最初,我有一个带日期字段的空数据框,后来我试图将其与新的数据框合并到一个for循环中。
com_df = pd.DataFrame(columns=['date'])
for i in data_dict.values():
response = requests.get('www.example.com/' + i + '?format=json')
data = json.loads(response.content.decode('utf-8'))
df = dataframe_format(data[1]) // convert list of dict to dataframe
com_df = pd.merge(com_df, df, on='date', how='outer')
所以现在的输出就像
date value_x value_y value_x value_y value
0 2017 1.722333e+13 8.711267e+12 3485.0 197.713256 46.030025
1 2016 1.829506e+13 7.320738e+12 3052.0 249.907289 -2.024998
2 2015 3.932602e+13 8.188019e+12 2827.0 480.287296 -6.007182
但是我希望列名成为下面字典的键,
data_dict = {'A': '1','B': '2','C': '3','D': '4','E': '5'}
即
date A B C D E
0 2017 1.722333e+13 8.711267e+12 3485.0 197.713256 46.030025
1 2016 1.829506e+13 7.320738e+12 3052.0 249.907289 -2.024998
2 2015 3.932602e+13 8.188019e+12 2827.0 480.287296 -6.007182
答案 0 :(得分:0)
如果您打算应用按其值排序的字典键,则可以执行以下操作:
@Autowired
private RedisClient redisClient;
private ObjectMapper objectMapper = new ObjectMapper();
public void saveAllVariants(Set<Variant> allVariantsSet) {
try {
final Pipeline pipeline = redisClient.getPipeline(4); // use redis DB4
for (Variant variant : allVariantsSet) {
String idKey = "variants:id:" + variant.getId();
String variantStr = objectMapper.writeValueAsString(variant); // convert to JSON object
pipeline.set(idKey.getBytes(), variantStr.getBytes());
}
pipeline.sync();
} catch (Exception e) {
log.error("Problem in saving variants in cache. Error: ", e.getMessage());
}
}
答案 1 :(得分:0)
我会将您的输入字典转换为一个到列名称的映射索引:
data_dict = {'A': '1','B': '2','C': '3','D': '4','E': '5'}
pos_col_dict = {int(v): k for k, v in data_dict.items()}
然后通过NumPy分配给列。您应该使用副本以避免产生副作用:
arr = df.columns.values
arr[list(pos_col_dict)] = list(pos_col_dict.values())
df.columns = arr