Fatal编程技术网
  • python/
  • Python 不区分大小写的替换（映射）

Python 不区分大小写的替换（映射）

python pandas dictionary dataframe

Python 不区分大小写的替换（映射）,python,pandas,dictionary,dataframe,case-insensitive,Python,Pandas,Dictionary,Dataframe,Case Insensitive,我发布了一个“第1部分”的问题，这让我找到了我所需要的函数的答案，但我认为这有理由提出自己的问题。如果没有，我将删除 我想对数据帧应用一个函数，该函数将全州名替换为缩写（newyork->NY）。然而，我在我的数据集中注意到，如果一个州被资本化，它显然不会与dicitonary相匹配。我试图解决它，但似乎无法破解代码： import pandas as pd import numpy as np dfp = pd.DataFrame({'A' : [np.NaN,np.NaN,3,4,5,5,3

我发布了一个“第1部分”的问题，这让我找到了我所需要的函数的答案，但我认为这有理由提出自己的问题。如果没有，我将删除

我想对数据帧应用一个函数，该函数将全州名替换为缩写（
newyork->NY
）。然而，我在我的数据集中注意到，如果一个州被资本化，它显然不会与dicitonary相匹配。我试图解决它，但似乎无法破解代码：

import pandas as pd
import numpy as np
dfp = pd.DataFrame({'A' : [np.NaN,np.NaN,3,4,5,5,3,1,5,np.NaN], 
                    'B' : [1,0,3,5,0,0,np.NaN,9,0,0], 
                    'C' : ['Pharmacy of IDAHO','NY Pharma','NJ Pharmacy','Idaho Rx','CA Herbals','Florida Pharma','AK RX','Ohio Drugs','PA Rx','USA Pharma'], 
                    'D' : [123456,123456,1234567,12345678,12345,12345,12345678,123456789,1234567,np.NaN],
                    'E' : ['Assign','Unassign','Assign','Ugly','Appreciate','Undo','Assign','Unicycle','Assign','Unicorn',]})

import us
statez = us.states.mapping('abbr', 'name')
inv_map = {v: k for k, v in statez.items()}

def replace_states(company):
    # find all states that exist in the string
    state_found = filter(lambda state: state.lower() in company.lower(), statez.values())

    # replace each state with its abbreviation
    for state in state_found:
        print(state, inv_map[state])
        company = company.replace(state, inv_map[state])
        print("---" , company)

    # return the modified string (or original if no states were found)
    return company

dfp['C'] = dfp['C'].map(replace_states)
产出：注意《爱达荷州药房》中缺少变化

有没有办法使这个函数不区分大小写？

我会找到它的索引，然后用它来替换它，不管大小写：

# replace each state with its abbreviation
    for state in state_found:
        print(state, inv_map[state])
        index = company.lower().find(state.lower())
        company = company.replace(company[index:index + len(state)], inv_map[state])
        print("---" , company)
这将保留字符串所有其他部分的大小写。

我将找到它的索引，然后使用该索引替换它，而不管大小写：

# replace each state with its abbreviation
    for state in state_found:
        print(state, inv_map[state])
        index = company.lower().find(state.lower())
        company = company.replace(company[index:index + len(state)], inv_map[state])
        print("---" , company)
这将保留字符串所有其他部分的大小写。

使用缩写替换状态名称（不区分大小写的矢量化解决方案）：

t1 = dfp.C.str.split(expand=True)
t2 = t1.stack().str.title().map(inv_map).unstack()
t1[t2.notnull()] = t2
dfp['new'] = t1.stack().groupby(level=0).agg(' '.join)

In [88]: dfp['state'] = dfp.C.str.extract(r'\b([A-Z]{2})\b', expand=False)

