Python KeyError, qr code barcode reader on raspberry pi - python

I'm a Korean. English translation may be wrong.
I am making a program that can output data in Python using a qr reader that is received as a usb input from a Raspberry Pi 4.
The code below raises KeyError:74 . What's the workaround?
ss += hid[int(ord(c))]
Below is the full code.
import sys
hid = {4: 'a', 5: 'b', 6: 'c', 7: 'd', 8: 'e', 9: 'f', 10: 'g', 11: 'h', 12: 'i', 13: 'j', 14: 'k', 15: 'l', 16: 'm',
17: 'n', 18: 'o', 19: 'p', 20: 'q', 21: 'r', 22: 's', 23: 't', 24: 'u', 25: 'v', 26: 'w', 27: 'x', 28: 'y',
29: 'z', 30: '1', 31: '2', 32: '3', 33: '4', 34: '5', 35: '6', 36: '7', 37: '8', 38: '9', 39: '0', 44: ' ',
45: '-', 46: '=', 47: '[', 48: ']', 49: '\\', 51: ';', 52: '\'', 53: '~', 54: ',', 55: '.', 56: '/'}
hid2 = {4: 'A', 5: 'B', 6: 'C', 7: 'D', 8: 'E', 9: 'F', 10: 'G', 11: 'H', 12: 'I', 13: 'J', 14: 'K', 15: 'L', 16: 'M',
17: 'N', 18: 'O', 19: 'P', 20: 'Q', 21: 'R', 22: 'S', 23: 'T', 24: 'U', 25: 'V', 26: 'W', 27: 'X', 28: 'Y',
29: 'Z', 30: '!', 31: '#', 32: '#', 33: '$', 34: '%', 35: '^', 36: '&', 37: '*', 38: '(', 39: ')', 44: ' ',
45: '_', 46: '+', 47: '{', 48: '}', 49: '|', 51: ':', 52: '"', 53: '~', 54: '<', 55: '>', 56: '?'}
fp = open('/dev/hidraw4', 'rb')
ss = ""
shift = False
done = False
while not done:
## Get the character from the HID
buffer = fp.read(8)
for c in buffer:
if ord(c) > 0:
## 40 is carriage return which signifies
## we are done looking for characters
if int(ord(c)) == 40:
done = True
break;
## If we are shifted then we have to
## use the hid2 characters.
if shift:
## If it is a '2' then it is the shift key
if int(ord(c)) == 2 :
shift = True
## if not a 2 then lookup the mapping
else:
ss += hid2[int(ord(c))]
shift = False
## If we are not shifted then use
## the hid characters
else:
## If it is a '2' then it is the shift key
if int(ord(c)) == 2 :
shift = True
## if not a 2 then lookup the mapping
else:
ss += hid[int(ord(c))]
print(ss)

A KeyError is raised when you try to access a key/value in a dict that does not contain that key. You probably want to re-check and update your mapping to contain the correct (ASCII) values as keys. The 74 comes from int(ord("J")).
You can avoid Key errors by changing hid[int(ord(c))] to hid.get(int(ord(c)) which would return None when the key does not exist.

