英文:
Counting unique zip codes with pandas
问题
我正在尝试计算每月时间段内每个邮政编码的数量。
```python
df1 = pd.DataFrame(df)
print(df[['Timestamp', 'zip']].value_counts())
我得到了以下输出：
Timestamp zip
Front Desk Call Log (Master Data) NaN 1
2023-07-21 12:22:47.697000 60191 1
2023-07-21 10:55:13.311000 NaN 1
2023-07-21 10:49:06.148000 60187 1
2023-07-21 10:29:08.396000 60189 1
..
2023-07-21 14:43:23.522000 60187 1
2023-07-21 14:45:12.332000 60440 1
2023-07-21 14:46:46.452000 NaN 1
2023-07-21 17:34:11.631000 60548 1
2023-07-21 17:39:36.358000 60133 1
Length: 8314, dtype: int64
我不需要邮政编码的数量，我需要每个邮政编码的计数。我期望的输出是：
60187 2
60542 1
60540 3
等等。
英文:
I'm trying to count the number of each zip codes within a monthly period of time frame.
df1 = pd.DataFrame(df)
print(df[['Timestamp', 'zip']].nunique())
I get an output of:
Timestamp 8314
zip 343
dtype: int64
I don't need the number of zip codes, I need the count of each zip code.
I was expecting:
60187 2
60542 1
60540 3
etc.
asked to post sample rows:
Timestamp zip
0 Front Desk Call Log (Master Data) NaN
1 2023-07-21 12:22:47.697000 60191
2 2023-07-21 10:55:13.311000 NaN
3 2023-07-21 10:49:06.148000 60187
4 2023-07-21 10:29:08.396000 60189
... ... ...
8309 2023-07-21 14:43:23.522000 60187
8310 2023-07-21 14:45:12.332000 60440
8311 2023-07-21 14:46:46.452000 NaN
8312 2023-07-21 17:34:11.631000 60548
8313 2023-07-21 17:39:36.358000 60133
[8314 rows x 2 columns]
答案1
得分: 1
使用groupby函数与value_counts结合的替代方法来实现相同的结果如下所示：
按“zip”分组并计算每个邮政编码的出现次数
df.groupby('zip').size()
英文:
An alternative solution to achieve the same result is by using the groupby function in combination with value_counts. Here's the code for the alternate solution:
Group by 'zip' and count the occurrences of each zip code
df.groupby('zip').size()
答案2
得分: 0
尝试：
df['zip'].value_counts()
如果您想要计算空值（Nulls）：
df['zip'].value_counts(dropna=False)
英文:
Try:
df['zip'].value_counts()
If you want to count the Nulls as well:
df['zip'].value_counts(dropna=False)
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