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How to expand the output of a truncated list to view more values of Pandas dataframe

Is there a specific way to display all the truncated data values of a list. The displayed values are as follows
v w
Row1 [0.1, 0.2, 0.3 .....1.0] [0.1, 0.2, 0.3 .....1.0]
Here are the option I tried
Option 1
p
pd.set_option('display.max_colwidth', None)
pd.set_option('display.max_seq_items', None)
z = pd.read_csv('a.csv')
Output:
it is still truncated
Option 2
for i, row in z.iterrows():
for j in row['w']:
print(j)
Output:
it is stll truncated
Any help on how to display all the truncated values and display the full list.

You can print after conversion to_string:
print(df.to_string())
output:
v w
Row1 [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0] [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0]
Used input:
L = list(np.arange(0, 1.1, 0.1).round(2))
df = pd.DataFrame({'v': [L], 'w': [L]}, index=['Row1'])
default print:
v \
Row1 [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, ...
w
Row1 [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, ...

Get indices greater than value and keep value

I have a 2D array that looks like this:
[[0.1, 0.2, 0.4, 0.6, 0.9]
[0.3, 0.7, 0.8, 0.3, 0.9]
[0.7, 0.9, 0.4, 0.6, 0.9]
[0.1, 0.2, 0.6, 0.6, 0.9]]
And I want to save the indices where the array is higher than 0.6 but I also want to keep the value of that position, so the output would be:
[0, 3, 0.6]
[0, 4, 0.9]
[1, 2, 0.7]
and so on.
To get the indices I did this:
x = np.where(PPCF> 0.6)
high_pc = np.asarray(x).T.tolist()
but how do I keep the value in a third position?

Simple, no loops:
x = np.where(PPCF > 0.6) # condition to screen values
vals = PPCF[x] # find values by indices
np.concatenate((np.array(x).T, vals.reshape(vals.size, 1)), axis = 1) # resulting array
Feel free to convert it to a list.

This should work :
x = np.where(PPCF> 0.6)
high_pc = np.asarray(x).T.tolist()
for i in high_pc:
i.append(float(PPCF[i[0],i[1]]))

You could just run a loop along the columns and rows and check if each element is greater than the threshold and save them in a list.
a = [[0.1, 0.2, 0.4, 0.6, 0.9],
[0.3, 0.7, 0.8, 0.3, 0.9],
[0.7, 0.9, 0.4, 0.6, 0.9],
[0.1, 0.2, 0.6, 0.6, 0.9]]
def find_ix(a, threshold = 0.6):
res_list = []
for i in range(len(a)):
for j in range(len(a[i])):
if a[i][j] >= threshold:
res_list.append([i, j, a[i][j]])
return res_list
print("Resulting list = \n ", find_ix(a))

import numpy as np
arr = np.array([[0.1, 0.2, 0.4, 0.6, 0.9],
[0.3, 0.7, 0.8, 0.3, 0.9],
[0.7, 0.9, 0.4, 0.6, 0.9],
[0.1, 0.2, 0.6, 0.6, 0.9]])
rows, cols = np.where(arr > 0.6) # Get rows and columns where arr > 0.6
values = arr[rows, cols] # Get all values > 0.6 in arr
result = np.column_stack((rows, cols, values)) # Stack three columns to create final array
"""
Result -
[ 0. 4. 0.9]
[ 1. 1. 0.7]
[ 1. 2. 0.8]
[ 1. 4. 0.9]
[ 2. 0. 0.7]
[ 2. 1. 0.9]
[ 2. 4. 0.9]
[ 3. 4. 0.9]]
"""
You can convert result into a list.

Why does random.shuffle fail on numpy lists?

I have an array of row vectors, upon which I run random.shuffle:
#!/usr/bin/env python
import random
import numpy as np
zzz = np.array([[0.1, 0.2, 0.3, 0.4, 0.5],
[0.6, 0.7, 0.8, 0.9, 1. ]])
iterations = 100000
f = 0
for _ in range(iterations):
random.shuffle(zzz)
if np.array_equal(zzz[0], zzz[1]):
print(zzz)
f += 1
print(float(f)/float(iterations))
Between 99.6 and 100% of the time, using random.shuffle on zzz returns a list with the same elements in it, e.g.:
$ ./test.py
...
[[ 0.1 0.2 0.3 0.4 0.5]
[ 0.1 0.2 0.3 0.4 0.5]]
0.996
Using numpy.random.shuffle appears to pass this test and shuffle row vectors correctly. I'm curious to know why random.shuffle fails.