Related

Sample dataframe maintaining multiple frequency distributions

I have an example pandas dataframe, df, below:
{'column_a': {0: 'b', 1: 'b', 2: 'a', 3: 'b', 4: 'd', 5: 'a', 6: 'b', 7: 'b', 8: 'c', 9: 'a', 10: 'a', 11: 'a', 12: 'a', 13: 'c', 14: 'c', 15: 'c', 16: 'b', 17: 'a', 18: 'a', 19: 'b', 20: 'd', 21: 'c', 22: 'a', 23: 'b', 24: 'c', 25: 'c', 26: 'c', 27: 'e', 28: 'e', 29: 'e', 30: 'e', 31: 'c', 32: 'e', 33: 'e', 34: 'd', 35: 'e', 36: 'd', 37: 'e', 38: 'd', 39: 'b', 40: 'd', 41: 'c', 42: 'b', 43: 'd', 44: 'c', 45: 'e', 46: 'd', 47: 'c', 48: 'e', 49: 'b', 50: 'c'}, 'column_b': {0: 'c', 1: 'b', 2: 'b', 3: 'd', 4: 'b', 5: 'a', 6: 'd', 7: 'c', 8: 'c', 9: 'd', 10: 'a', 11: 'a', 12: 'b', 13: 'a', 14: 'c', 15: 'd', 16: 'd', 17: 'c', 18: 'b', 19: 'd', 20: 'a', 21: 'a', 22: 'd', 23: 'b', 24: 'a', 25: 'c', 26: 'e', 27: 'd', 28: 'b', 29: 'c', 30: 'd', 31: 'b', 32: 'e', 33: 'b', 34: 'b', 35: 'c', 36: 'b', 37: 'b', 38: 'd', 39: 'c', 40: 'b', 41: 'a', 42: 'b', 43: 'e', 44: 'e', 45: 'c', 46: 'e', 47: 'c', 48: 'b', 49: 'b', 50: 'c'}, 'column_c': {0: 'b', 1: 'd', 2: 'b', 3: 'b', 4: 'd', 5: 'c', 6: 'b', 7: 'a', 8: 'a', 9: 'a', 10: 'a', 11: 'b', 12: 'd', 13: 'c', 14: 'b', 15: 'a', 16: 'a', 17: 'a', 18: 'b', 19: 'c', 20: 'a', 21: 'a', 22: 'b', 23: 'd', 24: 'd', 25: 'c', 26: 'd', 27: 'c', 28: 'c', 29: 'e', 30: 'd', 31: 'c', 32: 'd', 33: 'c', 34: 'b', 35: 'b', 36: 'd', 37: 'd', 38: 'd', 39: 'b', 40: 'c', 41: 'e', 42: 'e', 43: 'b', 44: 'b', 45: 'd', 46: 'd', 47: 'c', 48: 'e', 49: 'd', 50: 'b'}, 'column_d': {0: 'b', 1: 'c', 2: 'd', 3: 'd', 4: 'b', 5: 'b', 6: 'd', 7: 'd', 8: 'd', 9: 'b', 10: 'd', 11: 'c', 12: 'b', 13: 'a', 14: 'c', 15: 'c', 16: 'd', 17: 'c', 18: 'd', 19: 'a', 20: 'd', 21: 'b', 22: 'd', 23: 'b', 24: 'd', 25: 'e', 26: 'c', 27: 'c', 28: 'c', 29: 'd', 30: 'c', 31: 'e', 32: 'd', 33: 'd', 34: 'd', 35: 'b', 36: 'c', 37: 'e', 38: 'b', 39: 'e', 40: 'b', 41: 'c', 42: 'b', 43: 'e', 44: 'b', 45: 'c', 46: 'd', 47: 'c', 48: 'c', 49: 'b', 50: 'd'}, 'target': {0: 1, 1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1, 7: 1, 8: 1, 9: 1, 10: 1, 11: 1, 12: 1, 13: 1, 14: 1, 15: 1, 16: 1, 17: 1, 18: 1, 19: 1, 20: 1, 21: 1, 22: 1, 23: 1, 24: 1, 25: 0, 26: 0, 27: 0, 28: 0, 29: 0, 30: 0, 31: 0, 32: 0, 33: 0, 34: 0, 35: 0, 36: 0, 37: 0, 38: 0, 39: 0, 40: 0, 41: 0, 42: 0, 43: 0, 44: 0, 45: 0, 46: 0, 47: 0, 48: 0, 49: 0, 50: 0}}
What I am trying to accomplish is to select a subsample of this dataframe of an arbitrary length or percentage, but in doing so, I want to maintain (as closely as possible) the frequency distributions of each value for each class.
For example, if I want to simply subsample the dataframe, I can use .sample() method
smaller_df = df.sample(n=100) or smaller_df = df.sample(frac=0.1)
However, it could be the case that the distributions of each value in each column in each class are lost. I need to preserve these value densities while downsampling my dataset size.
I can see these frequency densities with:
for col in df.columns:
print(df.groupby(['target'])[col].value_counts(normalize=True))
That output looks like:
target column_a
0 e 0.384615
c 0.269231
d 0.230769
b 0.115385
1 a 0.360000
b 0.320000
c 0.240000
d 0.080000
I have seen this post on Stack Overflow which seemingly answers that for a single distribution, but not multiple.
Ideally, how can I downsample my dataframe to maintain each columns frequency distribution with less samples? My actual dataset is (8370994, 731)

