTTCCTTCACGCATACAGTTGTTCGTGCCGAACGAGTAAGCTTGCGACCAGCCCACCCGCTAGGGCTATGCAGCGGGTCATGGCTGGCGCCATACTGTGCGGACAACCCACGCTCTGGCAGAAAGCGTCTTGTGTTTTGTAGTAGCTCCAACGGTTAGACCTTCGATATCTATTCAGAGCGCGAGCGACCACTATTAGACGGCATGTAAACAATGTGTATTTGTTCGGCCCAACCGGTATATGGGTAAGACCGCGAAGGGCCTGCGCGAATACCAGCGTCCAAAAATTCCTCACCCGAGATATGCGGTTAGTACCCCTTGGGTAACGGTCCGCTACGGGTAGCGACGCGAGCCGGCCGCATCGGTTGGAGCCGAGTTGTCGGGCAGGCGAGTAACGTGTGCAATTTGATGGGCCCAAGCCTCCGGCACTATCCACCTCATACATCGACAAAAGCACCAAATATGGGGAAAAGCTGAGCGTCGATATGTACATCTACCCAGGAACCGGCCCGAACATTAGGCGGACGTGAATTTCCGACCTAGGTTCGGCTACATTTCTACGATCCAAGCACACGTGAAGGAGGAGGGGTGTTCCGACCGTAAATGAACGAGGTGCGCAGTGACCCGATGGCGTTTAGCGGATAGCCTTCCTATGCCGGCCTATGCTGTATGGTAGTTGGTTGGTGCCTCCAGAGCCACTGCACCCAATCATAGGGTCTACAGCAGCGTACTTATAAAATTGTACGGGTGACCCATATCCATTACGGGTTGCGACCAGTATAGGAGAGTATAACTGCGTGAACTAATGCGTTATGACGCTTCAGAGTTTGCTCGGGCCCGAGTTCTAGGGCTATAATGTGTTAGGGCGCAAGTATGCCAAGCTAAGATGTGGCGTGCACACTAGGAGTTGTGTTCCTCTGCAAGCAGACACGAGCACTCTGGCAGTAGTTTGACCACACCCGGGTATCACTGCTACTCCATTTCGAACAAGCTATTGGAGCGGACAAAATATGCTACTCAAGAGCATTAGTTATAGGTCTACGAGACAGAAGCAGTTACTGAGTCTGAATATTCGATATAAGTAGGCATGGAGGCGGAGCAAAACAACGTCTGCGATCAATCGTGTTGATGACGTATGGCGACTGGAAGGTAAGGACTATGGCCGGACGGAATGATTCATGTTCTGTTCAAAGCTATATTTCGAAGGGGTATATTAGCGGTCCTACACTTGGTTAGCACCCTCCCCCCTCTGGATCCTGCACTAATTCGAGCTGGCCTCCATCGGTATCAGTCCGGAAGCTCCACTCTCTATCGTAGTCCTAATCAACAGGGTGCCAGTTTGCTCACGTGGAAGTTTGAGGCCCTTTGTGCTCCATAGCCAATCACTAACCATGCACGCGCGACCCACTCTACGTCCAGATCGGCTATAATAGTTGCGCCCGGGACTGGCAGAGTAGACATGTAAGCTAGATAGAGCCCCGACATCGGCCAAGAGATCCTACGCTGCTTCCAGATAATGAGAGACATTCTAGCATTAGACATGCAAGTCGGCAGGGACTCCCCTTATCTAGTAATTTCGATGAATTGGTTTTTCGGCTAGCATCTAGTCTAGTCTAGTCTAGTCTAGTCTAGTCTAGTCTAGTCTAGTCTAGTCTAGTCTAGACCATGCCGACCTCATCATAGAAGGAATGCTCTAAACTTAGAGTGCTACTAGGAAAACTATTAATCAATGATCGTCCTGCTTACATAGCTGGACGGCGAAAGTTCTTATACTGCGGAGGTTGCTGACGTAGAGTGCGCTGGGTACAGCGGATAAGTTGATCAGGGTGGGGATAGGGTGGCTCACCGTTTATACTCATATAGATTCCTGGCGTCGACGCTGTGACAGGGTCGAGATCGAGGGGGAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGCGGAGCGGAGGGAAAATTATCACCAGAGGGTAGGGGCTCGCGACATTCTATTCAATGCATTTCAAGCTACTTACGTATTTCGGCACAGTGACTACTGCCTGCGCGGCAGCCGTAAGGTTTCCCGTCAATAGGTGGCACGTATCATTGATGAAAGTGTCAGCTAATCATTCAGGCCTTA"
x = 0 # Searching index.
dataSTR = { # All the STRs to seach for.
"AGATC":0,
"TTTTTTCT":0,
"AATG":0,
"TCTAG":0,
"GATA":0,
"TATC":0,
"GAAA":0,
"TCTG":0,
}
# This dict will hold all the count values of STR's in the text-file.
# Scanning STR's from the txt file.
total = len(subject)
limit = 8
while x < total:
currentString = subject[x:x+limit] # A temporary variable to hold the next few letters from the text-file at index x.
for STR in STRs:
if STR in currentString: # The STR is found within this set of letters?
lSTR = len(STR) - 1
if STR[0:lSTR] == currentString[0:lSTR]: # In order to minimise the risk of duplication...
dataSTR[STR] += 1 # ...the STR must be at the start of currentString.
#print(currentString, STR, x, dataSTR[STR])
x += lSTR # The index must be boosted each time a new STR is read. In the event that an STR is at the end of a stand...
x += 1 # The index counts up by 1 by default. (From above) ...so that no duplicates are added.
print(dataSTR.items())
print("The correct result is: AGATC - 22, TTTTTTCT - 33, AATG - 43, TCTAG - 12, GATA - 26, TATC - 18, GAAA - 47, TCTG - 41")
(Sorry its very long, it might be helpful to copy into a separate python file).
As you will see from running it, the result my program brings up from counting is incorrect. The correct results are in the final print statement of the program, but the program does not match this (yes I know that these results are 100% correct since this is part of a problem set from an online computer science course).
However, I cannot seem to find the bug or logic error that seems to be causing my program to count wrong and I have been trying for quite a while now. Does anyone know what the solution is?
Please feel free to ask me anything about the program, thank you all.

