Python 如何将索引转换为列表？
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how to convert Index into list?
提问by Sai Rajesh
My index:
我的指数：
Index([u'Newal', u'Saraswati Khera', u'Tohana'], dtype='object')
I have to convert this format into list with following format:
我必须将此格式转换为具有以下格式的列表：
['Newal','SaraswatiKhera','Tohana']
回答by jezrael
You can use tolistor list:
您可以使用tolist或list：
print df.index.tolist()
print list(df.index)
But the fastest solution is convert np.arryby valuestolist(thanks EdChum)
但是，最快的解决方法是转换np.arry的 （感谢EdChum）valuestolist
print df.index.values.tolist()
Sample:
样本：
import pandas as pd
idx = pd.Index([u'Newal', u'Saraswati Khera', u'Tohana'])
print idx
Index([u'Newal', u'Saraswati Khera', u'Tohana'], dtype='object')
print idx.tolist()
[u'Newal', u'Saraswati Khera', u'Tohana']
print list(idx)
[u'Newal', u'Saraswati Khera', u'Tohana']
If you need encode UTF-8:
如果您需要编码UTF-8：
print [x.encode('UTF8') for x in idx.tolist()]
['Newal', 'Saraswati Khera', 'Tohana']
Another solution:
另一种解决方案：
print [str(x) for x in idx.tolist()]
['Newal', 'Saraswati Khera', 'Tohana']
but it would fail if the unicode string characters do not lie in the ascii range.
但如果 unicode 字符串字符不在 ascii 范围内，它将失败。
Timings:
时间：
import pandas as pd
import numpy as np
#random dataframe
np.random.seed(1)
df = pd.DataFrame(np.random.randint(10, size=(3,3)))
df.columns = list('ABC')
df.index = [u'Newal', u'Saraswati Khera', u'Tohana']
print df
print df.index
Index([u'Newal', u'Saraswati Khera', u'Tohana'], dtype='object')
print df.index.tolist()
[u'Newal', u'Saraswati Khera', u'Tohana']
print list(df.index)
[u'Newal', u'Saraswati Khera', u'Tohana']
print df.index.values.tolist()
[u'Newal', u'Saraswati Khera', u'Tohana']
In [90]: %timeit list(df.index)
The slowest run took 37.42 times longer than the fastest. This could mean that an intermediate result is being cached
100000 loops, best of 3: 2.18 μs per loop
In [91]: %timeit df.index.tolist()
The slowest run took 22.33 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 1.75 μs per loop
In [92]: %timeit df.index.values.tolist()
The slowest run took 62.72 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 787 ns per loop