How to return previous element from group where some column is True

My data frame is as follows:
ex = {'group': {0: '0', 1: '0', 2: '0', 3: '0', 4: '0', 5: '0', 6: '0', 7: '0', 8: '0', 9: '0', 10: '0', 11: '0', 12: '0', 13: '0', 14: '0', 15: '0', 16: '0', 17: '0', 18: '0', 19: '0', 20: '0', 21: '1', 22: '1', 23: '1', 24: '1', 25: '1', 26: '1', 27: '1', 28: '1', 29: '1', 30: '1', 31: '1', 32: '1', 33: '1', 34: '1', 35: '1', 36: '1', 37: '1', 38: '1', 39: '1'}, 'order': {0: 0, 1: 1, 2: 2, 3: 3, 4: 4, 5: 5, 6: 6, 7: 7, 8: 8, 9: 9, 10: 10, 11: 11, 12: 12, 13: 13, 14: 14, 15: 15, 16: 16, 17: 17, 18: 18, 19: 19, 20: 20, 21: 0, 22: 1, 23: 2, 24: 3, 25: 4, 26: 5, 27: 6, 28: 7, 29: 8, 30: 9, 31: 10, 32: 11, 33: 12, 34: 13, 35: 14, 36: 15, 37: 16, 38: 17, 39: 18}, 'id': {0: '102', 1: '302', 2: '302', 3: '302', 4: '102', 5: '302', 6: '302', 7: '302', 8: '302', 9: '302', 10: '102', 11: '308', 12: '308', 13: '308', 14: '308', 15: '302', 16: '102', 17: '302', 18: '102', 19: '302', 20: '102', 21: '102', 22: '102', 23: '308', 24: '312', 25: '312', 26: '312', 27: '308', 28: '102', 29: '302', 30: '312', 31: '302', 32: '302', 33: '102', 34: '102', 35: '302', 36: '312', 37: '308', 38: '102', 39: '302'}, 'type': {0: 'A', 1: 'B', 2: 'C', 3: 'A', 4: 'D', 5: 'E', 6: 'D', 7: 'E', 8: 'A', 9: 'E', 10: 'E', 11: 'D', 12: 'A', 13: 'A', 14: 'A', 15: 'D', 16: 'D', 17: 'D', 18: 'A', 19: 'D', 20: 'A', 21: 'D', 22: 'F', 23: 'A', 24: 'D', 25: 'A', 26: 'E', 27: 'A', 28: 'E', 29: 'D', 30: 'E', 31: 'E', 32: 'G', 33: 'A', 34: 'D', 35: 'D', 36: 'H', 37: 'I', 38: 'A', 39: 'E'}, 'of_interest': {0: False, 1: False, 2: True, 3: False, 4: False, 5: True, 6: False, 7: True, 8: True, 9: True, 10: True, 11: True, 12: True, 13: False, 14: True, 15: True, 16: True, 17: True, 18: False, 19: False, 20: True, 21: False, 22: False, 23: False, 24: True, 25: False, 26: True, 27: True, 28: False, 29: True, 30: True, 31: False, 32: True, 33: True, 34: True, 35: True, 36: True, 37: False, 38: True, 39: False}}
ex.head()
group order id type of_interest
0 0 0 102 A False
1 0 1 302 B False
2 0 2 302 C True
3 0 3 302 A False
4 0 4 102 D False
I want to create a column that for each combination of group and id return previous type where of_interest == True.
My first attempt involved querying for of_interest == True, therefore returned value only for these rows:
ex['prev_type_of_interest'] = ex \
.query('of_interest == True') \
.groupby(['group', 'id'])['type'] \
.shift(1)
How can I return previous type of interest for every row?
I believe you need shift all rows per groups, then set missing values by Series.where and last replace missing values by previos non missing values by GroupBy.ffill:
ex1 = ex.groupby(['group', 'id']).shift()
ex['prev_type_of_interest'] = ex1['type'].where(ex1['of_interest'] == True)
ex['prev_type_of_interest'] = ex.groupby(['group', 'id'])['prev_type_of_interest'].ffill()
print (ex.head(10))
group order id type of_interest prev_type_of_interest
0 0 0 102 A False NaN
1 0 1 302 B False NaN
2 0 2 302 C True NaN
3 0 3 302 A False C
4 0 4 102 D False NaN
5 0 5 302 E True C
6 0 6 302 D False E
7 0 7 302 E True E
8 0 8 302 A True E
9 0 9 302 E True A

