#python Cookbook 2nd edition study (all recorded here)

#标题: Build lists with List resolution
#例1: For example, to add a number to each subkey in the list, as follows:
def myList (L):
return ([x+20 for X in L])
Print MyList ([1,2,3,4,5,6,7,8])
#例2: For example, to find a list larger than a certain number to build a new list, as follows
def myList (L):
return ([x for x in L if X>5])
Print MyList ([1,2,3,4,5,6,7,8])
#对例1和例2进行合并, as follows:
def myList (L):
return ([x+20 for x in L if X>5])
Print MyList ([1,2,3,4,5,6,7,8])
#说明: List resolution is the best way to consider changing a list, rather than creating a new one, such as the need to set a list L greater than 100 to 100, as follows
def myList1 (L):
L[:]=[min (x,100) for x in L]
Print L
MyList1 (list (range (90,102)))
#注意: It is not recommended to use list parsing when executing a loop!
#比如要复制一个列表, write directly to list (L), and when a function is called on each element, use map (f,l)
#例3: Summing a list subkey
def myList (L):
return sum ([X+20for x in L if X>5])
Print MyList (range (20))
#或者写成如下, without parentheses
def myList (L):
return sum (X+20for x in L if x>5)
Print myList (range) #这样做的原因是可以一次性将整个列表载入到内存

#给字典增加一个条目
#标题: Given a dictionary d, when K is the key to the dictionary, directly using d[k], if K is not the D key, then create this new entry, as follows
def addwork (index,word,page):
Index.setdefault (word,[]). Append (page)

#反转字典
#给定一个字典, this dictionary maps different keys to different values, and you want to create a reversed dictionary that reflects each value to the key
#通过dict列表解析式解决这个问题
def invert_dict (d):
    Return Dict ([(v,k) for k,v in D.iteritems ()])

Print invert_dict ({1:2, ' a ': 4})
#对于比较大的字典, you can use the Izip () of the Itertools module
From Itertools import Izip
def invert_dict (d):
    Return Dict (Izip (D.iteritems (), D.iterkeys ()))
Print invert_dict ({1:2, ' a ': 4})
"""
If the value in dictionary D is not unique, then D cannot be really reversed, i.e. there is no such dictionary, for any given key K, satisfies
Id[d[k]]==k, but for any of the dictionary D-value v,d[pd[v]]=v, given your original dictionary d, you can check if x is a D inverse dictionary or a pseudo-inverse dictionary: When and only Len (x) ==len (d), X is the true inverse Dictionary of D
Because if 2 different keys correspond to the same value, one of the 2 keys will definitely disappear for the 2 functions given by the solution.
Thus the generated pseudo-inverse dictionary length will be shorter than the original dictionary length, in any case, only if the value in D is hashed, then the previous function will work properly, otherwise the function would give a TypeError exception
This function obtains the Iteritems method to generate a pair of keys and other corresponding values K and V, wrapping them into (v,k) Order
and assigns the final generation sequence as a parameter to the dict, so that dict constructs a key for a value, and the original key becomes a new dictionary of the corresponding value,
"""

#python Cookbook 2nd edition study (all recorded here)