我有一个单词列表,如下所示。
wordlist = ['p1','p2','p3','p4','p5','p6','p7']
数据框如下所示。
df = pd.DataFrame({'id' : [1,2,3,4],
'path' : ["p1,p2,p3,p4","p1,p2,p1","p1,p5,p5,p7","p1,p2,p3,p3"]})
输出:
id path
1 p1,p2,p3,p4
2 p1,p2,p1
3 p1,p5,p5,p7
4 p1,p2,p3,p3
我想计算路径数据以获得以下输出。有没有可能得到这样的转变?
id p1 p2 p3 p4 p5 p6 p7
1 1 1 1 1 0 0 0
2 2 1 0 0 0 0 0
3 1 0 0 0 2 0 1
4 1 1 2 0 0 0 0
最佳答案
我认为这会很有效
# create Series with dictionaries
>>> from collections import Counter
>>> c = df["path"].str.split(',').apply(Counter)
>>> c
0 {u'p2': 1, u'p3': 1, u'p1': 1, u'p4': 1}
1 {u'p2': 1, u'p1': 2}
2 {u'p1': 1, u'p7': 1, u'p5': 2}
3 {u'p2': 1, u'p3': 2, u'p1': 1}
# create DataFrame
>>> pd.DataFrame({n: c.apply(lambda x: x.get(n, 0)) for n in wordlist})
p1 p2 p3 p4 p5 p6 p7
0 1 1 1 1 0 0 0
1 2 1 0 0 0 0 0
2 1 0 0 0 2 0 1
3 1 1 2 0 0 0 0
更新
另一种方法:
>>> dfN = df["path"].str.split(',').apply(lambda x: pd.Series(Counter(x)))
>>> pd.DataFrame(dfN, columns=wordlist).fillna(0)
p1 p2 p3 p4 p5 p6 p7
0 1 1 1 1 0 0 0
1 2 1 0 0 0 0 0
2 1 0 0 0 2 0 1
3 1 1 2 0 0 0 0
更新2
一些粗略的性能测试:
>>> dfL = pd.concat([df]*100)
>>> timeit('c = dfL["path"].str.split(",").apply(Counter); d = pd.DataFrame({n: c.apply(lambda x: x.get(n, 0)) for n in wordlist})', 'from __main__ import dfL, wordlist; import pandas as pd; from collections import Counter', number=100)
0.7363274283027295
>>> timeit('splitted = dfL["path"].str.split(","); d = pd.DataFrame({name : splitted.apply(lambda x: x.count(name)) for name in wordlist})', 'from __main__ import dfL, wordlist; import pandas as pd', number=100)
0.5305424618886718
# now let's make wordlist larger
>>> wordlist = wordlist + list(lowercase) + list(uppercase)
>>> timeit('c = dfL["path"].str.split(",").apply(Counter); d = pd.DataFrame({n: c.apply(lambda x: x.get(n, 0)) for n in wordlist})', 'from __main__ import dfL, wordlist; import pandas as pd; from collections import Counter', number=100)
1.765344003293876
>>> timeit('splitted = dfL["path"].str.split(","); d = pd.DataFrame({name : splitted.apply(lambda x: x.count(name)) for name in wordlist})', 'from __main__ import dfL, wordlist; import pandas as pd', number=100)
2.33328927599905
更新3
看完this topic我发现 Counter
真的很慢。您可以使用 defaultdict
对其进行一些优化:
>>> def create_dict(x):
... d = defaultdict(int)
... for c in x:
... d[c] += 1
... return d
>>> c = df["path"].str.split(",").apply(create_dict)
>>> pd.DataFrame({n: c.apply(lambda x: x[n]) for n in wordlist})
p1 p2 p3 p4 p5 p6 p7
0 1 1 1 1 0 0 0
1 2 1 0 0 0 0 0
2 1 0 0 0 2 0 1
3 1 1 2 0 0 0 0
和一些测试:
>>> timeit('c = dfL["path"].str.split(",").apply(create_dict); d = pd.DataFrame({n: c.apply(lambda x: x[n]) for n in wordlist})', 'from __main__ import dfL, wordlist, create_dict; import pandas as pd; from collections import defaultdict', number=100)
0.45942801555111146
# now let's make wordlist larger
>>> wordlist = wordlist + list(lowercase) + list(uppercase)
>>> timeit('c = dfL["path"].str.split(",").apply(create_dict); d = pd.DataFrame({n: c.apply(lambda x: x[n]) for n in wordlist})', 'from __main__ import dfL, wordlist, create_dict; import pandas as pd; from collections import defaultdict', number=100)
1.5798653213942089
关于python - Pandas 数据框计数行值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20369978/