Wrong Output when adding list to dict - python

I am trying to add lists l as values for different dict d keys. For the array a, having [6,12,18,24,30] I am trying to get the dict d to contain the following key-value pairs:
d[6] = [0, 0, 0.....0]
d[12] = [6, 0, 0, ..0]
d[18] = [6, 12, 0, ...0]
d[24] = [6, 12, 18, 0, ..0]
Where there are 59 elements in each of the lists above.
I use the code below to do this, but my output for the key 24 is:
{24: [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 12, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 18, 12, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 24, 18, 12, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]}
I am trying to understand where I am going wrong.
d = {}
l = []
a =numpy.array([6,12,18,24,30])
for x, value in numpy.ndenumerate(a):
months_to_maturity = value
for i in range(6, 354, 6):
if i <= months_to_maturity:
l.append(months_to_maturity - i)
else:
l.append(0)
d[months_to_maturity] = l

You're always appending to the same list. Thus all the dictionary values end up pointing to the same list. You want to append to a different list each time:
d = {}
a = numpy.array([6, 12, 18, 24, 30])
for months_to_maturity in a:
l = []
for i in range(6, 354, 6):
if i <= months_to_maturity:
l.append(months_to_maturity - i)
else:
l.append(0)
d[months_to_maturity] = l

Related

RuntimeError: Could not infer dtype of dict

RuntimeError: Could not infer dtype of dict
I'm getting above error when i'm using a pre-trained transformer model from hugging face
https://huggingface.co/SEBIS/code_trans_t5_base_code_comment_generation_java_transfer_learning_finetune
I want to fine-tune the model on one of the custom datasets that I have prepared.
Please access the dataset from Kaggle
https://www.kaggle.com/datasets/codkiller0911/kotlin-code-snippets
adding the dataset tot he notebook
import pandas as pd
train_dataset = pd.read_csv('kotlin_code_comment_train.tsv',sep = '\t')
val_dataset = pd.read_csv('kotlin_code_comment_test.tsv',sep = '\t')
Preparing the dataset for tokenisation
from transformers import AutoTokenizer, AutoModelForSeq2SeqLM
from transformers import TrainingArguments, Trainer
tokenizer = AutoTokenizer.from_pretrained("SEBIS/code_trans_t5_base_code_comment_generation_java_transfer_learning_finetune")
model = AutoModelForSeq2SeqLM.from_pretrained("SEBIS/code_trans_t5_base_code_comment_generation_java_transfer_learning_finetune")
train_encodings = tokenizer(train_dataset["Code_Function"].values.tolist(),truncation=True,padding=True,max_length=512)
train_message = tokenizer(train_dataset["Message"].values.tolist(),truncation=True,padding=True,max_length=512)
val_encodings = tokenizer(val_dataset["code_function"].values.tolist(),
truncation=True,
padding=True)
val_message = tokenizer(val_dataset["message"].values.tolist(),
truncation=True,
padding=True)
Loading the model and training args to start the training
import torch
from torch.utils.data import Dataset, DataLoader
class CustomTextDataset(Dataset):
def __init__(self, txt, labels):
self.labels = labels
self.text = txt
def __getitem__(self, idx):
text = {key: torch.tensor(val[idx]) for key, val in self.text.items()}
if self.labels:
text["labels"] = {key: torch.tensor(val[idx]) for key, val in self.labels.items()}
return text
def __len__(self):
return len(self.labels["input_ids"])
train_dataset = CustomTextDataset(train_encodings, train_message)
val_dataset = CustomTextDataset(val_encodings, val_message)
def compute_metrics(p):
print(type(p))
pred, labels = p
pred = np.argmax(pred, axis=1)
accuracy = accuracy_score(y_true=labels, y_pred=pred)
recall = recall_score(y_true=labels, y_pred=pred)
precision = precision_score(y_true=labels, y_pred=pred)
f1 = f1_score(y_true=labels, y_pred=pred)
return {"accuracy": accuracy, "precision": precision, "recall": recall, "f1": f1}
# Define Trainer
args = TrainingArguments(
output_dir="output",
max_steps=3000
)
trainer = Trainer(
model=model,
args=args,
train_dataset=train_dataset,
eval_dataset=val_dataset,
compute_metrics=compute_metrics
)
trainer.train()
After tokenization this is how the dataset looks tokenisation
{'input_ids': tensor([ 3800, 9146, 259, 588, 542, 219, 6698, 171, 9146, 259,
588, 542, 402, 12, 312, 587, 745, 14, 3199, 1330,
11, 204, 35, 540, 587, 745, 14, 3199, 1330, 29,
5, 6698, 35, 587, 554, 1233, 15571, 12, 32, 35,
3963, 11, 8338, 542, 1233, 219, 6698, 35, 3963, 402,
16, 32, 35, 1838, 11, 204, 35, 1838, 16, 1,
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'attention_mask': tensor([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
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'labels': {'input_ids': tensor([9401, 10, 15, 2406, 755, 41, 19, 60, 42, 204, 19, 113,
236, 42, 4, 1, 0, 0, 0, 0, 0, 0, 0, 0,
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0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
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'attention_mask': tensor([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
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0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0])}}
When I execute the training command I get the above-mentioned error.
Anyone have an solutions to this. Kind of banging my head on a wall for quite some time now in finding an solution to this.

