[英]JSON data convert to string data by python
我想将下面的 JSON 数据转换为与下面相同的字符串数据。
{
"Time Series (Daily)": {
"2020-04-02": {
"1. open": "1693.5300",
"2. high": "1726.7600",
"3. low": "1664.1300",
"4. close": "1724.8600",
"5. volume": "766313"
},
"2020-04-01": {
"1. open": "1737.2800",
"2. high": "1762.4399",
"3. low": "1685.3700",
"4. close": "1685.4600",
"5. volume": "1243600"
}
}
转换成这样
^KS11 |20200402|1664.1300 |1693.5300 |1726.7600 |1724.8600 |
^KS11 |20200401|1685.4600 |1737.2800 |1762.4399 |1685.3700 |
如果您以字符串形式提供 json 数据,我建议您使用json包,如下所示:
import json
jdata = """
{
"Time Series (Daily)": {
"2020-04-02": {
"1. open": "1693.5300",
"2. high": "1726.7600",
"3. low": "1664.1300",
"4. close": "1724.8600",
"5. volume": "766313"
},
"2020-04-01": {
"1. open": "1737.2800",
"2. high": "1762.4399",
"3. low": "1685.3700",
"4. close": "1685.4600",
"5. volume": "1243600"
}
}
}"""
def day_to_str(day, data):
return '|'.join(('^KS11 ',
day.replace('-', ''),
data['1. open'],
data['2. high'],
data['3. low'],
data['4. close'],
''))
time_series = json.loads(jdata)['Time Series (Daily)']
for day in time_series:
print(day_to_str(day, time_series[day]))
输出:
^KS11 |20200402|1693.5300|1726.7600|1664.1300|1724.8600|
^KS11 |20200401|1737.2800|1762.4399|1685.3700|1685.4600|
如果你不想得到volume
,它总是在最后。你可以试试这个:
yourDict = { # assume it is a dict.Json is also OK.
"Time Series (Daily)": {
"2020-04-02": {
"1. open": "1693.5300",
"2. high": "1726.7600",
"3. low": "1664.1300",
"4. close": "1724.8600",
"5. volume": "766313"
},
"2020-04-01": {
"1. open": "1737.2800",
"2. high": "1762.4399",
"3. low": "1685.3700",
"4. close": "1685.4600",
"5. volume": "1243600"
}
}
}
temp = "^KS11 |{} |\n"
result = ""
temp = "^KS11 |{} |\n" # put data into {}
result = "" # use string
# Easy string handle
for i in yourDict["Time Series (Daily)"]:
# get the value in each dict
numberList = list(yourDict["Time Series (Daily)"][i].values())[:-1] # extract them
oneLineNumber = " |".join(numberList)# make them together
oneLineStr = temp.format(oneLineNumber) # Add them
result += oneLineStr
print(result)
结果(多一个“\\n”):
^KS11 |20200402|1693.5300 |1726.7600 |1664.1300 |1724.8600 |
^KS11 |20200401|1737.2800 |1762.4399 |1685.3700 |1685.4600 |
或一行代码:
yourDict = {
"Time Series (Daily)": {
"2020-04-02": {
"1. open": "1693.5300",
"2. high": "1726.7600",
"3. low": "1664.1300",
"4. close": "1724.8600",
"5. volume": "766313"
},
"2020-04-01": {
"1. open": "1737.2800",
"2. high": "1762.4399",
"3. low": "1685.3700",
"4. close": "1685.4600",
"5. volume": "1243600"
}
}
}
print("\n".join(["^KS11 |{} |".format(i.replace("-","")+"|"+" |".join(list(yourDict["Time Series (Daily)"][i].values())[:-1])) for i in yourDict["Time Series (Daily)"]]))
这是(不再是“\\n”):
^KS11 |20200402|1693.5300 |1726.7600 |1664.1300 |1724.8600 |
^KS11 |20200401|1737.2800 |1762.4399 |1685.3700 |1685.4600 |
(也许有更好的方法)
import json
data = '''
{
"Time Series (Daily)": {
"2020-04-02": {
"1. open": "1693.5300",
"2. high": "1726.7600",
"3. low": "1664.1300",
"4. close": "1724.8600",
"5. volume": "766313"
},
"2020-04-01": {
"1. open": "1737.2800",
"2. high": "1762.4399",
"3. low": "1685.3700",
"4. close": "1685.4600",
"5. volume": "1243600"
}
}
}
'''
loaded_data = json.loads(data)
time_series_data = loaded_data["Time Series (Daily)"]
temp = "^KS11"
for date, date_data in time_series_data.items():
formatted_date = date.replace("-","")
low_data = date_data["3. low"]
open_data = date_data["1. open"]
high_data = date_data["2. high"]
close_data = date_data["4. close"]
print("{}\t|{}|{}\t|{}\t|{}\t|{}\t|\n".format(temp, formatted_date, low_data, open_data, high_data, close_data))
假设您知道如何使用 json 模块读取 json。
stocks = {
"Time Series (Daily)": {
"2020-04-02": {
"1. open": "1693.5300",
"2. high": "1726.7600",
"3. low": "1664.1300",
"4. close": "1724.8600",
"5. volume": "766313"
},
"2020-04-01": {
"1. open": "1737.2800",
"2. high": "1762.4399",
"3. low": "1685.3700",
"4. close": "1685.4600",
"5. volume": "1243600"
}
}
}
for k,v in stocks['Time Series (Daily)'].items():
print("^KS11\t|{}\t|{}\t|{}\t|{}\t|{}|".format(v['5. volume'],v['3. low'],v['1. open'],v['2. high'],v['4. close']))
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