Pandas：从每一行获取字符串的第二个字符
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Pandas: get second character of the string, from every row
提问by jjjayn
I've a array of data in Pandas and I'm trying to print second character of every string in col1. I can't figure out how to do it. I can easily print the second character of the each string individually, for example:
我在 Pandas 中有一组数据，我正在尝试打印 col1 中每个字符串的第二个字符。我不知道该怎么做。我可以轻松地单独打印每个字符串的第二个字符，例如：
array.col1[0][1]
However I'd like to print the second character from every row, so there would be a "list" of second characters.
但是我想打印每一行的第二个字符，所以会有一个第二个字符的“列表”。
I've tried
我试过了
array.col1[0:][1]
but that just returns the second line as a whole of col1.
但这只是将第二行作为整个 col1 返回。
Any advice?
有什么建议吗？
回答by Alex Riley
You can use strto access the string methods for the column/Series and then slice the strings as normal:
您可以使用str访问列/系列的字符串方法，然后像往常一样对字符串进行切片：
>>> df = pd.DataFrame(['foo', 'bar', 'baz'], columns=['col1'])
>>> df
col1
0 foo
1 bar
2 baz
>>> df.col1.str[1]
0 o
1 a
2 a
This strattribute also gives you access variety of very useful vectorised string methods, many of which are instantly recognisable from Python's own assortment of built-in string methods (split, replace, etc.).
该str属性也为您提供了访问各种非常有用的向量化字符串方法，其中有许多是瞬间从Python的内置字符串方法（分类识别split，replace等等）。
回答by jpp
As of Pandas 0.23.0, if your data is clean, you will find Pandas "vectorised" string methods via pd.Series.strwill generally underperformsimple iteration via a list comprehension or use of map.
由于大Pandas0.23.0，如果你的数据是干净的，你会发现大Pandas通过“矢量化”字符串方法pd.Series.str一般会弱于大盘通过列表理解或使用简单的迭代map。
For example:
例如：
from operator import itemgetter
df = pd.DataFrame(['foo', 'bar', 'baz'], columns=['col1'])
df = pd.concat([df]*100000, ignore_index=True)
%timeit pd.Series([i[1] for i in df['col1']]) # 33.7 ms
%timeit pd.Series(list(map(itemgetter(1), df['col1']))) # 42.2 ms
%timeit df['col1'].str[1] # 214 ms
A special case is when you have a large number of repeated strings, in which case you can benefit from converting your series to a categorical:
一种特殊情况是当您有大量重复的字符串时，在这种情况下，您可以从将系列转换为categorical 中受益：
df['col1'] = df['col1'].astype('category')
%timeit df['col1'].str[1] # 4.9 ms