Dictionary that maps ASCII keys to their corresponding values

I'm trying to get chr() output from 65 to 90:
I want to get a dictionary that looks like this:
{65: 'A', 66: 'B', ..... 90: 'Z'}
You just need a simple dict comprehension
>>> {x: chr(x) for x in range(65, 91)}
{65: 'A', 66: 'B', 67: 'C', 68: 'D', 69: 'E', 70: 'F', 71: 'G', 72: 'H', 73: 'I', 74: 'J', 75: 'K', 76: 'L', 77: 'M', 78: 'N', 79: 'O', 80: 'P', 81: 'Q', 82: 'R', 83: 'S', 84: 'T', 85: 'U', 86: 'V', 87: 'W', 88: 'X', 89: 'Y', 90: 'Z'}

How do i print my barcode output to a file?

I have set up a raspberry Pi with a USB barcode scanner for a little project. It works with my generated barcodes, it prints the output of the scanned code in the terminal. I really want to save this input to a txt file that doesn't overwrite itself. I have tried changing all the functions and i just cant get it to work. I'm just a novice in Python and i have been stuck on this for a long time now and i have looked all over the internet. If you can just point me to the specific place in code i need to change in order to print the output out i would be very appreciative.
Source: Instructables
!/usr/bin/python
import sys
import requests
import json
api_key = "" #https://upcdatabase.org/
def barcode_reader():
hid = {4: 'a', 5: 'b', 6: 'c', 7: 'd', 8: 'e', 9: 'f', 10: 'g', 11: 'h', 12: 'i', 13: 'j', 14: 'k', 15: 'l', 16: 'm',
17: 'n', 18: 'o', 19: 'p', 20: 'q', 21: 'r', 22: 's', 23: 't', 24: 'u', 25: 'v', 26: 'w', 27: 'x', 28: 'y',
29: 'z', 30: '1', 31: '2', 32: '3', 33: '4', 34: '5', 35: '6', 36: '7', 37: '8', 38: '9', 39: '0', 44: ' ',
45: '-', 46: '=', 47: '[', 48: ']', 49: '\\', 51: ';', 52: '\'', 53: '~', 54: ',', 55: '.', 56: '/'}
hid2 = {4: 'A', 5: 'B', 6: 'C', 7: 'D', 8: 'E', 9: 'F', 10: 'G', 11: 'H', 12: 'I', 13: 'J', 14: 'K', 15: 'L', 16: 'M',
17: 'N', 18: 'O', 19: 'P', 20: 'Q', 21: 'R', 22: 'S', 23: 'T', 24: 'U', 25: 'V', 26: 'W', 27: 'X', 28: 'Y',
29: 'Z', 30: '!', 31: '#', 32: '#', 33: '$', 34: '%', 35: '^', 36: '&', 37: '*', 38: '(', 39: ')', 44: ' ',
45: '_', 46: '+', 47: '{', 48: '}', 49: '|', 51: ':', 52: '"', 53: '~', 54: '<', 55: '>', 56: '?'}
fp = open('/dev/hidraw0', 'rb')
ss = ""
shift = False
done = False
while not done:
## Get the character from the HID
buffer = fp.read(8)
for c in buffer:
if ord(c) > 0:
## 40 is carriage return which signifies
## we are done looking for characters
if int(ord(c)) == 40:
done = True
break;
## If we are shifted then we have to
## use the hid2 characters.
if shift:
## If it is a '2' then it is the shift key
if int(ord(c)) == 2:
shift = True
## if not a 2 then lookup the mapping
else:
ss += hid2[int(ord(c))]
shift = False
## If we are not shifted then use
## the hid characters
else:
## If it is a '2' then it is the shift key
if int(ord(c)) == 2:
shift = True
## if not a 2 then lookup the mapping
else:
ss += hid[int(ord(c))]
return ss
def UPC_lookup(api_key,upc):
'''V3 API'''
url = "https://api.upcdatabase.org/product/%s/%s" % (upc, api_key)
headers = {
'cache-control': "no-cache",
}
response = requests.request("GET", url, headers=headers)
print("-----" * 5)
print(upc)
print(json.dumps(response.json(), indent=2))
print("-----" * 5 + "\n")
if __name__ == '__main__':
try:
while True:
UPC_lookup(api_key,barcode_reader())
except KeyboardInterrupt:
pass
If it is already printing to the console it means it's coming from this part of the code:
print("-----" * 5)
print(upc)
print(json.dumps(response.json(), indent=2))
print("-----" * 5 + "\n")
In order to save it to a file you can use the following:
with open('FILENAME.txt', 'a', encoding='utf-8') as file:
file.write('CONTENT THAT YOU WANT TO WRITE!\n')
Or in your particular case:
with open('FILENAME.txt', 'a', encoding='utf-8') as file:
file.write("-----" * 5)
file.write(upc)
file.write(json.dumps(response.json(), indent=2))
file.write("-----" * 5 + "\n")

