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When pandas are used for excel reading, the indexing begins with line 2.

Marcee last edited by
I use the pandas library for excel file. But somehow he ignores the first line in excel and allows us to contact the cells from the second.
```
data = pd.read_excel(path, engine='openpyxl')
print(data.iloc[0, 7])
```
In this case, it issues the value of the 2nd line of the 7th cell. How do you turn to line one?
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briley last edited by
In default, the first line is considered a headline. Tell me you don't have a title like:
```
data = pd.read_excel(path, header=None, engine='openpyxl')
```
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To remove and return the element you must use the method pop(index), del It only eliminates it but does not return anything.def deal_card(list1): return list1.pop(0) carta = deal_card(list1) print(carta) print(list1) Another recommendation is that to combine the cards do not use indices, it would be directly on the casts. It's more 'pythonic' and more efficient:def doing_deck(): value = ['a','2','3','4','5','6','7','8','9','t','j','q','k'] suits = ['H','D','S','C',] cards = [] for v in value: for x in suits: carlos = "".join((v,x)) cards.append(carlos) return cards Or use directly list compression:def doing_deck(): value = ['a','2','3','4','5','6','7','8','9','t','j','q','k'] suits = ['H','D','S','C',] return ["".join((v,x)) for x in suits for v in values] Another efficient option is to use itertools.product().The code could be like this:from random import shuffle from itertools import product def doing_deck(): value = ['a','2','3','4','5','6','7','8','9','t','j','q','k'] suits = ['H','D','S','C',] return ["".join((v,x)) for v, x in product(value, suits)] def get_shuffled_deck(cards): shuffle(cards) def deal_card(cards): return cards.pop(0) if name == 'main': print 'Obteniendo mazo de cartas' cards = doing_deck() print 'Barajando...' get_shuffled_deck(cards) carta1 = deal_card(cards) print 'Primera carta obtenida: {}.'.format(carta1) get_shuffled_deck(cards) print('Barajando...') carta2 = deal_card(cards) print 'Segunda carta obtenida: {}.'.format(carta2)
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Number of days timedelta can be obtained through its properties days - the value of this characteristic and return from function:def days_in_month(year, month): ... a=date1-date2 return a.days Then comparison day < days_in_month(year,month) I'll work.It can also be reversed. timedeltabut it is more logical to compare time periods rather than to check the day in time:from datetime import timedelta print(timedelta(days=10) < timedelta(days=20)) # True, т.е. период в 10 дней меньше периода в 20 дней
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Reply to question https://ru.stackoverflow.com/questions/601909/%d0%9a%d0%b0%d0%ba-%d1%81%d0%b4%d0%b5%d0%bb%d0%b0%d1%82%d1%8c-%d1%81%d0%bf%d0%b8%d1%81%d0%be%d0%ba-%d1%81-%d1%81%d1%82%d1%80%d0%be%d0%ba%d0%b8-%d0%a0%d0%b0%d0%b7%d0%b4%d0%b5%d0%bb%d0%b8%d1%82%d1%8c-%d1%81%d1%82%d1%80%d0%be%d0%ba%d1%83-%d0%b8%d1%81%d0%bf%d0%be%d0%bb%d1%8c%d0%b7%d1%83%d1%8f-%d1%80%d0%b5%d0%b3%d1%83%d0%bb%d1%8f%d1%80%d0%bd%d0%be%d0%b5-%d0%b2%d1%8b%d1%80%d0%b0%d0%b6%d0%b5%d0%bd%d0%b8%d0%b5#comment816464_601912 :If there's another proposal in the brackets in front of (1995), you can. I'd like to make sure that the program divides into the elements of the list. until (1995) and after? (sighs)For 1995 any 4-digit)Decision:In [7]: s = 'I (0) am a (smart) person (1995) Russia' In [8]: re.split(r'\s((\d{4}))\s', s) Out[8]: ['I (0) am a (smart) person', '(1995)', 'Russia'] https://regex101.com/r/VECOst/1
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k := Pos(find, ListBox1.Items.Text); if k = 1 then //совпадение с первого символа begin Listbox1.Items.Text := Copy(Listbox1.Items.Text, k + Length(find), Length(Listbox1.Items.Text)); end; You don't have to find the same thing in the line twice. And you can do without a temporary variable. Can be added to the number of copying symbols -Length(find)but not necessarily. There will be as many symbols as they say or until the line is over.
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http://ideone.com/6Y6cl0 :#!/usr/bin/env python3 import fileinput words = set() #NOTE: it is unordered but it doesn't affect the result tokens = (token for line in fileinput.input() for token in line.split()) for op, word in zip(*[tokens]2): if op == '+': words.add(word) elif op == '-': words.discard(word) elif op == '?': print(word, "Yes" if word in words else "No") @VladD asked:How does it work? [tokens]2zip([iterator]n) - it's (satisfactory, unusual) idiom bypass of the heterator. n the elements at a time. Official documentation is https://docs.python.org/3/library/itertools.html#itertools-recipes ♪ Look. https://stackoverflow.com/questions/434287/what-is-the-most-pythonic-way-to-iterate-over-a-list-in-chunks and https://ru.stackoverflow.com/q/594651/23044 The main idea is, zip() at every step, it's a big deal. next() function n one on the same terator:>>> from itertools import zip_longest >>> for a, b, c in zip_longest([iter(range(10))]*3, fillvalue='default'): ... print(a, b, c) ... 0 1 2 3 4 5 6 7 8 9 default default zip_longest() necessary only if the vent is not repeated n♪The terator is a simple object that generates elements if next() Called:>>> it = iter(range(2)) >>> next(it) 0 >>> next(it) 1 >>> next(it) Traceback (most recent call last): File "<stdin>", line 1, in <module> StopIteration >>> next(it) Traceback (most recent call last): File "<stdin>", line 1, in <module> StopIteration https://docs.python.org/3/glossary.html#term-iterator You can only go once. A new terator must be created to reboot the rounds. https://docs.python.org/3/glossary.html#term-generator-expression returns the terator.
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For this design:for i in my_list: i*=5 Python is creating a new variable i to which the lists are consistently listed. In the end, you change the variable values. ithe list remains unchanged.In [136]: my_list=[1,2,3,4,5,6] ...: for i in my_list: ...: i*=5 ...: print(i, my_list) ...: 5 [1, 2, 3, 4, 5, 6] 10 [1, 2, 3, 4, 5, 6] 15 [1, 2, 3, 4, 5, 6] 20 [1, 2, 3, 4, 5, 6] 25 [1, 2, 3, 4, 5, 6] 30 [1, 2, 3, 4, 5, 6] In the next line of the code, you attribute the variable. my_list_1 output list comprehension (list):my_list_1=[i*5 for i in my_list] Decision:my_list[:] = [i*5 for i in my_list] print(my_list) #[5, 10, 15, 20, 25, 30] PS can be read in more detail on the assignment of lists and cuts https://stackoverflow.com/questions/10623302/how-assignment-works-with-python-list-slice

When pandas are used for excel reading, the indexing begins with line 2.

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