I have this dataframe:
content
id
17 B
17 A
6 A
15 A
...
I want to count how many rows have the index 17 (in this case that would be 2). Is there a way to do that?
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3 Answers
You can try:
sum(df.index == 17)
df.index == 17 returns an array with boolean with True when index value matches else False. And while
using sum function True is equivalent to 1.
3 Comments
aabujamra
I'm getting results like [1 1 1 ..., 0 0 0]. I just want to know how many times 17 appears in index
niraj
@abutremutante do you mean while using
df.index == 17, you got [1 1 1 ..., 0 0 0]? Doesn't it work for the sample example provided?aabujamra
you're right, it does. since my df is huge I tried to simplify here but apparently wasn't the best approach. anyway, I solved in a not very pythonic way: a=0, for i in df.index: if i == 17: a+=1
You can groupby level
df.groupby(level=0).count()
Or reset_index()
df.reset_index().groupby('id').count()
Problem: How to count the quantity of index label?
Input: # Your DataFrame
test_dict = {'id': ['17', '17', '6', '15'], 'content': ['B', 'A', 'A', 'A']}
testd_df = pd.DataFrame.from_dict(test_dict) # create DataFrame from dict
testd_df.set_index('id', inplace=True) # set 'id' as index in inplace way
testd_df
Output:
|content
--------------
id |
-------------
17 | B
17 | A
6 | A
15 | A
Solution: Use api pandas.Index.value_counts
Based on the document, pandas.Index.value_counts will return object containing counts of unique values and return a pd.Series.
so now, I can select the specific index I want by using pandas.Series.loc (not get confused with .iloc)
# Solution
Input: index_count = pd.Index(testd_df.index).value_counts() # count value of unique value
index_count
Output: 17 2
15 1
6 1
dtype: int64
---------------------------------
Input: index_count.loc['17'] # select the information you care about
Output: 2
1 Comment
sanzoghenzo
There's no need to create a new, index, just use
testd_df.index.value_counts()