I have a dataset from which i need to take the index 0 and 1 then process the output then take the index value 2 and 3 then process the output and so on.
The code which i have tried takes the value of index 0 and 1 then 1 and 2 then 2 and 3 and so on.
for i,r in tqdm(gf.iterrows()):
lp = 0
for v in range(0, 10 + 1):
lp += r.length_10
ix.append(i)
basket.append(r.line.interpolate(lp))
The code must take the index value of o and 1 then 2 and 3 then 4 and 5 and so on....0
for v in range(0, 10+1,2):
print(v,v+1)
#prints
#0 1
#2 3
#4 5
#6 7
#8 9
Honestly, I couldn't understand very well your code. But, I think that the problem occurs here:
for v in range(0, 10 + 1):
this means that your iterator will go from 0 to 10 one by one, but try to change this line to:
for v in range(0, 10 + 1,2):
I think this one will do the job
Related
for k in range(8):
for i in range(2): #number of columns
for j in range(4): #number of row
print(k,j,i)
I want an output like this. no repetition of first for loop
k,j,i
-----
0 0 0
1 1 0
2 2 0
3 3 0
4 0 1
5 1 1
6 2 1
7 3 1
How I will achieve this?
Normally i would do
for i in range(8):
print(i, i%4, i%2)
Output:
0 0 0
1 1 1
2 2 0
3 3 1
4 0 0
5 1 1
6 2 0
7 3 1
But to reproduce your exact output:
for i in range(8):
print(i, i%4, int(i>3))
Output:
0 0 0
1 1 0
2 2 0
3 3 0
4 0 0
5 1 1
6 2 1
7 3 1
You can use if statements to say things like if k == 7 or something along those lines. This will only allow it to loop the first loop before moving on to the second loop.
Other answers have shown similar ways to produce your exact output, but this is another way to do it, and this would still work if you wanted the number of rows to be more than 8 and you wanted i to keep increasing
for i in range(8):
print(i, i%4, i//4)
I hope this is useful, sorry if it isn't
Looking at i and j, you have the cartesian product of {0,1} and {0,1,2,3}. You can compute that with itertools.product(range(2), range(4)), then use enumerate to number them for your k value.
from itertools import product
for k, (i, j) in enumerate(product(range(2), range(4))):
print(k, j, i)
Earlier arguments to product vary more slowly than later arguments.
I have a DataFrame that contains gas concentrations and the corresponding valve number. This data was taken continuously where we switched the valves back and forth (valves=1 or 2) for a certain amount of time to get 10 cycles for each valve value (20 cycles total). A snippet of the data looks like this (I have 2,000+ points and each valve stayed on for about 90 seconds each cycle):
gas1 valveW time
246.9438 2 1
247.5367 2 2
246.7167 2 3
246.6770 2 4
245.9197 1 5
245.9518 1 6
246.9207 1 7
246.1517 1 8
246.9015 1 9
246.3712 2 10
247.0826 2 11
... ... ...
My goal is to save the last N points of each valve's cycle. For example, the first cycle where valve=1, I want to index and save the last N points from the end before the valve switches to 2. I would then save the last N points and average them to find one value to represent that first cycle. Then I want to repeat this step for the second cycle when valve=1 again.
I am currently converting from Matlab to Python so here is the Matlab code that I am trying to translate:
% NOAA high
n2o_noaaHigh = [];
co2_noaaHigh = [];
co_noaaHigh = [];
h2o_noaaHigh = [];
ind_noaaHigh_end = zeros(1,length(t_c));
numPoints = 40;
for i = 1:length(valveW_c)-1
if (valveW_c(i) == 1 && valveW_c(i+1) ~= 1)
test = (i-numPoints):i;
ind_noaaHigh_end(test) = 1;
n2o_noaaHigh = [n2o_noaaHigh mean(n2o_c(test))];
co2_noaaHigh = [co2_noaaHigh mean(co2_c(test))];
co_noaaHigh = [co_noaaHigh mean(co_c(test))];
h2o_noaaHigh = [h2o_noaaHigh mean(h2o_c(test))];
end
end
ind_noaaHigh_end = logical(ind_noaaHigh_end);
This is what I have so far for Python:
# NOAA high
n2o_noaaHigh = [];
co2_noaaHigh = [];
co_noaaHigh = [];
h2o_noaaHigh = [];
t_c_High = []; # time
for i in range(len(valveW_c)):
# NOAA HIGH
if (valveW_c[i] == 1):
t_c_High.append(t_c[i])
n2o_noaaHigh.append(n2o_c[i])
co2_noaaHigh.append(co2_c[i])
co_noaaHigh.append(co_c[i])
h2o_noaaHigh.append(h2o_c[i])
Thanks in advance!
