pandas 如何从熊猫数据框中获取单个值作为字符串
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How to get a single value as a string from pandas data frame
提问by Oamar Kanji
I am querying a single value from my data frame which seems to be 'dtype: object'. I simply want to print the value as it is with out printing the index or other information as well. How do I do this?
我正在从我的数据框中查询一个似乎是“dtype: object”的值。我只是想按原样打印值，而不打印索引或其他信息。我该怎么做呢？
col_names = ['Host', 'Port']
df = pd.DataFrame(columns=col_names)
df.loc[len(df)] = ['a', 'b']
t = df[df['Host'] == 'a']['Port']
print(t)
OUTPUT:
输出：
EXPECTED OUTPUT: b
预期产出：b
回答by cs95
If you can guarantee only one result is returned, use locand call item:
如果您可以保证只返回一个结果，请使用loc并调用item：
>>> df.loc[df['Host'] == 'a', 'Port'].item()
'b'
Or, similarly,
或者，类似地，
>>> df.loc[df['Host'] == 'a', 'Port'].values[0]
'b'
...to get the firstvalue (similarly, .values[1]for the second). Which is better than df.loc[df['Host'] == 'a', 'Port'][0]because, if your DataFrame looks like this,
...获得第一个值（类似地，.values[1]第二个）。这比df.loc[df['Host'] == 'a', 'Port'][0]因为如果你的 DataFrame 看起来像这样更好，
Host Port
1 a b
Then "KeyError: 0" will be thrown—
然后会抛出“KeyError: 0”——
df.loc[df['Host'] == 'a', 'Port'][0]
---------------------------------------------------------------------------
KeyError Traceback (most recent call last)
Alternatively, use at:
或者，使用at：
>>> df.at[df['Host'].eq('a').idxmax(), 'Port']
'b'
The drawback is that if 'a' doesn't exist, idxmaxwill return the first index (and return an incorrect result).
缺点是如果 'a' 不存在，idxmax将返回第一个索引（并返回不正确的结果）。
回答by Karn Kumar
it should work simply..
它应该简单地工作..
>>> df
Host Port
0 a b
>>> df[df['Host'] == 'a']['Port'][0] # will choose the first index simply which is 'b'
'b'
OR, use with print which will strip off the surrounded single ticks.
或者，与 print 一起使用，这将去除被包围的单个刻度。
>>> print(df[df['Host'] == 'a']['Port'][0])
b
This will easier because you have just choose the desired Index even if you have Multiple values across Portcolumns
这会更容易，因为即使跨Port列有多个值，您也只需选择所需的索引
Example:
例子：
>>> df
Host Port
0 a b
1 c c
Looking for distinct a& cbased on Index:
寻找不同的a&c基于索引：
>>> df[df['Host'] == 'a']['Port'][0]
'b'
>>> df[df['Host'] == 'c']['Port'][1]
'c'
回答by PL200
As mentioned in my comment, using [1] should work afterwards, to pull the variable you're looking for.
正如我在评论中提到的，使用 [1] 应该可以在之后工作，以提取您正在寻找的变量。
t = df[df['Host'] == 'a']['Port'][1]