Your problem statement doesn't agree with the "correct results" given in your example code. Either you've misunderstood the problem, or you've taken the correct results from a different problem. (The "correct results" appear to be for the problem of finding the maximum number of consecutive repeats of each query string.) [The latter possibility is the point that Chris Charley makes in a comment on the original post.]
You can convince yourself by doing the problem "by hand": look at the subject string in a text editor, pick a query string, do a search on it, and step through the occurrences.
E.g., for the query string "GAAA", you'll count ~67 occurrences, but most of them are in a block of 47 repeats in subject[1449:1637]. (This is more obvious if you use a text editor that highlights all occurrences of the search string, as 188 characters of consecutive highlighting should jump out at you.) And 47 agrees with the "correct result" for GAAA.

Does this help?
count_results = dict()
STRs = ['AGATC','TTTTTTCT','AATG','TCTAG','GATA','TATC','GAAA','TCTG']
subject = "loooong string..."
for search_string in STRs:
count_results[search_string] = subject.count(search_string)
print(count_results)
{'AGATC': 28, 'TTTTTTCT': 33, 'AATG': 69, 'TCTAG': 18, 'GATA': 46, 'TATC': 36, 'GAAA': 67, 'TCTG': 60}
I realize the results are sometimes different to your expected counts, but I didn't go through the intricacies of your search algo and wonder if the expected output might be wrong? If not, check out the docs for the str.count() function, to see how & why it gets different output, and adapt what it does to your needs.

Try like this:
import re
# Define STRs and subject here
dic = {}
for x in STRs:
tv = len([m.start() for m in re.finditer(x,subject)])
tv += 1
dic[x] = tv
for y in dic.keys():
print(y,dic[y])

The results in the last print statement are incorrect. I checked it with python's built in method .count(), if you are allowed to use this method just use this one instead, but if not, I would recommend to do the following:
total = len(subject)
while x < total:
for STR in STRs:
limit = len(STR)
currentString = subject[x:x+limit]
if STR == currentString:
dataSTR[STR] += 1
x += 1
that way, you set the limit to the string's length so the STR is either exactly the string or not, so you don't have to check for duplicates. I don't know why your code didn't work, but I hope this will help you.

