Python 通过JSON解析

我有以下JSON转储。我喜欢提取术语（如“工作”）和计数（如“46”）

我喜欢以表格形式打印术语和计数

我有一个功能可以做到这一点：

print_two_cols(term1,term2)

---

但是，我不确定如何迭代上述JSON。我正在使用python 2.7

在

行上循环
键；每个条目都是一个带有
f
键的字典，包含带有
v
键的字典：

for row in result['rows']:
    job, count = (col['v'] for col in row['f'])
    print_two_cols(job, count)

演示：

这不是json。这与json没有任何关系……这是我从BigQueryIt得到的。它不是json字符串，但它看起来像是用json模块解析json时得到的。它是一个Python结构，可能是从json解析的。json只是文本（遵循上的规则）。如果您有一个Python对象图，那么这就是您所拥有的，而不是JSON——在这种情况下，请确保从问题中删除所有不必要的“JSON”讨论。接下来的任务不是“解析JSON”，而是根据一些规则“迭代”对象。工作得很有魅力。我是python新手，不知道我能做到这一点。
for row in result['rows']:
    job, count = (col['v'] for col in row['f'])
    print_two_cols(job, count)

>>> result = {u'kind': u'bigquery#queryResponse', u'rows': [{u'f': [{u'v': u'jobs'}, {u'v': u'46'}]},      {u'f': [{u'v': u'employment'}, {u'v': u'24'}]}, {u'f': [{u'v': u'Employment'}, {u'v': u'20'}]}, {u'f': [{u'v': u'donate furniture'}, {u'v': u'16'}]}, {u'f': [{u'v': u'feeding the homeless'}, {u'v': u'16'}]}, {u'f': [{u'v': u'Perth stores'}, {u'v': u'14'}]}, {u'f': [{u'v': u'bed'}, {u'v': u'14'}]}, {u'f': [{u'v': u'Jobs'}, {u'v': u'14'}]}, {u'f': [{u'v': u'food parcels'}, {u'v': u'14'}]}, {u'f': [{u'v': u'history'}, {u'v': u'14'}]}, {u'f': [{u'v': u'Adherent'}, {u'v': u'12'}]}, {u'f': [{u'v': u'volunteer'}, {u'v': u'12'}]}, {u'f': [{u'v': u'Historical society'}, {u'v': u'12'}]}, {u'f': [{u'v': u'Donate furniture'}, {u'v': u'12'}]}, {u'f': [{u'v': u'pick up'}, {u'v': u'12'}]}, {u'f': [{u'v': u'Careers'}, {u'v': u'12'}]}, {u'f': [{u'v': u'contact'}, {u'v': u'12'}]}, {u'f': [{u'v': u'helping with a funeral'}, {u'v': u'10'}]}, {u'f': [{u'v': u'Basketball'}, {u'v': u'10'}]}, {u'f': [{u'v': u'email'}, {u'v': u'10'}]}], u'jobReference': {u'projectId': u'2323', u'jobId': u'2323'}, u'jobComplete': True, u'totalRows': u'20', u'totalBytesProcessed': u'0', u'cacheHit': True, u'schema': {u'fields': [{u'type': u'STRING', u'name': u'Query', u'mode': u'NULLABLE'}, {u'type': u'INTEGER', u'name': u'Count', u'mode': u'NULLABLE'}]}}
>>> for row in result['rows']:
...     job, count = (col['v'] for col in row['f'])
...     print job, count
... 
jobs 46
employment 24
Employment 20
donate furniture 16
feeding the homeless 16
Perth stores 14
bed 14
Jobs 14
food parcels 14
history 14
Adherent 12
volunteer 12
Historical society 12
Donate furniture 12
pick up 12
Careers 12
contact 12
helping with a funeral 10
Basketball 10
email 10