Getting a non-optimal answer from KMeans algorithm run on one column

I am running KMeans on a singular column of a dataset containing the number of employees in companies. The graph of the data looks like this:
graph of dataset.
After looking at this and running KMeans on this particular dataset, it gives an output like this which is clearly not even close to correct:
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2, 2, 2,
2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
The code I've used for this is:
def perform_clustering_oneCol(data):
df2 = data.loc[:,"#Employees"]
print("hi: \n")
print(df2.head(15))
df2.dropna()
df2 = pd.get_dummies(df2)
arr = df2.to_numpy()
km = KMeans(n_clusters=3).fit(arr)
print(km.labels_.tolist())
I'm very new to this and would like to know how to get a more optimal result OR what other algorithms I can look at. Any help would be appreciated. Thanks.

How to create an assignment problem cost matrix

I have 250 projects and 50 supervisors. Each of our 130 students rank their preferred choice of projects from 1-5 (1 being favourite) and don't score the other 245. I would like to assign each student to a supervisor but a supervisor can only have up to 12 students.
I'm trying to make some dummy data but struggling to make the cost matrix.
Defining a matrix with random integers of size (5 x 5).
import random
def createMatrix(n):
firstRow = random.sample(range(n),n)
permutes = random.sample(range(n),n)
return list(firstRow[i:]+firstRow[:i] for i in permutes)
N = 5
m = createMatrix(N)
for i in m:
print(i)
[1, 0, 3, 4, 2]
[3, 4, 2, 1, 0]
[4, 2, 1, 0, 3]
[2, 1, 0, 3, 4]
[0, 3, 4, 2, 1]
But this is a Latin square matrix and I would like to allow some rows to contain duplicate values representing that two students have ranked a given same project the same value. Any thoughts would be much appreciated.
You have 130 students and want some duplicated choices. So generate f.e. 110 choices (already minor chance of a duplicate - but with 5 out of 250 not much).
Then choose some of the generated choices as dupe-candidates and and add some of them again until you got back up to 130 choices:
from random import sample,shuffle
projects = 250
students = 130
dup_choices = 20 # means we generate 110 choices and at least 20 will be dupes
per_dup = 5 # add up to this amount of dupes for each dupe candidate
un_duped = students - dup_choices
# random samples
student_choices = [ sample(range(projects), k = 5) for _ in range(un_duped)]
# select dupe candidates
dups = sample(student_choices, k = max(0, dup_choices // per_dup) + 1)
# add enough of each duplicate to statisfy your numbers
for d in dups:
student_choices.extend( (d.copy() for _ in range(per_dup) ) )
# integer rounding + 1 => you will overshoot - so trim back to number of students
student_choices = student_choices[:students]
# mix the dupes into the data
# shuffle(student_choices)
This will generate 130 choices of wich at least 20 are some kind of duplicate of some other.
You create your cost matrix from this data - the position of the projects number inside the students choice is its priority (0 == 1st, 1 = 2nd, ...):
costs = [[0 if pr not in choice else choice.index(pr) + 1
for pr in range(projects)]
for choice in student_choices]
# print first 3 of choices / costs
for sc in student_choices[:3]:
print(sc)
for c in costs[:3]:
print (c)
Output:
[124, 174, 43, 181, 63]
[158, 110, 129, 120, 149]
[226, 238, 183, 249, 90]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4]

