I have a exercise ,Input data will contain the total count of pairs to process in the first line.
The following lines will contain pairs themselves - one pair at each line.
Answer should contain the results separated by spaces.
My code:
n = int(raw_input())
sum = 0
for i in range(n):
y = raw_input().split(" ")
for i in y:
sum = sum + int(i)
print sum
With my code , I come the Sum together, but I will that the results to come separated by spaces . Thanks for yours help .
with your current code what you get is the total sum of all the given numbers, to get the sum per line you need to initialize your counter in the outer loop, and then print it, and as you want to print all it in the same line there are several ways to do it, like save it in a list or telling print that don't print a new, line which is done by adding a , at the end like print x, with that in mind then the changes needed are
n = int(raw_input())
for i in range(n):
pairs = raw_input().split() #by default split use spaces
pair_sum = 0
for p in pairs:
pair_sum += int(p) # a += b is the same as a = a + b
print pair_sum,
print "" # to print a new line so any future print is not done in the same line as the previous one
that was the version with print per line, next is the version using list
n = int(raw_input())
resul_per_line = []
for i in range(n):
pairs = raw_input().split() #by default split use spaces
pair_sum = 0
for p in pairs:
pair_sum += int(p) # a += b is the same as a = a + b
resul_per_line.append( str(pair_sum) ) #conver each number to a string to use with join bellow
print " ".join(resul_per_line)
with either of the above let said for example that the input data is
3
1 2
40 50
600 700
then the result would be
3 90 1300
some parts of the above code can be simplify by using built in functions like map and sum, for example this part
pair_sum = 0
for p in pairs:
pair_sum += int(p)
can become
pair_sum = sum( map(int,pairs) )
Uh oh, it looks like you're reusing the same variable i in the inner loop as the outer loop -- this is bad practice and can lead to bugs down the road.
What you're doing currently is adding both elements in each pair to sum and then printing that at the end, you can fix this in two different ways.
You can sum each pair, convert the sum to a string, and then concatenate that with your the rest of the sums as strings, or
You can print the sum of each pair immediately after summing them with print sum, which will print the number without the newline so that you can print all the results on a single line.
Related
After reading a text, I need to add 1 to a sum if I find a ( character, and subtract 1 if I find a ) character in the text. I can't figure out what I'm doing wrong.
This is what I tried at first:
file = open("day12015.txt")
sum = 0
up = "("
for item in file:
if item is up:
sum += 1
else:
sum -= 1
print(sum)
I have this long text like the following example (((())))((((( .... If I find a ), I need to subtract 1, if I find a (, I need to add 1. How can I solve it? I'm always getting 0 as output even if I change my file manually.
your for loop only gets all the string in the file so you have to loop through the string to get your desired output.
Example .txt
(((())))(((((
Full Code
file = open("Data.txt")
sum = 0
up = "("
for string in file:
for item in string:
if item is up:
sum += 1
else:
sum -= 1
print(sum)
Output
5
Hope this helps.Happy Coding :)
So you need to sum +1 for "(" character and -1 for ")".
Do it directly specifying what to occur when you encounter this character. Also you need to read the lines from a file as you're opening it. In your code, you are substracting one for every case that is not "(".
file = open("day12015.txt")
total = 0
for line in file:
for character in line:
if character == "(":
total += 1
elif character == ")":
total -= 1
print(sum)
That's simply a matter of counting each character in the text. The sum is the difference between those counts. Look:
from pathlib import Path
file = Path('day12015.txt')
text = file.read_text()
total = text.count('(') - text.count(')')
For the string you posted, for example, we have this:
>>> p = '(((())))((((('
>>> p.count('(') - p.count(')')
5
>>>
Just for comparison and out of curiosity, I timed the str.count() and a loop approach, 1,000 times, using a string composed of 1,000,000 randoms ( and ). Here is what I found:
import random
from timeit import timeit
random.seed(0)
p = ''.join(random.choice('()') for _ in range(1_000_000))
def f():
return p.count('(') - p.count(')')
def g():
a, b = 0, 0
for c in p:
if c == '(':
a = a + 1
else:
b = b + 1
return a - b
print('f: %5.2f s' % timeit(f, number=1_000))
print('g: %5.2f s' % timeit(g, number=1_000))
f: 8.19 s
g: 49.34 s
It means the loop approach is 6 times slower, even though the str.count() one is iterating over p two times to compute the result.
