From your expected result, it appears that each “group” is based on contiguous id values. For this, you can use the compare-cumsum-groupby pattern, and then use agg to get the min and max values. # Sample data. df = pd.DataFrame( {‘id’: [1, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, … Read more
Try looking at the diff function in numpy. That will give you the difference between each value and its right neighbor. You can then check if that value is smaller or larger than 0. 0 solved Python check if one value is greater than the next one in array/vector [duplicate]
The code you show does not represent the equation of the ellipse. It’s a bit hard to tell what exactly you were trying in that code, but the approach here is of course completely the same as in the linked question’s answer, namely to calculate the left hand side of the equation with a meshgrid … Read more
all the computationally expensive stuff can be done in opencv. so using python for the wrapper won’t cost you much computationally. 4 solved Image processing using Python and OpenCV on a Raspberry PI [closed]
If you only want groups of two (as opposed to groups of n), then you can hardcode n=2 and use a list comprehension to return a list of lists. This will also create a group of one at the end of the list if the length of the list is odd: some_list = [‘a’,’b’,’c’,’d’,’e’] [some_list[i:i+2] … Read more
You can use random.gauss to generate 6 numbers around a mean with a certain standard deviation (closeness) and as per @PatrickArtner suggestion construct the 7th one, e.g. using 3 standard devs from mean: In []: import random vals = [int(random.gauss(341/7, 3)) for _ in range(6)] vals.append(341 – sum(vals)) vals, sum(vals), sum(vals)/7 Out[]: ([47, 47, 48, … Read more
You were close. Try this: import pandas as pd df = pd.DataFrame({‘a’: [1, 2, 3], ‘b’: [ 4, 5, 2], ‘c’: [2, 4, 5]}) print(df) df = df[[x for _, x in sorted(zip(df.iloc[-1], df.columns), reverse=True)]] print(df) Starting DataFrame: a b c 0 1 4 2 1 2 5 4 2 3 2 5 Columns sorted … Read more
You have the # lines as a separation between your matrix. So if you loop line by line on your file, you can separate the matrix with this # lines, and built them : file = open(“file.txt”, “r”) lines = file.readlines() At this point, lines is a list. lines[i] is the line n° i+1 as … Read more
Check the shape of the two input arrays to plot in plt.plot(tscale1, np.array(np.hstack((0., np.abs(uamp), 0.))), …) solved python plot error of incompatible dimensions with numpy array [closed]
Yes, it is possible to find modulo (p) and sequence value (y) for given number x, but it requires using floating-arithmetics logarithm. pow = Floor(Log2(x+1)) p = 1 << pow y = 1 + (x + 1) % p For initial period s (3 in your second example) pow = Floor(Log2((x + s – 1) … Read more
>>> import numpy as np >>> x = np.arange(-1,1,0.1) >>> print(x) [ -1.00000000e+00 -9.00000000e-01 -8.00000000e-01 -7.00000000e-01 -6.00000000e-01 -5.00000000e-01 -4.00000000e-01 -3.00000000e-01 -2.00000000e-01 -1.00000000e-01 -2.22044605e-16 1.00000000e-01 2.00000000e-01 3.00000000e-01 4.00000000e-01 5.00000000e-01 6.00000000e-01 7.00000000e-01 8.00000000e-01 9.00000000e-01] >>> type(x[0]) <type ‘numpy.float64’> >>> print(x.tolist()) [-1.0, -0.9, -0.8, -0.7000000000000001, -0.6000000000000001, -0.5000000000000001, -0.40000000000000013, -0.30000000000000016, -0.20000000000000018, -0.1000000000000002, -2.220446049250313e-16, 0.09999999999999964, 0.19999999999999973, 0.2999999999999998, 0.3999999999999997, 0.49999999999999956, 0.5999999999999996, … Read more
It’s not so clear what you’re asking, but here are some options to get you started: Manual Min-max transform You could min-max transform so they are all in the range of [0,1], and then multiply by 255. Then, all your values are in the range of 0 to 255, proportional to your original array (i.e. … Read more
Something like this should work… df = pd.DataFrame({‘date’: [‘2017-01-01 01:01:01’, ‘2017-01-02 01:01:01’, ‘2017-01-03 01:01:01’, ‘2017-01-30 01:01:01’, ‘2017-01-31 01:01:01’], ‘value’: [99,98,97,95,94]}) df[‘date’] = pd.to_datetime(df[‘date’]) def get_list(row): subset = df[(row[‘date’] – df[‘date’] <= pd.to_timedelta(‘5 days’)) & (row[‘date’] – df[‘date’] >= pd.to_timedelta(‘0 days’))] return str(subset[‘value’].tolist()) df[‘list’] = df.apply(get_list, axis=1) Output: date value list 0 2017-01-01 01:01:01 99 [99] … Read more
It’s here: def preos(molecule, T, P, plotcubic=True, printresults=True): Tr = T / molecule.Tc # reduced temperature … preos(“methane”, 160, 10, “true”, “true”) You’re clearly passing “methane” into the preos function as a string, then trying to call .Tc on that string. The error is saying exactly that. This doesn’t have anything to do with IPython. … Read more
As I understand, img1 and img2 are matrices containing all the pixels of two images. Suppose: img1 = x x x x x x x x x img2 = o o o o o o o o Output is a matrix whose height is the highest between img1 and img2 and whose width is the … Read more