def bulk_insert():
"""测试bulk insert和一个个insert的性能区别:
结论:
Bulk insert的速度要远远快于一条条insert
注: Bulk insert支持生成器模式。
"""
list_of_documents1 = [{"name": fmter.tpl.randstr(8)} for i in range(1000)]
def document_generator():
for doc in list_of_documents1:
yield doc
timer.start()
users.insert(document_generator())
timer.timeup()
list_of_documents2 = [{"name": fmter.tpl.randstr(8)} for i in range(1000)]
timer.start()
for doc in list_of_documents2:
users.insert(doc)
timer.timeup()
print(users.find().count())
# bulk_insert()
def bulk_insert():
"""测试bulk insert和一个个insert的性能区别:
结论:
Bulk insert的速度要远远快于一条条insert
注: Bulk insert支持生成器模式。
"""
list_of_documents1 = [{"name": fmter.tpl.randstr(8)} for i in range(1000)]
def document_generator():
for doc in list_of_documents1:
yield doc
timer.start()
users.insert(document_generator())
timer.timeup()
list_of_documents2 = [{"name": fmter.tpl.randstr(8)} for i in range(1000)]
timer.start()
for doc in list_of_documents2:
users.insert(doc)
timer.timeup()
print(users.find().count())
# bulk_insert()
def can_key_be_other_than_string():
"""json file cannot have key other than string. So you cannot use integer as a key, even it
is valid in python dictionary.
In addition, choose key string wisely can save space.
"""
document = {1: "a"}
users.insert(document)
for doc in users.find():
print(doc)
def can_key_be_other_than_string():
"""json file cannot have key other than string. So you cannot use integer as a key, even it
is valid in python dictionary.
In addition, choose key string wisely can save space.
"""
document = {1: "a"}
users.insert(document)
for doc in users.find():
print(doc)
def date_and_datetime_type():
"""
mongodb doesn't support date object and only accept datetime. you have to convert
date to datetime using datetime.combine(date_object, datetime.min.time()) to
normalize to midnight.
"""
document = {"create_datetime": datetime.now(),
"create_date": datetime.combine(date.today(), datetime.min.time())}
users.insert(document)
for doc in users.find():
print(doc)
def bytes_type():
"""mongodb support bytes, which means you can use pickle to dump anything into mongodb.
But! Don't forget the maximum BSON document size is 16 megabytes
"""
documents = [
{"pickle": "hello world".encode("utf-8")},
{"pickle": obj2bytestr(set([1,2,3]))},
]
users.insert(documents)
for doc in users.find():
print(doc)
def list_and_set_type():
documents = [
{"list": [1,2,3]},
{"set": set([1,2,3])}, # this cannot be done
]
users.insert(documents)
for doc in users.find():
print(doc)
# list_and_set_type()
def date_and_datetime_type():
"""
mongodb doesn't support date object and only accept datetime. you have to convert
date to datetime using datetime.combine(date_object, datetime.min.time()) to
normalize to midnight.
"""
document = {
"create_datetime": datetime.now(),
"create_date": datetime.combine(date.today(), datetime.min.time())
}
users.insert(document)
for doc in users.find():
print(doc)
def insept_example():
"""insept的意思是: 首先尝试insert, 如果面临着_id重复问题, 则update
该逻辑可以用upsert实现。注: 有时候document是没有包含_id项的
"""
doc = {"_id": 1, "name": "obama", "new_field": 999}
try:
users.insert(doc)
except:
_id = doc["_id"]
del doc["_id"]
users.update({"_id": _id}, {"$set": doc}, upsert=True)
ppt(users.find({"name": "obama"})[0])
# insept_example()
def bytes_type():
"""mongodb support bytes, which means you can use pickle to dump anything into mongodb.
But! Don't forget the maximum BSON document size is 16 megabytes
"""
documents = [
{
"pickle": "hello world".encode("utf-8")
},
{
"pickle": obj2bytestr(set([1, 2, 3]))
},
]
users.insert(documents)
for doc in users.find():
print(doc)
def list_and_set_type():
documents = [
{
"list": [1, 2, 3]
},
{
"set": set([1, 2, 3])
}, # this cannot be done
]
users.insert(documents)
for doc in users.find():
print(doc)
# list_and_set_type()
def boolean_and_none_type():
"""{key: None} means key == None or key is not existing
"""
documents = [{"is_valid": True}, {"is_valid": False}, {"is_valid": None}]
users.insert(documents)
fmter.tpl._straightline("is_valid == True", 100)
for doc in users.find({"is_valid": True}):
print(doc)
fmter.tpl._straightline("is_valid == False", 100)
for doc in users.find({"is_valid": False}):
print(doc)
fmter.tpl._straightline("is_valid is null", 100)
for doc in users.find({"is_valid": None}):
print(doc)
fmter.tpl._straightline("is_valid not null", 100)
for doc in users.find({"is_valid": {"$ne": None}}):
print(doc)
def boolean_and_none_type():
"""{key: None} means key == None or key is not existing
"""
documents = [{"is_valid": True}, {"is_valid": False}, {"is_valid": None}]
users.insert(documents)
fmter.tpl._straightline("is_valid == True", 100)
for doc in users.find({"is_valid": True}):
print(doc)
fmter.tpl._straightline("is_valid == False", 100)
for doc in users.find({"is_valid": False}):
print(doc)
fmter.tpl._straightline("is_valid is null", 100)
for doc in users.find({"is_valid": None}):
print(doc)
fmter.tpl._straightline("is_valid not null", 100)
for doc in users.find({"is_valid": {"$ne": None}}):
print(doc)
def reserved_key_id():
"""如果用户不指定_id, 则系统会自动创建一个_id。问题是对于同样内容的文档, 自动生成的_id会不会
重复呢?
