我正在使用熊猫来分析一些选举结果。我有一个DF，结果，每个选区有一行，代表各方投票的栏目（超过100个）：
In[60]: Results.columns
Out[60]:
Index(['Constituency', 'Region', 'Country', 'ID', 'Type', 'Electorate',
'Total', 'Unnamed: 9', '30-50', 'Above',
...
'WP', 'WRP', 'WVPTFP', 'Yorks', 'Young', 'Zeb', 'Party', 'Votes',
'Share', 'Turnout'],
dtype='object', length=147)
因此...
In[63]: Results.head()
Out[63]:
Constituency Region Country ID Type \
PAID
1 Aberavon Wales Wales W07000049 County
2 Aberconwy Wales Wales W07000058 County
3 Aberdeen North Scotland Scotland S14000001 Burgh
4 Aberdeen South Scotland Scotland S14000002 Burgh
5 Aberdeenshire West & Kincardine Scotland Scotland S14000058 County
Electorate Total Unnamed: 9 30-50 Above ... WP WRP WVPTFP \
PAID ...
1 49821 31523 NaN NaN NaN ... NaN NaN NaN
2 45525 30148 NaN NaN NaN ... NaN NaN NaN
3 67745 43936 NaN NaN NaN ... NaN NaN NaN
4 68056 48551 NaN NaN NaN ... NaN NaN NaN
5 73445 55196 NaN NaN NaN ... NaN NaN NaN
Yorks Young Zeb Party Votes Share Turnout
PAID
1 NaN NaN NaN Lab 15416 0.489040 0.632725
2 NaN NaN NaN Con 12513 0.415052 0.662230
3 NaN NaN NaN SNP 24793 0.564298 0.648550
4 NaN NaN NaN SNP 20221 0.416490 0.713398
5 NaN NaN NaN SNP 22949 0.415773 0.751528
[5 rows x 147 columns]
每一方的每个选区成绩都在Results.ix[:, 'Unnamed: 9': 'Zeb']
我可以找到胜利方（即投票得票最高的一方）和使用的投票数：
RawResults = Results.ix[:, 'Unnamed: 9': 'Zeb']
Results['Party'] = RawResults.idxmax(axis=1)
Results['Votes'] = RawResults.max(axis=1).astype(int)
但是，我还需要知道第二方获得多少票（理想情况下它的索引/名称）。那么在pandas中有没有办法在每行的一组列中返回第二个最高值/索引？
答案 0 :(得分：19)
nlargest(int) 要获得列的最高值，可以使用nlargest()，即：
print( df['High'].nlargest(2) )
以上内容将打印列High
您也可以使用nsmallest()以相同的方式获取最低值。
答案 1 :(得分：11)
以下是NumPy解决方案：
In [120]: df
Out[120]:
a b c d e f g h
0 1.334444 0.322029 0.302296 -0.841236 -0.360488 -0.860188 -0.157942 1.522082
1 2.056572 0.991643 0.160067 -0.066473 0.235132 0.533202 1.282371 -2.050731
2 0.955586 -0.966734 0.055210 -0.993924 -0.553841 0.173793 -0.534548 -1.796006
3 1.201001 1.067291 -0.562357 -0.794284 -0.554820 -0.011836 0.519928 0.514669
4 -0.243972 -0.048144 0.498007 0.862016 1.284717 -0.886455 -0.757603 0.541992
5 0.739435 -0.767399 1.574173 1.197063 -1.147961 -0.903858 0.011073 -1.404868
6 -1.258282 -0.049719 0.400063 0.611456 0.443289 -1.110945 1.352029 0.215460
7 0.029121 -0.771431 -0.285119 -0.018216 0.408425 -1.458476 -1.363583 0.155134
8 1.427226 -1.005345 0.208665 -0.674917 0.287929 -1.259707 0.220420 -1.087245
9 0.452589 0.214592 -1.875423 0.487496 2.411265 0.062324 -0.327891 0.256577
In [121]: np.sort(df.values)[:,-2:]
Out[121]:
array([[ 1.33444404, 1.52208164],
[ 1.28237078, 2.05657214],
[ 0.17379254, 0.95558613],
[ 1.06729107, 1.20100071],
[ 0.86201603, 1.28471676],
[ 1.19706331, 1.57417327],
[ 0.61145573, 1.35202868],
[ 0.15513379, 0.40842477],
[ 0.28792928, 1.42722604],