In [89]: dfp
Out[89]:
     A    B                  C            D           E state
0  NaN  1.0  Pharmacy of IDAHO     123456.0      Assign   NaN
1  NaN  0.0          NY Pharma     123456.0    Unassign    NY
2  3.0  3.0        NJ Pharmacy    1234567.0      Assign    NJ
3  4.0  5.0           Idaho Rx   12345678.0        Ugly   NaN
4  5.0  0.0         CA Herbals      12345.0  Appreciate    CA
5  5.0  0.0     Florida Pharma      12345.0        Undo   NaN
6  3.0  NaN              AK RX   12345678.0      Assign    AK
7  1.0  9.0         Ohio Drugs  123456789.0    Unicycle   NaN
8  5.0  0.0              PA Rx    1234567.0      Assign    PA
9  NaN  0.0         USA Pharma          NaN     Unicorn   NaN

In [90]: dfp.C = dfp.C.replace(dfp.state.tolist(),
                               dfp.state.map(statez).tolist(),
                               regex=True)

In [91]: dfp
Out[91]:
     A    B                    C            D           E state
0  NaN  1.0    Pharmacy of IDAHO     123456.0      Assign   NaN
1  NaN  0.0      New York Pharma     123456.0    Unassign    NY
2  3.0  3.0  New Jersey Pharmacy    1234567.0      Assign    NJ
3  4.0  5.0             Idaho Rx   12345678.0        Ugly   NaN
4  5.0  0.0   California Herbals      12345.0  Appreciate    CA
5  5.0  0.0       Florida Pharma      12345.0        Undo   NaN
6  3.0  NaN            Alaska RX   12345678.0      Assign    AK
7  1.0  9.0           Ohio Drugs  123456789.0    Unicycle   NaN
8  5.0  0.0      Pennsylvania Rx    1234567.0      Assign    PA
9  NaN  0.0           USA Pharma          NaN     Unicorn   NaN
结果:

In [152]: x
Out[152]:
     A    B                  C            D           E             new
0  NaN  1.0  Pharmacy of IDAHO     123456.0      Assign  Pharmacy of ID
1  NaN  0.0          NY Pharma     123456.0    Unassign       NY Pharma
2  3.0  3.0        NJ Pharmacy    1234567.0      Assign     NJ Pharmacy
3  4.0  5.0           Idaho Rx   12345678.0        Ugly           ID Rx
4  5.0  0.0         CA Herbals      12345.0  Appreciate      CA Herbals
5  5.0  0.0     Florida Pharma      12345.0        Undo       FL Pharma
6  3.0  NaN              AK RX   12345678.0      Assign           AK RX
7  1.0  9.0         Ohio Drugs  123456789.0    Unicycle        OH Drugs
8  5.0  0.0              PA Rx    1234567.0      Assign           PA Rx
9  NaN  0.0         USA Pharma          NaN     Unicorn      USA Pharma
说明：

In [155]: t1 = dfp.C.str.split(expand=True)

In [156]: t1
Out[156]:
          0         1      2
0  Pharmacy        of  IDAHO
1        NY    Pharma   None
2        NJ  Pharmacy   None
3     Idaho        Rx   None
4        CA   Herbals   None
5   Florida    Pharma   None
6        AK        RX   None
7      Ohio     Drugs   None
8        PA        Rx   None
9       USA    Pharma   None

In [157]: t2 = t1.stack().str.title().map(inv_map).unstack()

In [158]: t2
Out[158]:
     0    1     2
0  NaN  NaN    ID
1  NaN  NaN  None
2  NaN  NaN  None
3   ID  NaN  None
4  NaN  NaN  None
5   FL  NaN  None
6  NaN  NaN  None
7   OH  NaN  None
8  NaN  NaN  None
9  NaN  NaN  None

