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Python needs to compare the increase over the last month's payroll.

Demir last edited by
Help me with the task. The user's injecting ten numbers. Write a program that checks whether they're being streamlined at age. If there was one case where the ep fell relative to the last month, that would suggest that it did not grow and that the message should be printed. I've got that code. But I can't figure it out if it falls.
```
curr = 0
for month in range(11):
  prev = int(input('Введите зарплату: '))
  if curr == 0:
    curr = prev
    continue
if prev > curr:
  print('Зарплата увеличилась ')
elif prev == curr:
  print('Зарплата не изменилась')
else:
  print('Зарплата уменьшилась')
```
Reply Quote 0
1 Reply Last reply

inna

You have to do this:

prev = -1
for month in range(10):

curr = int(input(f'Введите зарплату за месяц №{month + 1}: '))
if curr < prev:

print('Зарплата уменьшилась')

break
prev = curr

else:

print('Зарплата НЕ уменьшилась')

And you only have the first and the most recent elements in comparison.

P. S.

Such an approach may also be implemented (if payments are made through a gap):

months = (list(map(int, input().split())) + [0] * 10)[:10]
print("зарплата уменьшалась" if months != sorted(months) else "зарплата не уменьшалась")

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Replace the values of one column in one digit by the key of the second
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For reference dates:df55: Название Артикул Количество 0 Коврик для фитнесса голубой 10 SV-HK0067 5 1 Коврик для фитнесса 10 мм NBR фиолетовый SV-HK0068 7 2 Гиря 16 кг SV-HK0069 2 3 Блины 10кг SV-HK0070 4 4 Фибол SV-HK0071 6 df1C1: Название Артикул Склад 55 к-во 0 Коврик для фитнесса голубой 10 SV-HK0067 3 1 Коврик для фитнесса 10 мм NBR фиолетовый SV-HK0068 4 2 Гиря 16 кг SV-HK0069 5 3 Блины 10кг SV-HK0070 6 4 Фибол SV-HK0071 8 5 Перчатки SV-HK0072 0 6 Груши SV-HK0073 0 You can just...df1С1.update(df55.rename(columns={"Количество": "Склад 55 к-во"})) Look, I had to change the columns because they don't match your dates.Then... df1C1 It'll work. Название Артикул Склад 55 к-во 0 Коврик для фитнесса голубой 10 SV-HK0067 5 1 Коврик для фитнесса 10 мм NBR фиолетовый SV-HK0068 7 2 Гиря 16 кг SV-HK0069 2 3 Блины 10кг SV-HK0070 4 4 Фибол SV-HK0071 6 5 Перчатки SV-HK0072 0 6 Груши SV-HK0073 0 UPDATEIf I understood the author correctly, and updating the date-French requires a pair of keys matching the Artikulu-Colice, you can do it so (for illustration, I changed the first entries in df1C1:df55: Название Склад 55 к-во Артикул SV-HK0067 Коврик для фитнесса голубой 10 3 SV-HK0068 Коврик для фитнесса 10 мм NBR фиолетовый 4 SV-HK0069 Гиря 16 кг 5 SV-HK0070 Блины 10кг 6 SV-HK0071 Фибол 8 SV-HK0072 Перчатки 0 SV-HK0073 Груши 0 Название Склад 55 к-во df1C1: Артикул SV-HK0068 Коврик для фитнесса 10 мм NBR фиолетовый 5 SV-HK0067 Коврик для фитнесса голубой 10 7 SV-HK0069 Гиря 16 кг 2 SV-HK0070 Блины 10кг 4 SV-HK0071 Фибол 6 res = df1С1.set_index("Артикул") upd = df55.rename(columns={"Количество": "Склад 55 к-во"}).set_index("Артикул") res.update(upd) res = res.reset_index() Then res will: Артикул Название Склад 55 к-во 0 SV-HK0067 Коврик для фитнесса голубой 10 7 1 SV-HK0068 Коврик для фитнесса 10 мм NBR фиолетовый 5 2 SV-HK0069 Гиря 16 кг 2 3 SV-HK0070 Блины 10кг 4 4 SV-HK0071 Фибол 6 5 SV-HK0072 Перчатки 0 6 SV-HK0073 Груши 0
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Function len, get total items from the list contained in dictionary
Software Programming • python list dic languagetionaries • • Oba22