结论:
对于内存中不同的对象, mongodb是不会生成重复的_id的。但是如果是同样的对象, 则会生成重复的_id
test1:
每一个doc其实是生成了一个新字典。而每次生成了新字典的时候python就将变量名doc绑定
到新字典上。由于旧字典没有被reference, 那么系统就会自动垃圾回收释放内存了。所以每一次
doc其实是内存中不同的对象。所以_id不会冲突。
test2:
我们生成了一个document的列表。里面每一个元素在内存中其实是不同的。所以_id也不会冲突
test3:
我们重复调用了test2中的列表。由于里面每一个元素在内存中的地址是一样的, 所以生成了同样
的_id, 造成了冲突。
"""
# test 1
for i in range(10):
doc = {"text": "abcdefg"}
users.insert(doc)
print(users.find().count())
list_of_documents = [{"text": "abcdefg"} for i in range(10)]
# test 2
users.insert(list_of_documents)
print(users.find().count())
# test 3
for doc in list_of_documents:
users.insert(doc)
print(users.find().count())
def reserved_key_id():
"""如果用户不指定_id, 则系统会自动创建一个_id。问题是对于同样内容的文档, 自动生成的_id会不会
重复呢?
结论:
对于内存中不同的对象, mongodb是不会生成重复的_id的。但是如果是同样的对象, 则会生成重复的_id
test1:
每一个doc其实是生成了一个新字典。而每次生成了新字典的时候python就将变量名doc绑定
到新字典上。由于旧字典没有被reference, 那么系统就会自动垃圾回收释放内存了。所以每一次
doc其实是内存中不同的对象。所以_id不会冲突。
test2:
我们生成了一个document的列表。里面每一个元素在内存中其实是不同的。所以_id也不会冲突
test3:
我们重复调用了test2中的列表。由于里面每一个元素在内存中的地址是一样的, 所以生成了同样
的_id, 造成了冲突。
"""
# test 1
for i in range(10):
doc = {"text": "abcdefg"}
users.insert(doc)
print(users.find().count())
list_of_documents = [{"text": "abcdefg"} for i in range(10)]
# test 2
users.insert(list_of_documents)
print(users.find().count())
# test 3
for doc in list_of_documents:
users.insert(doc)
print(users.find().count())
def basic_insert_syntax():
"""
db.collection.insert(one_document) or db.collections.insert(list_of_documents)
BUT!
if any document in list_of_documents has _id conflict with existing document,
then that would be failed. you should use the following code:
for document in list_of_documents:
try:
db.collection.insert(document)
except:
pass
"""
documents1 = [
{"name": "Bill Gates", "lastname": "Bill", "firstname": "Gates",
"profile": {"year": 1955, "money": 700}},
{"name": "Steve Jobs", "lastname": "Steve", "firstname": "Jobs",
"profile": {"year": 1955, "money": 69}},
{"name": "Elon Musk", "lastname": "Elon", "firstname": "Musk",
"profile": {"year": 1971, "money": 103}},
]
documents2 = [
{"_id": 100, "name": "Obama", "nation": "USA", "money": None},
{"_id": 101, "name": "Churchill", "nation": "Egnland", "money": None},
{"_id": 101, "name": "Bin laden", "nation": "Pakistan", "money": None}, # 有重复
]
users.insert(documents1) # list of dict 一口气插入, 其中当然不能有_id重复
for doc in documents2: # 用 for loop 一个个插入
try:
users.insert(doc)
except Exception as e:
print(e)
for doc in users.find():
print(type(doc), doc) # 默认返回字典, 并非有序字典
def basic_insert_syntax():
"""
db.collection.insert(one_document) or db.collections.insert(list_of_documents)
BUT!
if any document in list_of_documents has _id conflict with existing document,
then that would be failed. you should use the following code:
for document in list_of_documents:
try:
db.collection.insert(document)
except:
pass
"""
documents1 = [
{
"name": "Bill Gates",
"lastname": "Bill",
"firstname": "Gates",
"profile": {
"year": 1955,
"money": 700
}
},
{
"name": "Steve Jobs",
"lastname": "Steve",
"firstname": "Jobs",
"profile": {
"year": 1955,
"money": 69
}
},
{
"name": "Elon Musk",
"lastname": "Elon",
"firstname": "Musk",
"profile": {
"year": 1971,
"money": 103
}
},
]
documents2 = [
{
"_id": 100,
"name": "Obama",
"nation": "USA",
"money": None
},
{
"_id": 101,
"name": "Churchill",
"nation": "Egnland",
"money": None
},
{
"_id": 101,
"name": "Bin laden",
"nation": "Pakistan",
"money": None
}, # 有重复
]
users.insert(documents1) # list of dict 一口气插入, 其中当然不能有_id重复
for doc in documents2: # 用 for loop 一个个插入
try:
users.insert(doc)
except Exception as e:
print(e)
for doc in users.find():
print(type(doc), doc) # 默认返回字典, 并非有序字典