Cannot understand how this Python code works

I found this question on HackerRank and I am unable to understand the code(solution) that is displayed in the discussions page.
The question is:
Consider a list (list = []). You can perform the following commands:
insert i e: Insert integer at position .
print: Print the list.
remove e: Delete the first occurrence of integer .
append e: Insert integer at the end of the list.
sort: Sort the list.
pop: Pop the last element from the list.
reverse: Reverse the list.
Even though I have solved the problem using if-else, I do not understand how this code works:
n = input()
slist = []
for _ in range(n):
s = input().split()
cmd = s[0]
args = s[1:]
if cmd !="print":
cmd += "("+ ",".join(args) +")"
eval("slist."+cmd)
else:
print slist

Well, the code takes advantage of Python's eval function. Many languages have this feature: eval, short for "evaluate", takes a piece of text and executes it as if it were part of the program instead of just a piece of data fed to the program. This line:
s = input().split()
reads a line of input from the user and splits it into words based on whitespace, so if you type "insert 1 2", s is set to the list ["insert","1","2"]. That is then transformed by the following lines into "insert(1,2)", which is then appended to "slist." and passed to eval, resulting in the method call slist.insert(1,2) being executed. So basically, this code is taking advantage of the fact that Python already has methods to perform the required functions, that even happen to have the same names used in the problem. All it has to do is take the name and arguments from an input line and transform them into Python syntax. (The print option is special-cased since there is no method slist.print(); for that case it uses the global command: print slist.)
In real-world code, you should almost never use eval; it is a very dangerous feature, since it allows users of your application to potentially cause it to run any code they want. It's certainly one of the easier features for hackers to use to break into things.

It's dirty code that's abusing eval.
Basically, when you enter, for example, "remove 1", it creates some code that looks like sList.remove(1), then gives the created code to eval. This has Python interpret it.
This is probably the worst way you could solve this outside of coding competitions though. The use of eval is entirely unnecessary here.

Actually I Find some error in the code, but I came to an understanding of how this code runs. here is it:
input :
3
1 2 3
cmd = 1 + ( 2 + 3)
then eval(cmd) i.e., eval("1 + (2 + 3)") which gives an output 6
another input:
4
4 5 6 2
cmd = 4 + ( 5 + 6 + 2)
eval(cmd)

if __name__ == '__main__':
N = int(raw_input())
lst=[]
for _ in range(N):
cmd, *line = input().split()
ele= list(map(str,line))
if cmd in dir(lst):
exec('lst.'+cmd+'('+','.join(ele)+')')
elif cmd == 'print':
print(lst)
else:
print('wrong command', cmd)

Python - turn some of the words in list/str to dots. len(list)?

I've started learning Python last week on codecademy and Google etc. but got stuck and couldn't find the answer anywhere so signed up on stackoverflow.com looking for your support.
I'm trying to build a program that only takes first 5 letters of any name and the remainder of the letter(s) to be shows as blank dot(s). e.g.
Adrian: "Adria."
Michael: "Micha.."
Alexander: "Alexa...." etc.
I tried to "fix" it with the "b" variable but that just prints three dots "..." regardless of how long the name is.
This is what I've got so far:
def namecheck():
name = raw_input("Name?")
if len(name) <=5:
print name
else:
if len(name) >5:
name = name[0:5]
b = ("...")
print name + b
namecheck()
I'm a total newbie so I apologise for any wrong spacing here, thank you for your support and patience.