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The result is the expected given that you do len({...}[1][1][1])A chained index. If we analyze the step-by-step code, it is clearer:{0: ['a', 'bcd'], 1: ['ef', 'ghi', 'k'], 2: ['1']}[1]. This returns the value associated with the key 1 of the dictionary, i.e. ['ef', 'ghi', 'k'].['ef', 'ghi', 'k'][1] this gets the item in the list index 1, the string 'ghi'.'ghi[1]': gets the character with the chain's index 1, so we get "h"len("h") returns the length of the chain, 1.To get the number of items together from all the dictionary lists you must apply len to each of them iterating about the dictionary and then adding their outputs:>>> d = {0: ['a', 'bcd'], 1: ['ef', 'ghi', 'k'], 2: ['1']} >>> print(sum(len(lista) for lista in d.values())) 6 or using functional programming:>>> print(sum(map(len, d.values()))) 6 Note that at no time is the value checked to be a list or even available to the method __len__ defined (like an integer). If your dictionary is heterogeneous you must have it in how much and act accordingly.
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Problem comparing numbers in Python
Software Programming • python if • • Kadir

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Please take into account all the comments you have made to your question as all of them are valuable to your learning process. Since you're learning I tried to reuse your code and modify it as little as possible because there are other more advanced ways that require less code to achieve your purpose, however, it works quite well the saying that prays: "First you have to learn to crawl and then you can walk"2 important comments for your code. The first as well told you @abulafia and @FJSevilla You're comparing chains and not numbers. Remember that when you use that data capture method you always save in string or string. That's why when you compare a with b and c He doesn't throw you the biggest number, but that of the three prevails alphabetically. The second is to be careful with the use of if and elif the way you set the comparison there is no need to use elif.Then I leave your code with the corrections that we have mentioned to you in the most didactic way that I found to clearly observe the conversion of types and thus work as you wish:a=input("Digite un numero entero de dos digitos:") b=input("Digite otro numero entero de dos digitos:") c=input("Digite otro numero entero:") a=int(a); b=int(b); c=int(c); if (a>b): if (a>c): print("El numero mayor es "+str(a)) if (b>a): if (b>c): print("El numero mayor es "+str(b)) if(c>a): if(c>b): print("El numero mayor es "+str(c)) I hope it's helpful. A greeting!
Python reboot service tor
Software Programming • python tor • • Marcee

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I'm using that code. I've been pulling from some github, but I don't remember where. from commands import getoutput from subprocess import call, check_call, CalledProcessError from json import load from urllib2 import urlopen, URLError from os import devnull from time import sleep from sys import stdout, stderr def restart_tor(): fnull = open(devnull, 'w') try: tor_restart = check_call( ["service", "tor", "restart"], stdout=fnull, stderr=fnull) if tor_restart is 0: print(" {0}".format( "[\033[92m+\033[0m] Anonymizer status \033[92m[ON]\033[0m")) print(" {0}".format( "[\033[92m*\033[0m] Getting public IP, please wait...")) retries = 0 my_public_ip = None while retries < 12 and not my_public_ip: retries += 1 try: my_public_ip = load(urlopen('http://jsonip.com/'))['ip'] except URLError: sleep(5) print(" [\033[93m?\033[0m] Still waiting for IP address...") print if not my_public_ip: my_public_ip = getoutput('wget -qO - v4.ifconfig.co') if not my_public_ip: exit(" \033[91m[!]\033[0m Can't get public ip address!") print(" {0}".format("[\033[92m+\033[0m] Your IP is \033[92m%s\033[0m" % my_public_ip)) except CalledProcessError as err: print("\033[91m[!] Command failed: %s\033[0m" % ' '.join(err.cmd)) if name == 'main': restart_tor()
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Problem with catching json data on python
Software Programming • python json integer • • jules