[ 0.48749578, 2.41126532]])
或作为熊猫数据框：
In [122]: pd.DataFrame(np.sort(df.values)[:,-2:], columns=['2nd-largest','largest'])
Out[122]:
2nd-largest largest
0 1.334444 1.522082
1 1.282371 2.056572
2 0.173793 0.955586
3 1.067291 1.201001
4 0.862016 1.284717
5 1.197063 1.574173
6 0.611456 1.352029
7 0.155134 0.408425
8 0.287929 1.427226
9 0.487496 2.411265
或faster solution from @Divakar：
In [6]: df
Out[6]:
a b c d e f g h
0 0.649517 -0.223116 0.264734 -1.121666 0.151591 -1.335756 -0.155459 -2.500680
1 0.172981 1.233523 0.220378 1.188080 -0.289469 -0.039150 1.476852 0.736908
2 -1.904024 0.109314 0.045741 -0.341214 -0.332267 -1.363889 0.177705 -0.892018
3 -2.606532 -0.483314 0.054624 0.979734 0.205173 0.350247 -1.088776 1.501327
4 1.627655 -1.261631 0.589899 -0.660119 0.742390 -1.088103 0.228557 0.714746
5 0.423972 -0.506975 -0.783718 -2.044002 -0.692734 0.980399 1.007460 0.161516
6 -0.777123 -0.838311 -1.116104 -0.433797 0.599724 -0.884832 -0.086431 -0.738298
7 1.131621 1.218199 0.645709 0.066216 -0.265023 0.606963 -0.194694 0.463576
8 0.421164 0.626731 -0.547738 0.989820 -1.383061 -0.060413 -1.342769 -0.777907
9 -1.152690 0.696714 -0.155727 -0.991975 -0.806530 1.454522 0.788688 0.409516
In [7]: a = df.values
In [8]: a[np.arange(len(df))[:,None],np.argpartition(-a,np.arange(2),axis=1)[:,:2]]
Out[8]:
array([[ 0.64951665, 0.26473378],
[ 1.47685226, 1.23352348],
[ 0.17770473, 0.10931398],
[ 1.50132666, 0.97973383],
[ 1.62765464, 0.74238959],
[ 1.00745981, 0.98039898],
[ 0.5997243 , -0.0864306 ],
[ 1.21819904, 1.13162068],
[ 0.98982033, 0.62673128],
[ 1.45452173, 0.78868785]])
答案 2 :(得分：2)
您可以对结果进行排序，以便第一行包含最大值。然后你可以简单地使用索引来获得前n个位置。
RawResults = Results.ix[:, 'Unnamed: 9': 'Zeb'].sort_values(by='votes', ascending=False)
RawResults.iloc[0, :] # First place
RawResults.iloc[1, :] # Second place
RawResults.iloc[n, :] # nth place
答案 3 :(得分：0)
这是使用最大功能的解决方案：
>>> df
a b c
0 4 20 2
1 5 10 2
2 3 40 5
3 1 50 10
4 2 30 15
>>> def give_largest(col,n):
... largest = col.nlargest(n).reset_index(drop = True)
... data = [x for x in largest]
... index = [f'{i}_largest' for i in range(1,len(largest)+1)]
... return pd.Series(data,index=index)
...
...
>>> def n_largest(df, axis, n):
... '''
... Function to return the n-largest value of each
... column/row of the input DataFrame.
... '''
... return df.apply(give_largest, axis = axis, n = n)
...
>>> n_largest(df,axis = 1, n = 2)
1_largest 2_largest
0 20 4
1 10 5
2 40 5
3 50 10
4 30 15
>>> n_largest(df,axis = 0, n = 2)
a b c
1_largest 5 50 15
2_largest 4 40 10
答案 4 :(得分：0)
这是一种有趣的方法。如果我们用最小值替换最大值并计算，该怎么办。虽然这是一个快速的技巧，但不建议使用！
first_highest_value_index = df.idxmax()
second_highest_value_index = df.replace(df.max(),df(min)).idxmax()
first_highest_value = df[first_highest_value_index]
second_highest_value = df[second_highest_value_index]