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This article introduces the relevant knowledge of "what are the useful Python skills". In the operation of practical cases, many people will encounter such a dilemma, so let the editor lead you to learn how to deal with these situations. I hope you can read it carefully and be able to achieve something!

Continuous function call

Def add (x): class AddNum (int): def _ call__ (self, x): return AddNum (self.numerator + x) return AddNum (x) print add (2) (3) (5) # 10 print add (2) (3) (4) (5) (6) (7) # 27 # javascript version var add = function (x) {var addNum = function (x) {return add (addNum + x) }; addNum.toString = function () {return x;} return addNum;} add (2) (3) (5) / / 10 add (2) (3) (3) (4) (5) (6) / / 27

Default value trap

> def evil (v = []): V.append (1)... Print v. > > evil () [1] > evil () [1,1]

Read and write csv files

Import csv with open ('data.csv',' rb') as f: reader = csv.reader (f) for row in reader: print row # writes import csv with open ('data.csv',' wb') as f: writer = csv.writer (f) writer.writerow (['name',' address'') to the csv file 'age']) # single-line data = [(' xiaoming', 'china','10'), (' Lily', 'USA',' 12')] writer.writerows (data) # multiple-line write

Number system conversion

> int ('1000, 2) 8 > int (' Aguilar, 16) 10

Format json

Echo' {"k": "v"}'| python-m json.tool

List flattening

List_ = [[1, 2, 3], [4, 5, 6], [7, 8, 9]] [k for i in list_ for k in i] # [1, 2, 3, 4, 5, 8, 9] import numpy as np print np.r_ [[1, 2, 3], [4, 5, 6], [7, 8, 9] import itertools print list (* [[1, 2, 3], [4, 5, 6] [7,8,9]) sum (list_, []) flatten = lambda x: [y for l in x for y in flatten (l)] if type (x) is list else [x] flatten (list_)

List merger

> a = [1,3,5,7,9] > > b = [2,3,4,5,6] > > c = [5,6,7,8,9] > > list (set (). Union (a, b, c)) [1, 2, 3, 4, 5, 6, 7, 8, 9]

The two letters that appear the most

From collections import Counter c = Counter ('hello world') print (c.most_common (2)) # [(' lumped, 3), ('oiled, 2)]

Use with caution

Eval ("_ _ import__ ('os'). System (' rm-rf /')", {})

Permutation matrix

Matrix = [[1,2,3], [4,5,6]] res = zip (* matrix) # res = [(1,4), (2,5), (3,6)]

List derivation

[item**2 for item in lst if item% 2] map (lambda item: item**2, filter (lambda item: item% 2, lst) > list (map (str, [1, 2, 3, 4, 5, 6, 7, 8, 9]) [1, 2, 3, 3, 9]) [1, 2, 3, 3, 4, 4, 5, 6, 7, 8, 9]

Permutation and combination

> for p in itertools.permutations ([1,2,3,4]):. Print''.join (str (x) for x in p)... 1234 1243 1324 1342 1423 1432 2134 2143 2314 2341 2413 2431 3142 3214 3241 3412 3421 4123 4132 4213 4231 4312 4321 > > for c in itertools.combinations ([1,2,3,4,5], 3):. Print'. Join (str (x) for x in c)... 123 124 125 134 135 145 234 235 245 345 > > for c in itertools.combinations_with_replacement ([1, 2, 3], 2): Print''.join (str (x) for x in c)... 11 12 13 22 23 33 > for p in itertools.product ([1,2,3], [4,5]): (1,4) (1,5) (2,4) (2,5) (3,4) (3,5)

Default dictionary

> m = dict () > > m ['a'] Traceback (most recent call last): File "", line 1 In KeyError:'a'> m = collections.defaultdict (int) > m ['a'] 0 > m ['b'] 0 > > m = collections.defaultdict (str) > m ['a']'> > m ['b'] + ='a' > > m ['b']'a' > > m = collections.defaultdict (lambda:'[default value]') > > m ['a']'[default value]'> > m ['b']'[default value]''

Reverse dictionary

> m = {'averse: 1,' baked: 2, 'caged: 3,' dashed: 4} > m {'k for k: 4, 'averse: 1,' baked: 2, 'crested: 3} > > {v: k for k, v in m.items ()} {1:' ajar, 2: 'baked, 3:' clocked, 4:'d'} what are the useful Python techniques? Thank you for your reading. If you want to know more about the industry, you can follow the website, the editor will output more high-quality practical articles for you!

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