If you look at the code of random.shuffle it performs swaps in the following way:
x[i], x[j] = x[j], x[i]
which for a numpy.array would fail, without raising any error. Example:
>>> zzz[1], zzz[0] = zzz[0], zzz[1]
>>> zzz
array([[0.1, 0.2, 0.3, 0.4, 0.5],
[0.1, 0.2, 0.3, 0.4, 0.5]])
The reason is that Python first evaluates the right hand side completely and then make the assignment (this is why with Python single line swap is possible) but for a numpy array this is not True.
numpy
>>> arr = np.array([[1],[1]])
>>> arr[0], arr[1] = arr[0]+1, arr[0]
>>> arr
array([[2],
[2]])
Python
>>> l = [1,1]
>>> l[0], l[1] = l[0]+1, l[0]
>>> l
[2, 1]

Try it like this :
#!/usr/bin/env python
import random
import numpy as np
zzz = np.array([[0.1, 0.2, 0.3, 0.4, 0.5],
[0.6, 0.7, 0.8, 0.9, 1. ]])
iterations = 100000
f = 0
for _ in range(iterations):
random.shuffle(zzz[0])
random.shuffle(zzz[1])
if np.array_equal(zzz[0], zzz[1]):
print(zzz)
f += 1
print(float(f)/float(iterations))

In [200]: zzz = np.array([[0.1, 0.2, 0.3, 0.4, 0.5],
...: [0.6, 0.7, 0.8, 0.9, 1. ]])
...:
In [201]: zl = zzz.tolist()
In [202]: zl
Out[202]: [[0.1, 0.2, 0.3, 0.4, 0.5], [0.6, 0.7, 0.8, 0.9, 1.0]]
random.random is probably using an in-place assignment like:
In [203]: zzz[0],zzz[1]=zzz[1],zzz[0]
In [204]: zzz
Out[204]:
array([[0.6, 0.7, 0.8, 0.9, 1. ],
[0.6, 0.7, 0.8, 0.9, 1. ]])
Note the replication.
But applied to a list of lists:
In [205]: zl[0],zl[1]=zl[1],zl[0]
In [206]: zl
Out[206]: [[0.6, 0.7, 0.8, 0.9, 1.0], [0.1, 0.2, 0.3, 0.4, 0.5]]
In [207]: zl[0],zl[1]=zl[1],zl[0]
In [208]: zl
Out[208]: [[0.1, 0.2, 0.3, 0.4, 0.5], [0.6, 0.7, 0.8, 0.9, 1.0]]
I tested zl = list(zzz) and still got the array behavior. This zl is a list with views of zzz. tolist makes a list of lists thats totally independent ofzzz`.
In short random.random cannot handle inplace modifications of a ndarray correctly. np.random.shuffle is designed to work with the 1st dim of an array, so it gets it right.
correct assignment for ndarray is:
In [211]: zzz = np.array([[0.1, 0.2, 0.3, 0.4, 0.5],
...: [0.6, 0.7, 0.8, 0.9, 1. ]])
...:
In [212]: zzz[[0,1]] = zzz[[1,0]]
In [213]: zzz
Out[213]:
array([[0.6, 0.7, 0.8, 0.9, 1. ],
[0.1, 0.2, 0.3, 0.4, 0.5]])
In [214]: zzz[[0,1]] = zzz[[1,0]]
In [215]: zzz
Out[215]:
array([[0.1, 0.2, 0.3, 0.4, 0.5],
[0.6, 0.7, 0.8, 0.9, 1. ]])

How can a tensor in tensorflow be sliced ​using elements of another array as an index?

I'm looking for a similar function to tf.unsorted_segment_sum, but I don't want to sum the segments, I want to get every segment as a tensor.
So for example, I have this code:
(In real, I have a tensor with shapes of (10000, 63), and the number of segments would be 2500)
to_be_sliced = tf.constant([[0.1, 0.2, 0.3, 0.4, 0.5],
[0.3, 0.2, 0.2, 0.6, 0.3],
[0.9, 0.8, 0.7, 0.6, 0.5],
[2.0, 2.0, 2.0, 2.0, 2.0]])
indices = tf.constant([0, 2, 0, 1])
num_segments = 3
tf.unsorted_segment_sum(to_be_sliced, indices, num_segments)
The output would be here
array([sum(row1+row3), row4, row2]
What I am looking for is 3 tensor with different shapes (maybe a list of tensors), first containing the first and third rows of the original (shape of (2, 5)), the second contains the 4th row (shape of (1, 5)), the third contains the second row, like this:
[array([[0.1, 0.2, 0.3, 0.4, 0.5],
[0.9, 0.8, 0.7, 0.6, 0.5]]),
array([[2.0, 2.0, 2.0, 2.0, 2.0]]),
array([[0.3, 0.2, 0.2, 0.6, 0.3]])]
Thanks in advance!

