I want to plot each list of tuples generated by groupby command.
import more_itertools as mit
df=pd.DataFrame({'a': [0,1,2,0,1,2,3], 'b':[2,10,24,56,90,1,3]})
for group in mit.consecutive_groups(zip(df['a'],df['b']),ordering=lambda t:t[0]):
print(list(group))
output:
[(0, 2), (1, 10),(2,24)]
[(0,56),(1,90),(2,1),(3,3)]
I want to plot first index of group [(0, 2), (1, 10),(2,24)] taking first element as x and second element of tuple as y ( x=0,y=2). The same applies to following list of tuples. I am still trying, but have not figured yet.
You are looking for:
df.assign(grp = df.a.diff().ne(1).cumsum()).groupby('grp').plot('a','b')
Related
I want an array that looks like this,
array([array([[1, 1], [2, 2]]), array([3, 3])], dtype=object)
I can make an empty array and then assign elements one by one like this,
z = [np.array([[1,1],[2,2]]), np.array([3,3])]
x = np.empty(shape=2, dtype=object)
x[0], x[1] = z
I thought if this possible then so should be this: x = np.array(z, dtype=object), but that gets me the error: ValueError: could not broadcast input array from shape (2,2) into shape (2).
So is the way given above the only way to make a ragged numpy array? Or, is there a nice one line constructor/function we can can call to make the array x from above.
I'm writing a python script to read in a netCDF file in Ugrid format. This requires reading in two 2D arrays:
x_coordinate = [[0,0],[1,200],[2,400],[3,600],[4,800]...]
y_coordinate = [[0,0],[1,5],[2,10],[3,15],[4,20]...]
and outputting an array:
coordinates = [[0,0],[200,5],[400,10],[600,15],[800,20]...]
so that I can then display it through mpl. Is there a way to do this efficiently without iterating through with comparative if statements?
In my opinion, zip() with list comprehension should solve your problem.
Example as follows:
>>>list(zip([el[1] for el in y_coordinate], [el[1] for el in x_coordinate]))
[(0, 0), (5, 200), (10, 400), (15, 600), (20, 800)]
Can you try:
[[x[0][1], x[1][1]] for x in zip(x_coordinate, y_coordinate)]
Say I have an image of 2x2 pixels named image_array, each pixel color is identified by a tuple of 3 entries (RGB), so the shape of image_array is 2x2x3.
I want to create an np.array c which has the shape 2x2x1 and which last coordinate is an empty set.
I tried this:
import numpy as np
image = (((1,2,3), (1,0,0)), ((1,1,1), (2,1,2)))
image_array = np.array(image)
c = np.empty(image_array.shape[:2], dtype=set)
c.fill(set())
c[0][1].add(124)
print(c)
I get:
[[{124} {124}]
[{124} {124}]]
And instead I would like the return:
[[{} {124}]
[{} {}]]
Any idea ?
The object array has to be filled with separate set() objects. That means creating them individually, as I do with a list comprehension:
In [279]: arr = np.array([set() for _ in range(4)]).reshape(2,2)
In [280]: arr
Out[280]:
array([[set(), set()],
[set(), set()]], dtype=object)
That construction should highlight that fact that this array is closely related to a list, or list of lists.
Now we can do a set operation on one of those elements:
In [281]: arr[0,1].add(124) # more idiomatic than arr[0][1]
In [282]: arr
Out[282]:
array([[set(), {124}],
[set(), set()]], dtype=object)
Note that we cannot operate on more than one set at a time. The object array offers few advantages compared to a list.
This is a 2d array; the sets don't form a dimension. Contrast that with
In [283]: image = (((1,2,3), (1,0,0)), ((1,1,1), (2,1,2)))
...: image_array = np.array(image)
...:
In [284]: image_array
Out[284]:
array([[[1, 2, 3],
[1, 0, 0]],
[[1, 1, 1],
[2, 1, 2]]])
While it started with tuples, it made a 3d array of integers.