I want to be able to generate 12 character long chain, of hexadecimal, BUT with no more than 2 identical numbers duplicate in the chain: 00 and not 000
Because, I know how to generate ALL possibilites, including 00000000000 to FFFFFFFFFFF, but I know that I won't use all those values, and because the size of the file generated with ALL possibilities is many GB long, I want to reduce the size by avoiding the not useful generated chains.
So my goal is to have results like 00A300BF8911 and not like 000300BF8911
Could you please help me to do so?
Many thanks in advance!
if you picked the same one twice, remove it from the choices for a round:
import random
hex_digits = set('0123456789ABCDEF')
result = ""
pick_from = hex_digits
for digit in range(12):
cur_digit = random.sample(hex_digits, 1)[0]
result += cur_digit
if result[-1] == cur_digit:
pick_from = hex_digits - set(cur_digit)
else:
pick_from = hex_digits
print(result)
Since the title mentions generators. Here's the above as a generator:
import random
hex_digits = set('0123456789ABCDEF')
def hexGen():
while True:
result = ""
pick_from = hex_digits
for digit in range(12):
cur_digit = random.sample(hex_digits, 1)[0]
result += cur_digit
if result[-1] == cur_digit:
pick_from = hex_digits - set(cur_digit)
else:
pick_from = hex_digits
yield result
my_hex_gen = hexGen()
counter = 0
for result in my_hex_gen:
print(result)
counter += 1
if counter > 10:
break
Results:
1ECC6A83EB14
D0897DE15E81
9C3E9028B0DE
CE74A2674AF0
9ECBD32C003D
0DF2E5DAC0FB
31C48E691C96
F33AAC2C2052
CD4CEDADD54D
40A329FF6E25
5F5D71F823A4
You could also change the while true loop to only produce a certain number of these based on a number passed into the function.
I interpret this question as, "I want to construct a rainbow table by iterating through all strings that have the following qualities. The string has a length of 12, contains only the characters 0-9 and A-F, and it never has the same character appearing three times in a row."
def iter_all_strings_without_triplicates(size, last_two_digits = (None, None)):
a,b = last_two_digits
if size == 0:
yield ""
else:
for c in "0123456789ABCDEF":
if a == b == c:
continue
else:
for rest in iter_all_strings_without_triplicates(size-1, (b,c)):
yield c + rest
for s in iter_all_strings_without_triplicates(12):
print(s)
Result:
001001001001
001001001002
001001001003
001001001004
001001001005
001001001006
001001001007
001001001008
001001001009
00100100100A
00100100100B
00100100100C
00100100100D
00100100100E
00100100100F
001001001010
001001001011
...
Note that there will be several hundred terabytes' worth of values outputted, so you aren't saving much room compared to just saving every single string, triplicates or not.
import string, random
source = string.hexdigits[:16]
result = ''
while len(result) < 12 :
idx = random.randint(0,len(source))
if len(result) < 3 or result[-1] != result[-2] or result[-1] != source[idx] :
result += source[idx]
You could extract a random sequence from a list of twice each hexadecimal digits:
digits = list('1234567890ABCDEF') * 2
random.shuffle(digits)
hex_number = ''.join(digits[:12])
If you wanted to allow shorter sequences, you could randomize that too, and left fill the blanks with zeros.
import random
digits = list('1234567890ABCDEF') * 2
random.shuffle(digits)
num_digits = random.randrange(3, 13)
hex_number = ''.join(['0'] * (12-num_digits)) + ''.join(digits[:num_digits])
print(hex_number)
You could use a generator iterating a window over the strings your current implementation yields. Sth. like (hex_str[i:i + 3] for i in range(len(hex_str) - window_size + 1)) Using len and set you could count the number of different characters in the slice. Although in your example it might be easier to just compare all 3 characters.
You can create an array from 0 to 255, and use random.sample with your list to get your list
Here is my question
count += 1
num = 0
num = num + 1
obs = obs_%d%(count)
mag = mag_%d%(count)
while num < 4:
obsforsim = obs + mag
mylist.append(obsforsim)
for index in mylist:
print index
The above code gives the following results
obs1 = mag1
obs2 = mag2
obs3 = mag3
and so on.
obsforrbd = parentV = {0},format(index)
cmds.dynExpression(nPartilce1,s = obsforrbd,c = 1)
However when i run the code above it only gives me
parentV = obs3 = mag3
not the whole list,it only gives me the last element of the list why is that..??
Thanks.
I'm having difficulty interpreting your question, so I'm just going to base this on the question title.