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The problem is that you don't write the keys with the same lowercases.This works:for dato in datos["images"]: print(f'Progress={dato["progress"]}') print(f'minDisk={dato["minDisk"]}') print(f'minRam={dato["minRam"]}') produces:Progress=100 minDisk=10 minRam=20 Progress=100 minDisk=30 minRam=40 Note: change the values in the JSON to check that they were well recovered. The JSON stays like this:datos = { "images": [ { "OS-EXT-IMG-SIZE:size": 13167616, "created": "2021-07-13T09:35:56Z", "id": "15f490a5-9b64-4315-a592-3933e91356a5", ... "links": [ "metadata": {}, "minDisk": 10, "minRam": 20, "updated": "2021-07-13T09:35:57Z" }, { "OS-EXT-IMG-SIZE:size": 1020264448, "created": "2021-07-11T15:14:03Z", "id": "ba417b6f-edde-4862-8425-6d3b1d545dcc", ... "metadata": {}, "minDisk": 30, "minRam": 40, "name": "centos", "progress": 100, "status": "ACTIVE", "updated": "2021-07-11T15:14:22Z" } ] }
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calculate square root of 2 with precision in Python
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As the method math.sqrt() is from another independent module, you may be calculating without using advanced precision. Luckily, the https://docs.python.org/3/library/decimal.html has an integrated method to calculate the square root with the given accuracy.>>> import decimal >>> decimal.getcontext().prec = 65 >>> decimal.Decimal(2).sqrt() Decimal('1.4142135623730950488016887242096980785696718753769480731766797380')
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Decision algorithm:draw up a list of lists from the baseline data: [['apple', 'malum'], ['apple', 'pomum'], ..., ['punishment', 'multa']] pass through the script of paragraph 1 and draw the vocabulary you need using the second element of the internal list as the key and the first as the value.PS code won't be given until you've seen your own attempt on the question.
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I need to make a Python entry by storing it as a list, all in the same line
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If the values are related to each other, do not store them in distinct structures, this will only increase the complexity of your application. It will be easier for you to have only one list and store all the values in it, in one tuple. For example:veiculos = [] for i in range(10): identificador, quilometragem, consumo = raw_input().split() veiculos.append((int(identificador), int(quilometragem), float(consumo))) This way your list will look like:veiculos = [ (1001, 231, 59.2), (1002, 496, 60.4) ] Where the identifier is index 0 of the tuple, mileage index 1 and consumption index 2. To calculate yield, simply divide mileage by consumption. You can set a function for this:def calcular_rendimento(veiculo): return veiculo[1] / veiculo[2] And thus display the performance of vehicles:for veiculo in veiculos: rendimento = calcular_rendimento(veiculo) print 'Veículo {} teve um rendimento de {:.2f} km/L'.format(veiculo[0], rendimento) Which would generate a similar output:Veículo 1001 teve um rendimento de 3.90 km/L Veículo 1002 teve um rendimento de 8.21 km/L For vehicles with lower yield, read about the native Python function min and its parameter key.But for a more readable and idiomatic solution, I suggest you read this https://pt.stackoverflow.com/a/290835/5878 .
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I need to go line to line a python file
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What I told you in the comment is that the opening of the file in which you will write the result do it inside the loop foropening a different file every time. That is, something like this (I have removed parts of the code to leave only the relevant):def comandos(): # Omito el inicio que quedaría igual archivo1 = open("listaip.txt","r") archivo2 = open("comandos.txt","r") terminales = archivo1.readlines() comandoss = archivo2.readlines() # Ya que no vuelves a leer de archivo1 ni archivo2 podemos # cerrarlos ya aqui archivo1.close() archivo2.close() for host in terminales: host = host.rstrip() # El archivo3 se abre aquí, y su nombre es el del host, con extensión txt archivo3 = open("{}.txt".format(host), "w") for command in comandoss: # El interior de este bucle queda exactamente como estaba # Al salir de ese bucle, cerramos archivo3 archivo3.close()
How can I compare lists and remove duplicates on Python?
Software Programming • python list • • Alberto

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You have a list of lists and you want to validate if the first one is equal to the second by removing the last one (and penultimate?) element.Simplifying, if we have:d = [ [1, 2, 3, 4, 5], [1, 2, 3, 4, 6] ] Then we get all the elements from the first list, except the last one, saying:>>> d[0][:-1] [1, 2, 3, 4] Or all but the last two with:>>> d[0][:-2] [1, 2, 3] Then it is a question of comparing these sublists:if d[0][:-1] == d[1][:-1]: # si esta sublista es igual if d[0][-1] < d[1][-1]: # el último elemento es menor en la 1.ª lista d.remove(d[1]) # por tanto, eliminamos la 2.ª lista else: d.remove(d[0]) # en caso contrario, eliminamos la 1.ª lista In my example, the result would be:>>> d [[1, 2, 3, 4, 5]] However, in the comments we saw that what you have is something a little more complex: a series of lists you want to group taking into account all the elements less the last; once done that, choose the list whose last value is less.To do this we must use a way of making more generic and it occurs to me like this:creates a dictionary whose indices are all the possible "carried" lists, that is, without the last element.then go through all lists and take the lowest value for each type of list. Save the value in that dictionary.In code:listas = {tuple(x[:-1]): x for x in datos} for elemento in datos: indice = elemento[:-1] listas[tuple(indice)][-1] = min(indice[-1], listas[tuple(indice)][-1]) And the result is in:listas.values() For example:>>> datos = [[1, 2, 3], [1, 2, 2], [7, 8, 9], [7, 8, 8], [5, 6, 6]] >>> listas = {tuple(x[:-1]): x for x in datos} >>> listas {(1, 2): [1, 2, 2], (7, 8): [7, 8, 8], (5, 6): [5, 6, 6]} >>> for elemento in datos: indice = elemento[:-1] listas[tuple(indice)][-1] = min(indice[-1], listas[tuple(indice)][-1]) >>> listas {(1, 2): [1, 2, 2], (7, 8): [7, 8, 8], (5, 6): [5, 6, 6]}
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Python Unsupported operand type(s) for *: 'int' and 'range'
Software Programming • python operator integer • • courtlanda

