Develop Reference

Python KeyError, qr code barcode reader on raspberry pi

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.

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)

One hot encoding of custom built vocabolary

I have charset as follows.
charset =set([ '$', '^', '#', '(', ')', '-', '.', '/', '1', '2', '3', '4', '5', '6', '7', '=', 'Br',
'C', 'Cl', 'F', 'I', 'N', 'O', 'P', 'S', '[2H]', '[Br-]', '[C##H]', '[C##]', '[C#H]', '[C#]',
'[Cl-]', '[H]', '[I-]', '[N+]', '[N-]', '[N#+]', '[N##+]', '[NH+]', '[NH2+]', '[NH3+]', '[N]',
'[Na+]', '[O-]', '[P+]', '[S+]', '[S-]', '[S#+]', '[S##+]', '[SH]', '[Si]', '[n+]', '[n-]',
'[nH+]', '[nH]', '[o+]', '[se]', '\\', 'c', 'n', 'o', 's', '!', 'E'])
On the basis of this charset, I create char_to_int as follows.
char_to_int = dict((c,i) for i,c in enumerate(charset))
{'[nH]': 0, '[2H]': 1, '2': 2, 'N': 3, 'Cl': 4, 'c': 5, '$': 6,
'O': 7, '(': 8, '6': 9, 's': 10, '[S#+]': 11, '[C##H]': 12, 'C':
13, '[nH+]': 14, '/': 15, '[NH+]': 16, '[Br-]': 17, '[Si]': 18,
'4': 19, '[N#+]': 20, '[se]': 21, 'P': 22, '[SH]': 23, '[N+]':
24, '[N]': 25, '^': 26, '5': 27, '7': 28, 'n': 29, '!': 30,
'\': 31, '[n-]': 32, 'S': 33, '[NH3+]': 34, '#': 35, 'I': 36,
'[O-]': 37, '1': 38, '[NH2+]': 39, '[S##+]': 40, 'Br': 41, 'F':
42, '[Na+]': 43, 'E': 44, '[S-]': 45, '.': 46, ')': 47, '[C#]':
48, '=': 49, '3': 50, '-': 51, '[C#H]': 52, '[Cl-]': 53, '[I-]':
54, '[H]': 55, '[P+]': 56, '[S+]': 57, 'o': 58, '[N##+]': 59,
'[N-]': 60, '[n+]': 61, '[o+]': 62, '[C##]': 63}
and int_to_char as follows.
int_to_char = dict((i,c) for i,c in enumerate(charset))
{0: '[nH]', 1: '[2H]', 2: '2', 3: 'N', 4: 'Cl', 5: 'c', 6: '$',
7: 'O', 8: '(', 9: '6', 10: 's', 11: '[S#+]', 12: '[C##H]', 13:
'C', 14: '[nH+]', 15: '/', 16: '[NH+]', 17: '[Br-]', 18: '[Si]',
19: '4', 20: '[N#+]', 21: '[se]', 22: 'P', 23: '[SH]', 24:
'[N+]', 25: '[N]', 26: '^', 27: '5', 28: '7', 29: 'n', 30: '!',
31: '\', 32: '[n-]', 33: 'S', 34: '[NH3+]', 35: '#', 36: 'I',
37: '[O-]', 38: '1', 39: '[NH2+]', 40: '[S##+]', 41: 'Br', 42:
'F', 43: '[Na+]', 44: 'E', 45: '[S-]', 46: '.', 47: ')', 48:
'[C#]', 49: '=', 50: '3', 51: '-', 52: '[C#H]', 53: '[Cl-]', 54:
'[I-]', 55: '[H]', 56: '[P+]', 57: '[S+]', 58: 'o', 59: '[N##+]',
60: '[N-]', 61: '[n+]', 62: '[o+]', 63: '[C##]'}
I have a string which I want to convert to one hot encoding on the basis of char_to_int and int_to_char.
string = 'N[C#H]1C[C##H](N2Cc3nn4cccnc4c3C2)CC[C##H]1c1cc(F)c(F)cc1F'
Is there any efficient way which uses the self defined char_to_int and int_to_char to convert a string to one hot vector?

from itertools import chain, repeat, islice
import re
string = 'N[C#H]1C[C##H](N2Cc3nn4cccnc4c3C2)CC[C##H]1c1cc(F)c(F)cc1F'
items_list=[ '$', '^', '#', '(', ')', '-', '.', '/', '1', '2', '3', '4', '5', '6', '7', '=', 'Br',
'C', 'Cl', 'F', 'I', 'N', 'O', 'P', 'S', '[2H]', '[Br-]', '[C##H]', '[C##]', '[C#H]', '[C#]',
'[Cl-]', '[H]', '[I-]', '[N+]', '[N-]', '[N#+]', '[N##+]', '[NH+]', '[NH2+]', '[NH3+]', '[N]',
'[Na+]', '[O-]', '[P+]', '[S+]', '[S-]', '[S#+]', '[S##+]', '[SH]', '[Si]', '[n+]', '[n-]',
'[nH+]', '[nH]', '[o+]', '[se]', '\\', 'c', 'n', 'o', 's', '!', 'E']
charset = set(items_list)
char_to_int = dict((c,i) for i,c in enumerate(charset))
pattern = '|'.join(re.escape(item) for item in items_list)
tokens = re.findall(pattern, string)
x=[char_to_int[k] for k in tokens]
Here, xis one hot encoded.
x=[3, 52, 38, 13, 12, 8, 3, 2, 13, 5, 50, 29, 29, 19, 5, 5, 5, 29, 5, 19, 5, 50, 13, 2, 47, 13, 13, 12, 38, 5, 38, 5, 5, 8, 42, 47, 5, 8, 42, 47, 5, 5, 38, 42]

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'}

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