super simple cipher function in python. comparing a list to keys in a dictionary

I am working on ways to get me thinking in python. I have a simple idea that will take a number and give the corresponding "value" from a dictionary.
So basically I would like to have a number or numbers given, and then turn those numbers into a word.
The overall all view is to have a dictionary with keys ranging for 1 to 26 with values going from a to z. So 1 would equal "a" and 26 would equal "z".
I have a variable d = 1, and need to get the output of 'a'. Then increase size of this function for a list like (1,2,3,4) which output would be a, b, c, d.
Here is what I have so far.
d = 1
def code_scram(x):
c = {1: 'a', 2: 'b', 3: 'c', 4: 'd', 5: 'e', 6: 'f', 7: 'g', 8: 'h', 9: 'i', 10: 'j', 11: 'k', 12: 'l', 13: 'm', 14: 'n', 15: 'o', 16: 'p', 17: 'q', 18: 'r', 19: 's', 20: 't', 21: 'u', 22: 'v', 23: 'w', 24: 'x', 25: 'y', 26: 'z'}
scram = ""
for i in d:
if i in c:
scram += c[i]
return scram
print code_scram(d)
However, its not working out as planned.
Your for loop should iterate through x, not d.
def code_scram(x):
c = {1: 'a', 2: 'b', 3: 'c', 4: 'd', 5: 'e', 6: 'f', 7: 'g', 8: 'h', 9: 'i', 10: 'j', 11: 'k', 12: 'l', 13: 'm', 14: 'n', 15: 'o', 16: 'p', 17: 'q', 18: 'r', 19: 's', 20: 't', 21: 'u', 22: 'v', 23: 'w', 24: 'x', 25: 'y', 26: 'z'}
scram = ""
for i in x:
if i in c:
scram += c[i]
return scram
print code_scram([1,2,3,4])
Result:
abcd
The function only works for lists, so passing in the integer d won't work. Pass in a list instead.
d = [1]
print code_scram(d)
If you want the function to work for lists and lone integers, you can perform a type check, and convert as necessary.
def code_scram(x):
if isinstance(x, int):
x = [x]
c = {1: 'a', 2: 'b', 3: 'c', 4: 'd', 5: 'e', 6: 'f', 7: 'g', 8: 'h', 9: 'i', 10: 'j', 11: 'k', 12: 'l', 13: 'm', 14: 'n', 15: 'o', 16: 'p', 17: 'q', 18: 'r', 19: 's', 20: 't', 21: 'u', 22: 'v', 23: 'w', 24: 'x', 25: 'y', 26: 'z'}
scram = ""
for i in x:
if i in c:
scram += c[i]
return scram
d = 1
print code_scram(d)
Result:
a

Categories

Resources