Let's say you have a list of items (they could be anything, numbers, strings, characters, etc)
myList = [1,2,3,4,"abcd"]
If you do something like:
for i in myList:
print(i)
you will get:
1
2
3
4
"abcd"
If you want to convert this to a string:
myString = ' '.join(myList)
should have:
print(myString)
>"1 2 3 4 abcd"
Now for some explanation:
' ' is a string in python, and strings have certain methods associated with them (functions that can be applied to strings). In this instance, we're calling the .join() method. This method takes a list as an argument, and extracts each element of the list, converts it to a string representation and 'joins' it based on ' ' as a separator. If you wanted a comma separated list representation, just replace ' ' with ','.
I think your indentations wrong ... it should be
while num < 4:
obsforsim = obs + mag
mylist.append(obsforsim)
for index in mylist:
but Im not sure if thats your problem or not
the reason it did not work before is
while num < 4:
obsforsim = obs + mag
#does all loops before here
mylist.append(obsforsim) #appends only last
The usual pythonic way to spit out a list of numbered items would be either the range function:
results = []
for item in range(1, 4):
results.append("obs%i = mag_%i" % (item, item))
> ['obs1 = mag_1', 'obs2 = mag_2', 'ob3= mag_3']
and so on (note in this example you have to pass in the item variable twice to get it to register twice.
If that's to be formatted into something like an expression you could use
'\n'.join(results)
as in the other example to create a single string with the obs = mag pairs on their own lines.
Finally, you can do all that in one line with a list comprehension.
'\n'.join([ "obs%i = mag_%i" % (item, item) for item in range (1, 4)])
As other people have pointed out, while loops are dangerous - its easier to use range
I have a nested list comprehension which has created a list of six lists of ~29,000 items. I'm trying to parse this list of final data, and create six separate dictionaries from it. Right now the code is very unpythonic, I need the right statement to properly accomplish the following:
1.) Create six dictionaries from a single statement.
2.) Scale to any length list, i.e., not hardcoding a counter shown as is.
I've run into multiple issues, and have tried the following:
1.) Using while loops
2.) Using break statements, will break out of the inner most loop, but then does not properly create other dictionaries. Also break statements set by a binary switch.
3.) if, else conditions for n number of indices, indices iterate from 1-29,000, then repeat.
Note the ellipses designate code omitted for brevity.
# Parse csv files for samples, creating a dictionary of key, value pairs and multiple lists.
with open('genes_1') as f:
cread_1 = list(csv.reader(f, delimiter = '\t'))
sample_1_values = [j for i, j in (sorted([x for x in {i: float(j)
for i, j in cread_1}.items()], key = lambda v: v[1]))]
sample_1_genes = [i for i, j in (sorted([x for x in {i: float(j)
for i, j in cread_1}.items()], key = lambda v: v[1]))]
...
# Compute row means.
mean_values = []
for i, (a, b, c, d, e, f) in enumerate(zip(sample_1_values, sample_2_values, sample_3_values, sample_4_values, sample_5_values, sample_6_values)):
mean_values.append((a + b + c + d + e + f)/6)
# Provide proper gene names for mean values and replace original data values by corresponding means.
sample_genes_list = [i for i in sample_1_genes, sample_2_genes, sample_3_genes, sample_4_genes, sample_5_genes, sample_6_genes]
sample_final_list = [sorted(zip(sg, mean_values)) for sg in sample_genes_list]
# Create multiple dictionaries from normalized values for each dataset.
class BreakIt(Exception): pass
try:
count = 1
for index, items in enumerate(sample_final_list):
sample_1_dict_normalized = {}
for index, (genes, values) in enumerate(items):
sample_1_dict_normalized[genes] = values
count = count + 1
if count == 29595:
raise BreakIt
except BreakIt:
pass
...
try:
count = 1
for index, items in enumerate(sample_final_list):
sample_6_dict_normalized = {}
for index, (genes, values) in enumerate(items):
if count > 147975:
sample_6_dict_normalized[genes] = values
count = count + 1
if count == 177570:
raise BreakIt
except BreakIt:
pass
# Pull expression values to qualify overexpressed proteins.
print 'ERG values:'
print 'Sample 1:', round(sample_1_dict_normalized.get('ERG'), 3)
print 'Sample 6:', round(sample_6_dict_normalized.get('ERG'), 3)
Your code is too long for me to give exact answer. I will answer very generally.
First, you are using enumerate for no reason. if you don't need both index and value, you probably don't need enumerate.