I'm not sure if I understood correctly, but I guess this is what you are looking for:
# First we create a column to show cycles:
df['cycle'] = (df.valveW.diff() != 0).cumsum()
print(df)
gas1 valveW time cycle
0 246.9438 2 1 1
1 247.5367 2 2 1
2 246.7167 2 3 1
3 246.677 2 4 1
4 245.9197 1 5 2
5 245.9518 1 6 2
6 246.9207 1 7 2
7 246.1517 1 8 2
8 246.9015 1 9 2
9 246.3712 2 10 3
10 247.0826 2 11 3
Now you can use groupby method to get the average for the last n points of each cycle:
n = 3 #we assume this is n
df.groupby('cycle').apply(lambda x: x.iloc[-n:, 0].mean())
Output:
cycle 0
1 246.9768
2 246.6579
3 246.7269
Let's call your DataFrame df; then you could do:
results = {}
for k, v in df.groupby((df['valveW'].shift() != df['valveW']).cumsum()):
results[k] = v
print(f'[group {k}]')
print(v)
Shift(), as it suggests, shifts the column of the valve cycle allows to detect changes in number sequences. Then, cumsum() helps to give a unique number to each of the group with the same number sequence. Then we can do a groupby() on this column (which was not possible before because groups were either of ones or twos!).
which gives e.g. for your code snippet (saved in results):
[group 1]
gas1 valveW time
0 246.9438 2 1
1 247.5367 2 2
2 246.7167 2 3
3 246.6770 2 4
[group 2]
gas1 valveW time
4 245.9197 1 5
5 245.9518 1 6
6 246.9207 1 7
7 246.1517 1 8
8 246.9015 1 9
[group 3]
gas1 valveW time
9 246.3712 2 10
10 247.0826 2 11
Then to get the mean for each cycle; you could e.g. do:
df.groupby((df['valveW'].shift() != df['valveW']).cumsum()).mean()
which gives (again for your code snippet):
gas1 valveW time
valveW
1 246.96855 2.0 2.5
2 246.36908 1.0 7.0
3 246.72690 2.0 10.5
where you wouldn't care much about the time mean but the gas1 one!
Then, based on results you could e.g. do:
n = 3
mean_n_last = []
for k, v in results.items():
if len(v) < n:
mean_n_last.append(np.nan)
else:
mean_n_last.append(np.nanmean(v.iloc[len(v) - n:, 0]))
which gives [246.9768, 246.65796666666665, nan] for n = 3 !
If your dataframe is sorted by time you could get the last N records for each valve like this.
N=2
valve1 = df[df['valveW']==1].iloc[-N:,:]
valve2 = df[df['valveW']==2].iloc[-N:,:]
If it isn't currently sorted you could easily sort it like this.
df.sort_values(by=['time'])
I'm very new to learning python, though I understand the basics of the looping, I am unable to understand the method in which output is arrived at.
In particular, how does the mapping of all three for loops happen to give the desired output, as I finding it impossible to understand the logic to be applied, when I try to write the output on paper without referring to IDE.
Code:
n = 4
a = 3
z = 2
for i in range(n):
for j in range(a):
for p in range(z):
print(i, j, p)
Output is:
0 0 0
0 0 1
0 1 0
0 1 1
0 2 0
0 2 1
1 0 0
1 0 1
1 1 0
1 1 1
1 2 0
1 2 1
2 0 0
2 0 1
2 1 0
2 1 1
2 2 0
2 2 1
3 0 0
3 0 1
3 1 0
3 1 1
3 2 0
3 2 1
The first loop iterates four times.
The second loop iterates three times. However since it is embedded inside the first loop, it actually iterates twelve times (4 * 3.)
The third loop iterates two times. However since it is embedded inside the first and second loops, it actually iterates twenty-four times (4 * 3 * 2).
while I was working on the Python practice, I found a question that I cannot solve by myself.
The question is,
Input one integer(n), and then write the codes that make a triangle using 1 to 'n'. Use the following picture. You should make only one function, and call that function various times to solve the question. The following picture is the result that you should make in the codes.
Receive one integer as an argument, print the number from 1 to the integer received as a factor in a single line, and then print the line break character at the end. Once this function is called, only one line of output should be printed.
So by that question, I found that this is a question that requires the
recursion since I have to call your function only once.
