Hen*_*ton 27 json gzip python-3.x
对于Python3,我跟着@Martijn Pieters的代码:
import gzip
import json
# writing
with gzip.GzipFile(jsonfilename, 'w') as fout:
for i in range(N):
uid = "whatever%i" % i
dv = [1, 2, 3]
data = json.dumps({
'what': uid,
'where': dv})
fout.write(data + '\n')
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但这会导致错误:
Traceback (most recent call last):
...
File "C:\Users\Think\my_json.py", line 118, in write_json
fout.write(data + '\n')
File "C:\Users\Think\Anaconda3\lib\gzip.py", line 258, in write
data = memoryview(data)
TypeError: memoryview: a bytes-like object is required, not 'str'
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关于发生了什么的任何想法?
Tom*_*lak 69
你在这里有四个转变步骤.
让我们逐一采取这些步骤.
import gzip
import json
data = []
for i in range(N):
uid = "whatever%i" % i
dv = [1, 2, 3]
data.append({
'what': uid,
'where': dv
}) # 1. data
json_str = json.dumps(data) + "\n" # 2. string (i.e. JSON)
json_bytes = json_str.encode('utf-8') # 3. bytes (i.e. UTF-8)
with gzip.GzipFile(jsonfilename, 'w') as fout: # 4. gzip
fout.write(json_bytes)
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请注意,"\n"这里添加完全是多余的.它不会破坏任何东西,但除此之外它没有任何用处.
阅读完全相反:
with gzip.GzipFile(jsonfilename, 'r') as fin: # 4. gzip
json_bytes = fin.read() # 3. bytes (i.e. UTF-8)
json_str = json_bytes.decode('utf-8') # 2. string (i.e. JSON)
data = json.loads(json_str) # 1. data
print(data)
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当然,这些步骤可以合并:
with gzip.GzipFile(jsonfilename, 'w') as fout:
fout.write(json.dumps(data).encode('utf-8'))
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和
with gzip.GzipFile(jsonfilename, 'r') as fin:
data = json.loads(fin.read().decode('utf-8'))
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Jan*_*rik 21
此处提到的解决方案(感谢@Rafe)有一个很大的优势:由于编码是即时完成的,因此您无需为生成的 json 创建两个完整的中间字符串对象。对于大对象,这可以节省内存。
with gzip.open(jsonfilename, 'wt', encoding='UTF-8') as zipfile:
json.dump(data, zipfile)
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此外,读取和解码也很简单:
with gzip.open(jsonfilename, 'rt', encoding='UTF-8') as zipfile:
my_object = json.load(zipfile)
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