As an alternative to sequence multiplication (one which is somewhat more self-documenting, and hopefully less confusing to maintainers), just use str.ljust to do your padding:
def namecheck():
name = raw_input("Name?")
# Reduce to first five (or less) characters, then pad with .s to original length
# with str.ljust
print name[:5].ljust(len(name), '.')
print name[:5] + '.' * (len(name) - 5) works fine, it's just a bit arcane (and also involves more temporary values, though in practice, the lack of actual method calls makes it faster on CPython).

you can try to use the function replace().
name = 'abcdefg'
name.replace(name[5:], '.' * len(name[5:]))
output: 'abcde..'
name='randy12345'
name.replace(name[5:],'.' * len(name[5:]))
output: 'randy.....'
name[5:] means get all the element starting 6 (5+1 because it start with 0)
'.' * len(name[5:] then this code count it and multiply it by dot
name.replace(name[5:],'.' * len(name[5:])) then use replace function to replace the excess element with dots

The most concise way I can think of:
def namecheck():
name = raw_input("Name?")
print(name[0:5] + '.' * (len(name) - 5))
namecheck()

Try something like this:
def namecheck():
name = raw_input("Name?")
if len(name) <= 5:
print name
else:
print name[0:5] + '.' * (len(name)-5)
namecheck()

make a global condition break

allow me to preface this by saying that i am learning python on my own as part of my own curiosity, and i was recommended a free online computer science course that is publicly available, so i apologize if i am using terms incorrectly.
i have seen questions regarding this particular problem on here before - but i have a separate question from them and did not want to hijack those threads. the question:
"a substring is any consecutive sequence of characters inside another string. The same substring may occur several times inside the same string: for example "assesses" has the substring "sses" 2 times, and "trans-Panamanian banana" has the substring "an" 6 times. Write a program that takes two lines of input, we call the first needle and the second haystack. Print the number of times that needle occurs as a substring of haystack."
my solution (which works) is:
first = str(input())
second = str(input())
count = 0
location = 0
while location < len(second):
if location == 0:
location = str.find(second,first,0)
if location < 0:
break
count = count + 1
location = str.find(second,first,location +1)
if location < 0:
break
count = count + 1
print(count)
if you notice, i have on two separate occasions made the if statement that if location is less than 0, to break. is there some way to make this a 'global' condition so i do not have repetitive code? i imagine efficiency becomes paramount with increasing program sophistication so i am trying to develop good practice now.
how would python gurus optimize this code or am i just being too nitpicky?

I think Matthew and darshan have the best solution. I will just post a variation which is based on your solution:
first = str(input())
second = str(input())
def count_needle(first, second):
location = str.find(second,first)
if location == -1:
return 0 # none whatsoever
else:
count = 1
while location < len(second):
location = str.find(second,first,location +1)
if location < 0:
break
count = count + 1
return count
print(count_needle(first, second))
Idea:
use function to structure the code when appropriate
initialise the variable location before entering the while loop save you from checking location < 0 multiple times

Check out regular expressions, python's re module (http://docs.python.org/library/re.html). For example,
import re
first = str(input())
second = str(input())
regex = first[:-1] + '(?=' + first[-1] + ')'
print(len(re.findall(regex, second)))

As mentioned by Matthew Adams the best way to do it is using python'd re module Python re module.
For your case the solution would look something like this:
import re
def find_needle_in_heystack(needle, heystack):
return len(re.findall(needle, heystack))
Since you are learning python, best way would be to use 'DRY' [Don't Repeat Yourself] mantra. There are lots of python utilities that you can use for many similar situation.
For a quick overview of few very important python modules you can go through this class:
Google Python Class
which should only take you a day.

even your aproach could be imo simplified (which uses the fact, that find returns -1, while you aks it to search from non existent offset):
>>> x = 'xoxoxo'
>>> start = x.find('o')
>>> indexes = []
>>> while start > -1:
... indexes.append(start)
... start = x.find('o',start+1)
>>> indexes
[1, 3, 5]

needle = "ss"
haystack = "ssi lass 2 vecess estan ss."
print 'needle occurs %d times in haystack.' % haystack.count(needle)

Here you go :
first = str(input())
second = str(input())
x=len(first)
counter=0
for i in range(0,len(second)):
if first==second[i:(x+i)]:
counter=counter+1
print(counter)

Answer
needle=input()
haystack=input()
counter=0
for i in range(0,len(haystack)):
if(haystack[i:len(needle)+i]!=needle):
continue
counter=counter+1
print(counter)