[Python] strip vs. replace

Here are some examples showing the differences between the function of "strip()" and "replace()" in Python.

	#EX.1
	>>> x = "citt""
	>>> x.strip(""")
	'ci'
	>>> x.replace(""","")
	'citt'
	#EX.2
	>>> x = ""city" is"
	>>> x.strip(""")
	'city" is'
	>>> x.replace(""","")
	'city is'

view raw strip_replace.py hosted with ❤ by GitHub

Be careful! strip() will delete any character in the beginning or end of the word that matches "any" character in the word we put in the strip function. For example, we put "& quot;" as a parameter into strip(), then any character matches ('&', 'q', 'u', 'o', 't', ';') will be deleted. It doesn't have to be exactly the same "string."

Install CMake command line tools on MacOS

The following video is a good demonstration.

[Python] Find part of speech (POS) by using NLTK

	>>> import nltk
	#remember to use list
	>>> x = nltk.pos_tag(['New'])
	>>> print(x)
	[('New', 'NNP')]
	>>> print(x[0][1])
	NNP
	#if you don't use list, then the pos_tag function will find the pos for each character
	>>> x = nltk.pos_tag('New')
	>>> print(x)
	[('N', 'NNP'), ('e', 'VBP'), ('w', 'NN')]

view raw pos.py hosted with ❤ by GitHub

[Python] Performance enhance by not using range() for comparison

When I was trying to enhance the performance of my Python code, the frist thing I tried is reducing the usage of functions. After I substituted the range() function, the performance had improved a lot, from 404ms to 106ms. The following is the difference between the two versions of the "ZigZag Conversion" code.

404ms

	class Solution:
	# @return a string
	def convert(self, s, nRows):
	row_len = 2 * nRows - 2
	string = ["" for _ in range(nRows)]
	if nRows <= 2:
	for i in range(len(s)):
	string[i % nRows] += s[i]
	else:
	for i in range(len(s)):
	res = i % row_len
	if res in range(nRows, row_len):
	string[row_len - res] += s[i]
	else:
	string[res] += s[i]
	output = ""
	for i in range(nRows):
	output += string[i]
	return output

view raw ZigZag_404.py hosted with ❤ by GitHub

106ms

	class Solution:
	# @return a string
	def convert(self, s, nRows):
	row_len = 2 * nRows - 2
	string = ["" for _ in range(nRows)]
	if nRows <= 2:
	for i in range(len(s)):
	string[i % nRows] += s[i]
	else:
	for i in range(len(s)):
	res = i % row_len
	if res >= nRows and res < row_len: # I just replaced the range() function here
	string[row_len - res] += s[i]
	else:
	string[res] += s[i]
	output = ""
	for i in range(nRows):
	output += string[i]
	return output

view raw ZigZag_106.py hosted with ❤ by GitHub

Although the performance result depends on the test data, this substitution really improves the entire performance.

[Python] Find synonyms by using WordNet, NLTK

WordNet is a great reference when you want to find synonyms of a word. For more information, please see the website: http://wordnet.princeton.edu/

The following Python code is a simple example.

	#Python 3
	#Find synonyms of a word
	>>> from nltk.corpus import wordnet as wn
	>>> syn = wn.synsets("car")
	>>> syn
	[Synset('car.n.01'), Synset('car.n.02'), Synset('car.n.03'), Synset('car.n.04'), Synset('cable_car.n.01')]
	>>> syn[0].lemma_names()
	['car', 'auto', 'automobile', 'machine', 'motorcar']
	#Note that synonyms are always in singular form.
	#If you want to compare words, need to deal with the words that are in plural form.

view raw WordNet.py hosted with ❤ by GitHub

WordNet Interface: http://www.nltk.org/howto/wordnet.html

[Python] try... except

	#python
	try: #prevent unexpected crash
	ofile = open(filename)
	except:
	print "file cannot be openned:", filename
	exit()
	#read one line at a time
	for line in ofile:
	line = line.rstrip() #delete "/n" from original file line
	print line
	#read "whole" file at a time
	wholeFile = ofile.read()

view raw read_file.py hosted with ❤ by GitHub

Apply BFS to special MST problems

Given a undirected and connected graph, and the edges have weights of either 1 or 2 (actually it is also applicable for weight larger than 2). How can we find a MST of the graph by using BFS?

Ans (I guess):
First, we can view an edge with weight of 2 as two edges with weight of 1 connected together. Then we can simply apply BFS to solve this problem. After finish the BFS algorithm, we check all the edges originally with weight of 2. If both parts of an edge are not both chosen as a path of the MST, then we delete this path. Finally, the result is a MST.

Use pdb to debug Python code

use pdb to debug python code:

$ python -m pdb filename.py

view raw pdb hosted with ❤ by GitHub

some (not all) commands of pdb:
h: see the entire help list

l: l(ist) [first [,last]]

List source code for the current file.

Without arguments, list 11 lines around the current line or continue the previous listing.

With one argument, list 11 lines starting at that line.

With two arguments, list the given range; if the second argument is less than the first, it is a count. 

s:

Execute the current line, stop at the first possible occasion

(either in a function that is called or in the current function).

c: 
Continue execution, only stop when a breakpoint is encountered.

b: b(reak) ([file:]lineno | function) [, condition]

p: p expression

Print the value of the expression.

for more information: https://docs.python.org/2/library/pdb.html