Python: Find the second most repeated word in a given string - w3resource

from PIL import Image
import glob
image_list = []
for filename in glob.glob('yourpath/*.gif'): #assuming gif
    im=Image.open(filename)
    image_list.append(im)

percent_missing = df.isnull().sum() * 100 / len(df)
missing_value_df = pd.DataFrame({'column_name': df.columns,
                                 'percent_missing': percent_missing})

days = 0
week = [‘Monday’, ‘Tuesday’, ‘Wednesday’, ‘Thursday’, ‘Friday’, ‘Saturday’, 3.‘Sunday’]
while day < 7:
print(“Today is” + week[days])
days += 1
                                
                                

>>> ["foo", "bar", "baz"].index("bar")
1

This method gets vowels (‘a’, ‘e’, ‘i’, ‘o’, ‘u’) found in a string.
   
#make a function:
def get_vowels(string):

#return is the keyword which means function have to return value: 
 return [each for each in string if each in 'aeiou']


#assign the words and function will return vowels words.
get_vowels('foobar') # ['o', 'o', 'a']


get_vowels('gym') # []

options = webdriver.ChromeOptions()
options.add_experimental_option("excludeSwitches", ["enable-automation"])
options.add_experimental_option('useAutomationExtension', False)
options.add_argument("--disable-blink-features=AutomationControlled")
driver = webdriver.Chrome(options=options)

import numpy as np

def pagerank(M, num_iterations=100, d=0.85):
    N = M.shape[1]
    v = np.random.rand(N, 1)
    v = v / np.linalg.norm(v, 1)
    iteration = 0
    while iteration < num_iterations:
        iteration += 1
        v = d * np.matmul(M, v) + (1 - d) / N
    return v

import pandas as pd

sheets_dict = pd.read_excel('Book1.xlsx', sheetname=None)

full_table = pd.DataFrame()
for name, sheet in sheets_dict.items():
    sheet['sheet'] = name
    sheet = sheet.rename(columns=lambda x: x.split('\n')[-1])
    full_table = full_table.append(sheet)

full_table.reset_index(inplace=True, drop=True)

print full_table

from datetime import datetime

datetime_object = datetime.strptime('Jun 1 2005  1:33PM', '%b %d %Y %I:%M%p')

from selenium import webdriver
from selenium.webdriver.chrome.options import Options
from fake_useragent import UserAgent

options = Options()
ua = UserAgent()
userAgent = ua.random
print(userAgent)
options.add_argument(f'user-agent={userAgent}')
driver = webdriver.Chrome(chrome_options=options, executable_path=r'C:\WebDrivers\ChromeDriver\chromedriver_win32\chromedriver.exe')
driver.get("https://www.google.co.in")
driver.quit()

def read_csv_pgbar(csv_path, chunksize, usecols, dtype=object):


    # print('Getting row count of csv file')

    rows = sum(1 for _ in open(csv_path, 'r')) - 1 # minus the header
    # chunks = rows//chunksize + 1
    # print('Reading csv file')
    chunk_list = []

    with tqdm(total=rows, desc='Rows read: ') as bar:
        for chunk in pd.read_csv(csv_path, chunksize=chunksize, usecols=usecols, dtype=dtype):
            chunk_list.append(chunk)
            bar.update(len(chunk))

    df = pd.concat((f for f in chunk_list), axis=0)
    print('Finish reading csv file')

    return df

keys, values)) # {'a': 2, 'c': 4, 'b': 3}
 
 
#make a function: def is the keyword for the function:
def to_dictionary(keys, values):
 
 
#return is the keyword that tells program that function has to return value   
return dict(zip(keys, values))
 
  
 
# keys and values are the lists:
 
keys = ["a", "b", "c"]   
 
values = [2, 3, 4]
                                
                                

>>> from enum import Enum
>>> class Build(Enum):
...   debug = 200
...   build = 400
... 

Build['debug']

my_list = [27, 5, 9, 6, 8]

def RemoveValue(myVal):
    if myVal not in my_list:
        raise ValueError("Value must be in the given list")
    else:
        my_list.remove(myVal)
    return my_list

print(RemoveValue(27))
print(RemoveValue(27))