continue the loop if the element does not exist in other arrays in a 3d array

I have a 3d array x [k][n][m] with shape (10, 10, 100) and i need to verify if element m in range(0, 100) does not exist in other arrays x[k-1][n-1] then it can access the loop otherwise it needs to pass to m+1
This is what I did
def main(self):
self.xam = [[[0 for m in range(0, self.Vt)] for n in range(0, self.N)] for x in range(0, self.rn)]
for k in range(0, self.rn):
for n in range(0, self.N):
for m in range(0, self.Vt):
self.iteration = 0
if (self.xam[k-1][n-1][m] != 1):
while self.iteration < self.Iter_max:
if self.NU[k][n][m] == self.AU[k][n][m] == 1:
self.xam[k][n][m] = 1
print("station", k, "channel", n, "user", m, self.xam[k][n][m])
self.iteration += 1
the problem that i had is that it does not take the condition into account:
station 7 channel 9 [0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
station 8 channel 0 [0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

How should we pad text sequence in keras using pad_sequences?

I have coded a sequence to sequence learning LSTM in keras myself using the knowledge gained from the web tutorials and my own intuitions. I converted my sample text to sequences and then padded using pad_sequence function in keras.
from keras.preprocessing.text import Tokenizer,base_filter
from keras.preprocessing.sequence import pad_sequences
def shift(seq, n):
n = n % len(seq)
return seq[n:] + seq[:n]
txt="abcdefghijklmn"*100
tk = Tokenizer(nb_words=2000, filters=base_filter(), lower=True, split=" ")
tk.fit_on_texts(txt)
x = tk.texts_to_sequences(txt)
#shifing to left
y = shift(x,1)
#padding sequence
max_len = 100
max_features=len(tk.word_counts)
X = pad_sequences(x, maxlen=max_len)
Y = pad_sequences(y, maxlen=max_len)
After a carefully inspection I found my padded sequence looks like this
>>> X[0:6]
array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7]], dtype=int32)
>>> X
array([[ 0, 0, 0, ..., 0, 0, 1],
[ 0, 0, 0, ..., 0, 0, 3],
[ 0, 0, 0, ..., 0, 0, 2],
...,
[ 0, 0, 0, ..., 0, 0, 13],
[ 0, 0, 0, ..., 0, 0, 12],
[ 0, 0, 0, ..., 0, 0, 14]], dtype=int32)
Is the padded sequence suppose to look like this? Except the last column in the array the rest are all zeros. I think I made some mistake in padding the text to sequence and if so can you tell me where I made the error?
If you want to tokenize by char, you can do it manually, it's not too complex:
First build a vocabulary for your characters:
txt="abcdefghijklmn"*100
vocab_char = {k: (v+1) for k, v in zip(set(txt), range(len(set(txt))))}
vocab_char['<PAD>'] = 0
This will associate a distinct number for every character in your txt. The character with index 0 should be preserved for the padding.
Having the reverse vocabulary will be usefull to decode the output.
rvocab = {v: k for k, v in vocab.items()}
Once you have this, you can first split your text into sequences, say you want to have sequences of length seq_len = 13 :
[[vocab_char[char] for char in txt[i:(i+seq_len)]] for i in range(0,len(txt),seq_len)]
your output will look like :
[[9, 12, 6, 10, 8, 7, 2, 1, 5, 13, 11, 4, 3],
[14, 9, 12, 6, 10, 8, 7, 2, 1, 5, 13, 11, 4],
...,
[2, 1, 5, 13, 11, 4, 3, 14, 9, 12, 6, 10, 8],
[7, 2, 1, 5, 13, 11, 4, 3, 14]]
Note that the last sequence doesn't have the same length, you can discard it or pad your sequence to max_len = 13, it will add 0's to it.
You can build your targets Y the same way, by shifting everything by 1. :-)
I hope this helps.
The problem is in this line:
tk = Tokenizer(nb_words=2000, filters=base_filter(), lower=True, split=" ")
When you set such split (by " "), due to nature of your data, you'll get each sequence consisting of a single word. That's why your padded sequences have only one non-zero element. To change that try:
txt="a b c d e f g h i j k l m n "*100
The argument padding controls padding either before or after each sequence. Use like this:
X = pad_sequences(x, maxlen=max_len, padding='post')
Y = pad_sequences(y, maxlen=max_len, padding='post')

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