You can do that like this:
import tensorflow as tf
to_be_sliced = tf.constant([[0.1, 0.2, 0.3, 0.4, 0.5],
[0.3, 0.2, 0.2, 0.6, 0.3],
[0.9, 0.8, 0.7, 0.6, 0.5],
[2.0, 2.0, 2.0, 2.0, 2.0]])
indices = tf.constant([0, 2, 0, 1])
num_segments = 3
result = [tf.boolean_mask(to_be_sliced, tf.equal(indices, i)) for i in range(num_segments)]
with tf.Session() as sess:
print(*sess.run(result), sep='\n')
Output:
[[0.1 0.2 0.3 0.4 0.5]
[0.9 0.8 0.7 0.6 0.5]]
[[2. 2. 2. 2. 2.]]
[[0.3 0.2 0.2 0.6 0.3]]

For your case, you can do Numpy slicing in Tensorflow. So this will work:
sliced_1 = to_be_sliced[:3, :]
# [[0.4 0.5 0.5 0.7 0.8]
# [0.3 0.2 0.2 0.6 0.3]
# [0.3 0.2 0.2 0.6 0.3]]
sliced_2 = to_be_sliced[3, :]
# [0.3 0.2 0.2 0.6 0.3]
Or a more general option, you can do it in the following way:
to_be_sliced = tf.constant([[0.1, 0.2, 0.3, 0.4, 0.5],
[0.3, 0.2, 0.2, 0.6, 0.3],
[0.9, 0.8, 0.7, 0.6, 0.5],
[2.0, 2.0, 2.0, 2.0, 2.0]])
first_tensor = tf.gather_nd(to_be_sliced, [[0], [2]])
second_tensor = tf.gather_nd(to_be_sliced, [[3]])
third_tensor = tf.gather_nd(to_be_sliced, [[1]])
concat = tf.concat([first_tensor, second_tensor, third_tensor], axis=0)

Fast way to iterate over three dictionaries?

I am dealing with very large three dictionaries which looks like this:
dict_a = { ( 't','e' ) : [0.5,0.1,0.6], ( 'a','b' ) : [0.2,0.3,0.9] }
dict_b = { ( 'a','b' ) : [0.1,0.5,0.3] , ( 't','e' ) : [0.6,0.1,0.6] }
dict_c = { ( 'a','b' ) : [0.1,0.5,0.3] , ( 't','e' ) : [0.6,0.5,0.6] }
I am looking for the output like this :
name first_value second_value third_value
0 (t, e) [0.5, 0.1, 0.6] [0.6, 0.1, 0.6] [0.6, 0.5, 0.6]
1 (a, b) [0.2, 0.3, 0.9] [0.1, 0.5, 0.3] [0.1, 0.5, 0.3]
What I've tried is :
final_dict = {'name': [] , 'first_value' : [] ,'second_value': [] , 'third_value': [] }
for a,b in dict_a.items():
for c,d in dict_b.items():
for e,f in dict_c.items():
if a==c==e:
final_dict['name'].append(a)
final_dict['first_value'].append(b)
final_dict['second_value'].append(d)
final_dict['third_value'].append(f)
Which is really not efficient and optimize way to do this task. I was thinking to use pandas.
How can I do this task in minimal time complexity?
Thank you !

Because these are dictionaries, you only need to iterate over one. You can use the key to get the corresponding value from the others.
Example:
for key, value in dict_a.items():
final_dict['name'].append(key)
final_dict['first_value'].append(value)
final_dict['second_value'].append(dict_b[key])
final_dict['third_value'].append(dict_c[key])

Try this way:-
df = pd.DataFrame([dict_a, dict_b, dict_c], index = ['first_value',
'second_value', 'third_value']).T
df['names'] = df.index
df.index = [0, 1]
print(df)
Output:-
first_value second_value third_value names
0 [0.2, 0.3, 0.9] [0.1, 0.5, 0.3] [0.1, 0.5, 0.3] (a, b)
1 [0.5, 0.1, 0.6] [0.6, 0.1, 0.6] [0.6, 0.5, 0.6] (t, e)

How about:
pd.DataFrame({i:d for i,d in enumerate([dict_a,dict_b,dict_c])} )
Output:
0 1 2
a b [0.2, 0.3, 0.9] [0.1, 0.5, 0.3] [0.1, 0.5, 0.3]
t e [0.5, 0.1, 0.6] [0.6, 0.1, 0.6] [0.6, 0.5, 0.6]

Here is one way
pd.concat([pd.Series(x) for x in [dict_a,dict_b,dict_c]],axis=1)
Out[332]:
0 1 2
a b [0.2, 0.3, 0.9] [0.1, 0.5, 0.3] [0.1, 0.5, 0.3]
t e [0.5, 0.1, 0.6] [0.6, 0.1, 0.6] [0.6, 0.5, 0.6]