I tried to work on the codes that I made many times, but I couldn't solve it.
global a
a = 1
def printLine(n):
global a
if (n == 0):
return
for i in range(1, a + 1):
print(i, end=" ")
print()
a += 1
for k in range(1, n+1):
print(k, end=" ")
print()
printLine(n - 1)
n = int(input())
printLine(n)
Then I wrote some codes to solve this question, but the ascending and descending part is kept overlapping. :(
What I need to do is to break two ascending and descending parts separately in one function, but I really cannot find how can I do that. So which part should I have to put the recursive function call?
Or is there another way can divide the ascending and descending part in the function?
Any ideas, comments, or solutions are appreciated.
Thx
You can use the below function:
def create_triangle(n, k: int = 1, output: list = []):
if n == 1:
output.append(n)
return output
elif k >= n:
output.append(" ".join([str(i) for i in range(1, n + 1)]))
return create_triangle(n - 1, k)
else:
output.append(" ".join([str(i) for i in range(1, n + 1)[:k]]))
return create_triangle(n, k + 1)
for i in create_triangle(5):
print(i)
Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1
# function to print all the numbers from 1 to n with spaces
def printLine(k):
# create a range. if k is 4, will create the range: 1, 2, 3, 4
rng = range(1, k + 1)
# convert each number to string
str_rng = map(lambda x: str(x), rng)
# create one long string with spaces
full_line = " ".join(str_rng)
print(full_line)
# capture input
n = int(input())
# start from 1, and up to n, printing the first half of the triangle
for i in range(1, n):
printLine(i)
# now create the bottom part, by creating a descending range
for i in range(n, 0, -1):
printLine(i)
Using default parameter as a dict, you can manipulate it as your function variables, so in that way, you can have a variable in your function that keeps the current iteration you are at and if your function is ascending or descending.
def triangle_line(n, config={'max':1, 'ascending':True}):
print(*range(1, config['max'] + 1))
if config['ascending']:
config['max'] += 1
else:
config['max'] -= 1
if config['max'] > n:
config['ascending'] = False
config['max'] = n
elif config['max'] == 0:
config['ascending'] = True
config['max'] = 1
Each call you make will return one iteration.
>>> triangle_line(4)
1
>>> triangle_line(4)
1 2
>>> triangle_line(4)
1 2 3
>>> triangle_line(4)
1 2 3 4
>>> triangle_line(4)
1 2 3 4
>>> triangle_line(4)
1 2 3
>>> triangle_line(4)
1 2
>>> triangle_line(4)
1
Or you can run on a loop, two times your input size.
>>> n = 4
>>> for i in range(0,n*2):
... triangle_line(n)
...
1
1 2
1 2 3
1 2 3 4
1 2 3 4
1 2 3
1 2
1
I'm trying to print a half pyramid that stars on the left side in python.
So far, this is my code
for i in range(1,12):
for j in range(12 - i):
print(" ", end = " ")
for j in range(1, i):
print(j, end = " " )
print("\n")
and my output is
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
1 2 3 4 5 6
1 2 3 4 5 6 7
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9 10
However, my output is meant to be in the opposite order:
1
2 1
3 2 1
4 3 2 1
5 4 3 2 1
How can I make this change?
Just reverse the second loop -- the one that prints that actual numbers:
for j in range(i-1, 0, -1):
The last parameter controls the "step", or how much the variable changes on each loop iteration. Output:
1
2 1
3 2 1
4 3 2 1
5 4 3 2 1
...
Reverse the range by adding the third argument (-1). Also format your numbers to use 2 places, so 10 is not pushing the last line to the right. Finally, the last print should probably not have \n, since that is already the default ending character of print:
for i in range(1,12):
for j in range(12 - i):
print(" ", end = "")
for j in range(i-1, 0,-1):
print(str(j).rjust(2), end = "" )
print()
You could just reverse the range that you print out as numbers
for i in range(1,12):
for j in range(12 - i):
print(" ", end = " ")
for j in reversed(range(1, i)):
print(j, end = " " )
print("\n")
The problem is in your second for loop, as you are looping from 1 to i, meaning you start off with 1 being printed first, and every following number until (not including) i.
Fortunately, for loops are able to go in reverse. So, instead of:
for j in range(1, i)
You could write:
for j in range((i-1), 0, -1)
Where the first parameter is signifies where the loop starts, the second is where the loop finishes, and the third signifies how large our jumps are going to be, in this case negative. The reason we are starting at i-1 and finishing at 0 is because loops start at exactly the first given number, and loop until just before the second given number, so in your given code the loop stops just before i, and so this one starts just before i as well, although you could remove the -1 if you wish to include 12 in the pyramid.