我希望在Python中使用Pandas数据框实现R中match()函数的相当功能，而不使用for循环。
在R中，match()返回第一个参数在第二个参数中（第一次）匹配位置的向量。
假设我有两个数据框A和B，它们都包含列C。其中
A$C = c('a','b')
B$C = c('c','c','b','b','c','b','a','a')
match(A$C,B$C) = c(7,3)
在Python中，有没有一种与pandas数据框的列相当的方法，而不需要循环遍历值。
这里有一个一行代码：
B.reset_index().groupby('C')['index'].first()[A.C].values
这个解决方案以与输入A相同的顺序返回结果，就像R中的match一样。
完整示例：
import pandas as pd
A = pd.DataFrame({'C':['a','b']})
B = pd.DataFrame({'C':['c','c','b','b','c','b','a','a']})
B.reset_index().groupby('C')['index'].first()[A.C].values
Output array([6, 2])
编辑（2023-04-12）：在较新版本的pandas中，.loc匹配所有符合条件的行。因此，之前的解决方案（B.reset_index().set_index('c').loc[A.c, 'index'].values）将返回所有匹配项，而不仅仅是第一个。
drop_duplicates，然后使用布尔索引、isin或合并。Python从0开始计数，所以输出结果时需要加上1。A = pd.DataFrame({'c':['a','b']})
B = pd.DataFrame({'c':['c','c','b','b','c','b','a','a']})
B = B.drop_duplicates('c')
print (B)
c
0 c
2 b
6 a
print (B[B.c.isin(A.c)])
c
2 b
6 a
print (B[B.c.isin(A.c)].index)
Int64Index([2, 6], dtype='int64')
print (pd.merge(B.reset_index(), A))
index c
0 2 b
1 6 a
print (pd.merge(B.reset_index(), A)['index'])
0 2
1 6
Name: index, dtype: int64
R 中，match 的输出顺序对应于输入：('a', 'b') 对应于 c(7,3)，但是这里的答案首先返回了 b 的结果，然后是 a。请参阅我的答案，其中提供了一种保持顺序的单行选项。 - toto_ticoimport pandas as pd
df1 = pd.DataFrame({'C': ['a','b']})
print df1
C
0 a
1 b
df2 = pd.DataFrame({'C': ['c','c','b','b','c','b','a','a']})
print df2
C
0 c
1 c
2 b
3 b
4 c
5 b
6 a
7 a
match = df2['C'].isin(df1['C'])
print [i for i in range(match.shape[0]) if match[i]]
#[2, 3, 5, 6, 7]
这里有多种实现方法。在不使用 C 或 C++ 的情况下，最快的方法似乎是使用 datatable：
def match(x, y, method = "dt"):
'''
x and y are two numpy 1d arrays containing only finite values.
method = 'dt': use datatable
method = 'pandas': use pandas
method = 'numpy': use numpy
method = 'dict': use hashing.
'''
if method == 'dt': # Use datatable
xdf = datatable.Frame({'val': x})
ydf = datatable.Frame({'val': y, 'ind': np.arange(y.shape[0]) })[
:, datatable.min(datatable.f.ind), datatable.by(datatable.f.val)]
ydf.key = 'val'
rst = xdf[:, :, datatable.join(ydf)]['ind'].to_numpy()
return rst.filled(-1 - y.shape[0]).ravel()
if method == 'pandas': # Use pandas dataframe.
xdf = pd.DataFrame({'val': x})
ydf = pd.DataFrame({'val': y, 'ind': np.arange(y.shape[0]) }).groupby(
['val']).min()
joined = xdf.join(ydf, on = 'val', lsuffix = '_x', rsuffix = '_y')
rst = joined['ind'].to_numpy()
rst[np.isnan(rst)] = -1 - y.shape[0]
return rst.astype(int)
rst = np.zeros(x.shape[0], dtype = np.int32) - (y.shape[0] + 1)
if method == 'numpy':
yorder = y.argsort()
ysorted = y[yorder]
ind = np.searchsorted(ysorted, x)
outofBound = ind >= y.shape[0]
ind[outofBound] = 0
eq = ysorted[ind] == x
eq[outofBound] = False
rst[eq] = yorder[ind][eq]
else: # Hashing.
D = dict(zip(y[::-1], np.arange(y.shape[0] - 1, -1, -1)))
for i, u in enumerate(x):
val = D.get(u)
if val is not None: rst[i] = val
return rst
测试代码：
import datatable
import pandas
import time
import numpy as np
N = int(1e9)
k = int(1e7)
x = np.random.choice(N, k)
y = np.random.choice(N, k)
timeCosts = {}
st = time.time()
ind = match(x, y, "dt")
timeCosts['datatable'] = time.time() - st
np.all(x[ind >= 0] == y[ind[ind >= 0]])
st = time.time()
ind = match(x, y, "pandas")
timeCosts['pandas'] = time.time() - st
np.all(x[ind >= 0] == y[ind[ind >= 0]])
st = time.time()
ind = match(x, y, "numpy")
timeCosts['numpy'] = time.time() - st
np.all(x[ind >= 0] == y[ind[ind >= 0]])
st = time.time()
ind = match(x, y, "hashing")
timeCosts['hashing'] = time.time() - st
np.all(x[ind >= 0] == y[ind[ind >= 0]])
时间花费（秒）：
{'datatable': 1.55, 'pandas': 8.01, 'numpy': 14.91, 'hashing': 6.04}
但最快的仍然比R的match慢：1.05s
R一定使用了类似于基数排序中的哈希技术..
import numpy as np
import pandas as pd
def match(needles,haystack): # parameters are python lists or 1 dimensional np arrays.
# Copyleft 2023 David A. Kra Creative Commons License: BY-SA Attribution-ShareAlike
# Derived from method3 in https://www.statology.org/numpy-find-index-of-value/
# which is Copyright 2021 Zach Bobbitt# match(needles,haystack) returns indices to use
#find index location of first occurrence of each value of interest
# returns an np 1 dimensional array of indices. notfound values get None
# tresult=np.array(needles.size*[2*1024*1024*1024], dtype='i4', copy=True,) # dtype 'intp' is for indexing, as an alternative to 'i4' or 'int32'
sorter = np.argsort( nphaystack:=np.array(haystack) )
tresult=sorter[np.searchsorted(nphaystack, (npneedles:=np.array(needles)), sorter=sorter)]
# if a needle value is greater than any value in in the haystack, will throw an IndexError, such as
# IndexError: index 6 is out of bounds for axis 0 with size 6
# If a needle is missing from the haystack, the index returned is of the next higher needle that is in the haystack.
# Instead, turn it into None.
return [tresult[i] if nphaystack[ tresult[i] ] == npneedles[i] else None for i in range(tresult.size) ]
# usage:
npHaystackID=pdHaystack['ID_BUC'].to_numpy()
npNeedlesID=pdNeedles['ID_BUC'].to_numpy()
rowsOfNeedlesInBUC=match(npNeedlesID,npHaystackID)