In [159]: t1[t2.notnull()] = t2

In [160]: t1
Out[160]:
          0         1     2
0  Pharmacy        of    ID
1        NY    Pharma  None
2        NJ  Pharmacy  None
3        ID        Rx  None
4        CA   Herbals  None
5        FL    Pharma  None
6        AK        RX  None
7        OH     Drugs  None
8        PA        Rx  None
9       USA    Pharma  None
将状态缩写替换为其名称（不区分大小写的矢量化解决方案）：

t1 = dfp.C.str.split(expand=True)
t2 = t1.stack().str.title().map(inv_map).unstack()
t1[t2.notnull()] = t2
dfp['new'] = t1.stack().groupby(level=0).agg(' '.join)

In [88]: dfp['state'] = dfp.C.str.extract(r'\b([A-Z]{2})\b', expand=False)

In [89]: dfp
Out[89]:
     A    B                  C            D           E state
0  NaN  1.0  Pharmacy of IDAHO     123456.0      Assign   NaN
1  NaN  0.0          NY Pharma     123456.0    Unassign    NY
2  3.0  3.0        NJ Pharmacy    1234567.0      Assign    NJ
3  4.0  5.0           Idaho Rx   12345678.0        Ugly   NaN
4  5.0  0.0         CA Herbals      12345.0  Appreciate    CA
5  5.0  0.0     Florida Pharma      12345.0        Undo   NaN
6  3.0  NaN              AK RX   12345678.0      Assign    AK
7  1.0  9.0         Ohio Drugs  123456789.0    Unicycle   NaN
8  5.0  0.0              PA Rx    1234567.0      Assign    PA
9  NaN  0.0         USA Pharma          NaN     Unicorn   NaN

In [90]: dfp.C = dfp.C.replace(dfp.state.tolist(),
                               dfp.state.map(statez).tolist(),
                               regex=True)

In [91]: dfp
Out[91]:
     A    B                    C            D           E state
0  NaN  1.0    Pharmacy of IDAHO     123456.0      Assign   NaN
1  NaN  0.0      New York Pharma     123456.0    Unassign    NY
2  3.0  3.0  New Jersey Pharmacy    1234567.0      Assign    NJ
3  4.0  5.0             Idaho Rx   12345678.0        Ugly   NaN
4  5.0  0.0   California Herbals      12345.0  Appreciate    CA
5  5.0  0.0       Florida Pharma      12345.0        Undo   NaN
6  3.0  NaN            Alaska RX   12345678.0      Assign    AK
7  1.0  9.0           Ohio Drugs  123456789.0    Unicycle   NaN
8  5.0  0.0      Pennsylvania Rx    1234567.0      Assign    PA
9  NaN  0.0           USA Pharma          NaN     Unicorn   NaN
用缩写替换状态名称（不区分大小写的矢量化解决方案）：

t1 = dfp.C.str.split(expand=True)
t2 = t1.stack().str.title().map(inv_map).unstack()
t1[t2.notnull()] = t2
dfp['new'] = t1.stack().groupby(level=0).agg(' '.join)

In [88]: dfp['state'] = dfp.C.str.extract(r'\b([A-Z]{2})\b', expand=False)

In [89]: dfp
Out[89]:
     A    B                  C            D           E state
0  NaN  1.0  Pharmacy of IDAHO     123456.0      Assign   NaN
1  NaN  0.0          NY Pharma     123456.0    Unassign    NY
2  3.0  3.0        NJ Pharmacy    1234567.0      Assign    NJ
3  4.0  5.0           Idaho Rx   12345678.0        Ugly   NaN
4  5.0  0.0         CA Herbals      12345.0  Appreciate    CA
5  5.0  0.0     Florida Pharma      12345.0        Undo   NaN
6  3.0  NaN              AK RX   12345678.0      Assign    AK
7  1.0  9.0         Ohio Drugs  123456789.0    Unicycle   NaN
8  5.0  0.0              PA Rx    1234567.0      Assign    PA
9  NaN  0.0         USA Pharma          NaN     Unicorn   NaN

In [90]: dfp.C = dfp.C.replace(dfp.state.tolist(),
                               dfp.state.map(statez).tolist(),
                               regex=True)