Try this:
import numpy as np
x = np.empty((2, 2), dtype=np.object)
x[0, 0] = set(1, 2, 3)
print(x)
[[{1, 2, 3} None]
[None None]]
For non-number types in numpy you should use np.object.
whenever you do fill(set()), this will fill the array with exactly same set, as they refer to the same set. To fix this, just make a set if there isnt one everytime you need to add to the set
c = np.empty(image_array.shape[:2], dtype=set)
if not c[0][1]:
c[0,1] = set([124])
else:
c[0,1].add(124)
print (c)
# [[None {124}]
# [None None]]
Try changing your line c[0][1].add to this.
c[0][1] = 124
print(c)
I am converting a user input list of strings into tuples. The user inputs a list of fractions ie: (Please no "import fractions" suggestions)
fractions = ["1/2","3/5","4/3","3/8","1/9","4/7"]
I normally would use the following code that works:
user_input = 0
list_frac = []
print('Enter fractions into a list until you type "stop" in lower case:')
while user_input != 'stop':
user_input = input('Enter a fraction ie: "1/2" >>>')
list_frac.append(user_input)
list_frac.pop() # pop "stop" off the list
result = []
for i in list_frac:
result.append(tuple(i.split('/')))
print(result)
The result is a list of tuples:
fractions = [('1','2'),('3','5')('4','3'),('3','8'),('1','9'),('4','7')]
I want to change the values in the tuples to integers as well and I dont know how
However I also wish to learn lambda functions so I am practicing on simple code like this. This is my attempt at the same code using lambda function syntax:
tup_result = tuple(map(lambda i: result.append(i.split('/')), result))
But the result is an empty list and no error codes to help me.
The Question: How to change the strings in the list of tuples to ints, and then accomplish all this with the lambda function one liner.
Any suggestions, I have the general concept pf a lambda function down but actually implementing this is a little confusing, thanks for the help folks!
I used comprehensitions to solve the task:
fractions = ["1/2","3/5","4/3","3/8","1/9","4/7"]
print([(int(x),int(y)) for (x,y) in [k.split('/') for k in fractions]])
>>>[(1, 2), (3, 5), (4, 3), (3, 8), (1, 9), (4, 7)]
I started with python not long time ago myself and was confused how to use lambda in the beginning as well. Then I read, that Guido van Rossum had suggested, that lambda forms would disappear in Python3.0 AlternateLambdaSyntax, since then I have not used lambda at all and have no problem with the issue at all. You have to understand how it works when you see it in some code, but you can almost always can write more readable code without using lambda (though I can be wrong). I hope, it helped.
Update
there is a solution solution with map() and lambda, though I would not wish to see it in my code on my worst enemy:
print([(int(x),int(y)) for [x,y] in list(map(lambda frac: frac.split('/'),fractions))])
>>>[(1, 2), (3, 5), (4, 3), (3, 8), (1, 9), (4, 7)]
I'm currently using np.loadtxt to load some mixed data into a structured numpy array. I do some calculations on a few of the columns to output later. For compatibility reasons I need to maintain a specific output format so I'd like to insert those columns at specific points and use np.savetxt to export the array in one shot.
A simple setup:
import numpy as np
x = np.zeros((2,),dtype=('i4,f4,a10'))
x[:] = [(1,2.,'Hello'),(2,3.,'World')]
newcol = ['abc','def']
For this example I'd like to make newcol the 2nd column. I'm very new to Python (coming from MATLAB). From my searching all I've been able to find so far are ways to append newcol to the end of x to make it the last column, or x to newcol to make it the first column. I also turned up np.insert but it doesn't seem to work on a structured array because it's technically a 1D array (from my understanding).
What's the most efficient way to accomplish this?
EDIT1:
I investigated np.savetxt a little further and it seems like it can't be used with a structured array, so I'm assuming I would need to loop through and write each row with f.write. I could specify each column explicitly (by field name) with that approach and not have to worry about the order in my structured array, but that doesn't seem like a very generic solution.
For the above example my desired output would be:
1, abc, 2.0, Hello
2, def, 3.0, World
This is a way to add a field to the array, at the position you require:
from numpy import zeros, empty
def insert_dtype(x, position, new_dtype, new_column):
if x.dtype.fields is None:
raise ValueError, "`x' must be a structured numpy array"
new_desc = x.dtype.descr
new_desc.insert(position, new_dtype)
y = empty(x.shape, dtype=new_desc)
for name in x.dtype.names:
y[name] = x[name]
y[new_dtype[0]] = new_column
return y
x = zeros((2,), dtype='i4,f4,a10')
x[:] = [(1, 2., 'Hello'), (2, 3., 'World')]
new_dt = ('my_alphabet', '|S3')
new_col = ['abc', 'def']
x = insert_dtype(x, 1, new_dt, new_col)
Now x looks like
array([(1, 'abc', 2.0, 'Hello'), (2, 'def', 3.0, 'World')],
dtype=[('f0', '<i4'), ('my_alphabet', 'S3'), ('f1', '<f4'), ('f2', 'S10')])
The solution is adapted from here.
To print the recarray to file, you could use something like:
from matplotlib.mlab import rec2csv
rec2csv(x,'foo.txt')