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WriteQ=range(int(q1+q2)) rule1=math.exp(beta*q1)//sum(math.exp(beta*Q)) where Q It's a rangeOne. unchangeable sequence. It is not a list or a climb, but something else, thought to use in iterations.So, on the second line, you want to do beta * Qwhere the first variable is an integer. That's not valid or makes sense.Looking at the rest of the expression sum(math.exp(beta*Q)) Looks like he expects a list of values, but math.exp returns a scalar value and also expects a scalar value as a parameter, so nothing fits: doesn't make sense sum() of a single value.The logic behind your calculations is unclear.
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To write a function that finds semi-completed numbers before the limit
Software Programming • python python 3.x functions • • guilherme

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The semi-completed number is the natural number, the sum of all or some of its own proprietors matches the number itself.import math def devis(n): # список для записи делителей res = [] for i in range(1, round(math.sqrt(n))+1): if n%i == 0: res.append(i) # не допускам повторов делителей if i*i != n: res.append(n//i) res.sort() return res Список первых нескольких полусовершенных чисел: for num in range(2, 300): aa = devis(num) s = num for i in aa[-2::-1]: if s-i>0: s = s - i elif s-i==0: print(num, sep=' ', end=', ') break 6, 12, 18, 20, 24, 28, 30, 36, 40, 42, 48, 54, 56, 60, 66, 72, 78, 80, 84, 88, 90, 100, 102, 104, 108, 112, 120, 126, 132, 138, 140, 144, 150, 156, 160, 162, 168, 174, 176, 180, 186, 192, 196, 216, 204,
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The discord on the python adds roles, but doesn't remove them.
Software Programming • python python 3.x discord discord.py • • amarantha

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Try this:@client.event async def on_raw_reaction_remove(payload): message_id = payload.message_id if message_id == 822401200942350336: guild_id = payload.guild_id guild = discord.utils.find(lambda g: g.id == guild_id, client.guilds) if payload.emoji.name == '': role = discord.utils.get(guild.roles, id=788080179624149002) elif payload.emoji.name == '': role = discord.utils.get(guild.roles, id=788080230182682634) elif payload.emoji.name == '': role = discord.utils.get(guild.roles, id=790238512774512660) else: role = None if role is not None: member = utils.get(message.guild.members, id=payload.user_id) if member is not None: await member.remove_roles(role) print("done") else: print("Member not found") else: print("Role not found")
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search all records of the last 12 months including zero
Software Programming • mysql sql • • kalena

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Try to take the FROM_UNIXTIME:SELECT y, m, Count(bud_quotations.created_at) FROM ( SELECT y, m FROM (SELECT YEAR(CURDATE()) y UNION ALL SELECT YEAR(CURDATE())-1) years, (SELECT 1 m UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9 UNION ALL SELECT 10 UNION ALL SELECT 11 UNION ALL SELECT 12) months) ym LEFT JOIN bud_quotations ON ym.y = YEAR(bud_quotations.created_at) AND ym.m = MONTH(bud_quotations.created_at) WHERE (y=YEAR(CURDATE()) AND m<=MONTH(CURDATE())) OR (y<YEAR(CURDATE()) AND m>MONTH(CURDATE())) GROUP BY y, m;
We need to get a n*m matrix and fill it with zeros, python.
Software Programming • python • • carriann

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Renumber the list, not the element: https://ideone.com/y3Ox6i n, m = map(int, input().split()) a = [[0] * m] * n print(a) There's only a catch: when multiplied, the list is duplicated by reference, so it's not going to change the lists separately and after.a[0][0] = 1 print(a) It'll work. https://ideone.com/VdKotg [[1, 0, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0]] To have new lists in each element, please refer to @splash58: https://ideone.com/ja4iy6 n, m = map(int, input().split()) a = [[0] * m for _ in range(n)] print(a) a[0][0] = 1 print(a)

Python needs to compare the increase over the last month's payroll.

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