Python 如何计算DataFrame中字符串中的单词数？
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How to calculate number of words in a string in DataFrame?
提问by Sergei
Suppose we have simple Dataframe
假设我们有简单的 Dataframe
df = pd.DataFrame(['one apple','banana','box of oranges','pile of fruits outside', 'one banana', 'fruits'])
df.columns = ['fruits']
how to calculate number of words in keywords, similar to:
如何计算关键字中的字数，类似于：
1 word: 2
2 words: 2
3 words: 1
4 words: 1
回答by EdChum
IIUC then you can do the following:
IIUC 然后您可以执行以下操作：
In [89]:
count = df['fruits'].str.split().apply(len).value_counts()
count.index = count.index.astype(str) + ' words:'
count.sort_index(inplace=True)
count
Out[89]:
1 words: 2
2 words: 2
3 words: 1
4 words: 1
Name: fruits, dtype: int64
Here we use the vectorised str.splitto split on spaces, and then applylento get the count of the number of elements, we can then call value_countsto aggregate the frequency count.
这里我们使用向量化str.split在空间上进行分割，然后获取元素数量的计数，然后我们可以调用聚合频率计数。applylenvalue_counts
We then rename the index and sort it to get the desired output
然后我们重命名索引并对其进行排序以获得所需的输出
UPDATE
更新
This can also be done using str.lenrather than applywhich should scale better:
这也可以使用str.len而不是apply哪个应该更好地扩展：
In [41]:
count = df['fruits'].str.split().str.len()
count.index = count.index.astype(str) + ' words:'
count.sort_index(inplace=True)
count
Out[41]:
0 words: 2
1 words: 1
2 words: 3
3 words: 4
4 words: 2
5 words: 1
Name: fruits, dtype: int64
Timings
时间安排
In [42]:
%timeit df['fruits'].str.split().apply(len).value_counts()
%timeit df['fruits'].str.split().str.len()
1000 loops, best of 3: 799 μs per loop
1000 loops, best of 3: 347 μs per loop
For a 6K df:
对于 6K df：
In [51]:
%timeit df['fruits'].str.split().apply(len).value_counts()
%timeit df['fruits'].str.split().str.len()
100 loops, best of 3: 6.3 ms per loop
100 loops, best of 3: 6 ms per loop
回答by Zero
You could use str.countwith space ' 'as delimiter.
您可以使用str.count空格' '作为分隔符。
In [1716]: count = df['fruits'].str.count(' ').add(1).value_counts(sort=False)
In [1717]: count.index = count.index.astype('str') + ' words:'
In [1718]: count
Out[1718]:
1 words: 2
2 words: 2
3 words: 1
4 words: 1
Name: fruits, dtype: int64
Timings
时间安排
str.countis marginally faster
str.count稍微快一点
Small
小的
In [1724]: df.shape
Out[1724]: (6, 1)
In [1725]: %timeit df['fruits'].str.count(' ').add(1).value_counts(sort=False)
1000 loops, best of 3: 649 μs per loop
In [1726]: %timeit df['fruits'].str.split().apply(len).value_counts()
1000 loops, best of 3: 840 μs per loop
Medium
中等的
In [1728]: df.shape
Out[1728]: (6000, 1)
In [1729]: %timeit df['fruits'].str.count(' ').add(1).value_counts(sort=False)
100 loops, best of 3: 6.58 ms per loop
In [1730]: %timeit df['fruits'].str.split().apply(len).value_counts()
100 loops, best of 3: 6.99 ms per loop
Large
大的
In [1732]: df.shape
Out[1732]: (60000, 1)
In [1733]: %timeit df['fruits'].str.count(' ').add(1).value_counts(sort=False)
1 loop, best of 3: 57.6 ms per loop
In [1734]: %timeit df['fruits'].str.split().apply(len).value_counts()
1 loop, best of 3: 73.8 ms per loop