This part:
with open('genes.csv') as f:
cread_1 = list(csv.reader(f, delimiter = '\t'))
sample_1_dict = {i: float(j) for i, j in cread_1}
sample_1_list = [x for x in sample_1_dict.items()]
sample_1_values_sorted = sorted(sample_1_list, key=lambda expvalues: expvalues[1])
sample_1_genes = [i for i, j in sample_1_values_sorted]
sample_1_values = [j for i, j in sample_1_values_sorted]
sample_1_graph_raw = [float(j) for i, j in cread_1]
should be (a) using a list named samples and (b) much shorter, since you don't really need to extract all this information from sample_1_dict and move it around right now. It can be something like:
samples = [None] * 6
for k in range(6):
with open('genes.csv') as f: #but something specific to k
cread = list(csv.reader(f, delimiter = '\t'))
samples[k] = {i: float(j) for i, j in cread}
after that, calculating the sum and mean will be way more natural.
In this part:
class BreakIt(Exception): pass
try:
count = 1
for index, items in enumerate(sample_final_list):
sample_1_dict_normalized = {}
for index, (genes, values) in enumerate(items):
sample_1_dict_normalized[genes] = values
count = count + 1
if count == 29595:
raise BreakIt
except BreakIt:
pass
you should be (a) iterating of the samples list mentioned earlier, and (b) not using count at all, since you can iterate naturally over samples or sample[i].list or something like that.
Your code has several problems. You should put your code in functions that preferably do one thing each. Than you can call a function for each sample without repeating the same code six times (I assume that is what the ellipsis is hiding.). Give each function a self-describing name and a doc string that explains what it does. There is quite a bit unnecessary code. Some of this might become obvious once you have it in functions. Since functions take arguments you can hand in your 29595, for example.
What is wrong in the method end in the code?
The method end returns always 1 although it should return 0 with the current data.
# return 1 if the sum of four consecutive elements equal the sum over other sum of the other three sums
# else return 0
# Eg the current sums "35 34 34 34" should return 0
data = "2|15|14|4|12|6|7|9|8|10|11|5|13|3|2|16"
arra = data.split("|");
def do_row ( arra, n ):
return arra[4*n:4 + 4*n]
def row_summa (row):
return sum(map(int,row))
def end ( summat ): # problem here!
equality = 1
for i in summat[2:5]:
print "Comparing: ", summat[1], " and ", i, ".\n"
if summat[1] != i:
equality = 0
print equality
for i in range(0,4):
summat = []
summat.append( row_summa( do_row(arra,i) ) )
print row_summa ( do_row(arra,i) )
summa = 0
end(summat)
I can't really tell what you're trying to do here, but I can certainly say why end() returns 1 instead of 0. In your last for loop, you reset summat to [] at the start of the loop, so at the end, summat only contains a single value (the one you most recently appended on). So when you ask for summat[2:5] on a list of a single item, Python returns an empty list (as there are no values in that range) - in which case there are no chances for equality to be set to zero because the loop in end never runs.
I think you may have an off-by-one error. Remember that array indexes in Python start at 0, not 1. So where you do this:
for i in summat[2:5]:
print "Comparing: ", summat[1], " and ", i, ".\n"
if summat[1] != i:
equality = 0
you are not looking at summat[0] at all. Try perhaps:
for i in summat[1:4]:
print "Comparing: ", summat[0], " and ", i, ".\n"
if summat[0] != i:
equality = 0
You should also study this piece of code
data = "2|15|14|4|12|6|7|9|8|10|11|5|13|3|2|16"
arra = map(int,data.split("|"))
summat = [sum(arra[i:i+4]) for i in range(0,len(arra),4)]
print summat
print len(set(summat))==1
First off, end doesn't return 1. It returns None. It prints 1. Kind of deceptive if you're running it from the command line.
Second, when you call end, summat is equal to [34]. So this:
for i in summat[2:5]:
never even executes. It won't do anything unless summat contains at least 3 elements.
You have two problems. Initialising summat to [] inside the loop, also the off by one error Greg mentioned
data = "2|15|14|4|12|6|7|9|8|10|11|5|13|3|2|16"
arra = data.split("|");
def do_row ( arra, n ):
return arra[4*n:4 + 4*n]
def row_summa (row):
return sum(map(int,row))
def end ( summat ): # problem here!
equality = 1
for i in summat[1:]: # 1 <=== IS THE SECOND ELEMENT
print "Comparing: ", summat[0], " and ", i, ".\n"
if summat[0] != i:
equality = 0
print equality
summat = [] # <=== DO THIS BEFORE THE LOOP
for i in range(0,4):
summat.append( row_summa( do_row(arra,i) ) )
print row_summa ( do_row(arra,i) )
summa = 0
end(summat)