In [91]: dfp
Out[91]:
     A    B                    C            D           E state
0  NaN  1.0    Pharmacy of IDAHO     123456.0      Assign   NaN
1  NaN  0.0      New York Pharma     123456.0    Unassign    NY
2  3.0  3.0  New Jersey Pharmacy    1234567.0      Assign    NJ
3  4.0  5.0             Idaho Rx   12345678.0        Ugly   NaN
4  5.0  0.0   California Herbals      12345.0  Appreciate    CA
5  5.0  0.0       Florida Pharma      12345.0        Undo   NaN
6  3.0  NaN            Alaska RX   12345678.0      Assign    AK
7  1.0  9.0           Ohio Drugs  123456789.0    Unicycle   NaN
8  5.0  0.0      Pennsylvania Rx    1234567.0      Assign    PA
9  NaN  0.0           USA Pharma          NaN     Unicorn   NaN
结果:

In [152]: x
Out[152]:
     A    B                  C            D           E             new
0  NaN  1.0  Pharmacy of IDAHO     123456.0      Assign  Pharmacy of ID
1  NaN  0.0          NY Pharma     123456.0    Unassign       NY Pharma
2  3.0  3.0        NJ Pharmacy    1234567.0      Assign     NJ Pharmacy
3  4.0  5.0           Idaho Rx   12345678.0        Ugly           ID Rx
4  5.0  0.0         CA Herbals      12345.0  Appreciate      CA Herbals
5  5.0  0.0     Florida Pharma      12345.0        Undo       FL Pharma
6  3.0  NaN              AK RX   12345678.0      Assign           AK RX
7  1.0  9.0         Ohio Drugs  123456789.0    Unicycle        OH Drugs
8  5.0  0.0              PA Rx    1234567.0      Assign           PA Rx
9  NaN  0.0         USA Pharma          NaN     Unicorn      USA Pharma
说明：

In [155]: t1 = dfp.C.str.split(expand=True)

In [156]: t1
Out[156]:
          0         1      2
0  Pharmacy        of  IDAHO
1        NY    Pharma   None
2        NJ  Pharmacy   None
3     Idaho        Rx   None
4        CA   Herbals   None
5   Florida    Pharma   None
6        AK        RX   None
7      Ohio     Drugs   None
8        PA        Rx   None
9       USA    Pharma   None

In [157]: t2 = t1.stack().str.title().map(inv_map).unstack()

In [158]: t2
Out[158]:
     0    1     2
0  NaN  NaN    ID
1  NaN  NaN  None
2  NaN  NaN  None
3   ID  NaN  None
4  NaN  NaN  None
5   FL  NaN  None
6  NaN  NaN  None
7   OH  NaN  None
8  NaN  NaN  None
9  NaN  NaN  None

In [159]: t1[t2.notnull()] = t2

In [160]: t1
Out[160]:
          0         1     2
0  Pharmacy        of    ID
1        NY    Pharma  None
2        NJ  Pharmacy  None
3        ID        Rx  None
4        CA   Herbals  None
5        FL    Pharma  None
6        AK        RX  None
7        OH     Drugs  None
8        PA        Rx  None
9       USA    Pharma  None
将状态缩写替换为其名称（不区分大小写的矢量化解决方案）：

t1 = dfp.C.str.split(expand=True)
t2 = t1.stack().str.title().map(inv_map).unstack()
t1[t2.notnull()] = t2
dfp['new'] = t1.stack().groupby(level=0).agg(' '.join)

In [88]: dfp['state'] = dfp.C.str.extract(r'\b([A-Z]{2})\b', expand=False)

In [89]: dfp
Out[89]:
     A    B                  C            D           E state
0  NaN  1.0  Pharmacy of IDAHO     123456.0      Assign   NaN
1  NaN  0.0          NY Pharma     123456.0    Unassign    NY
2  3.0  3.0        NJ Pharmacy    1234567.0      Assign    NJ
3  4.0  5.0           Idaho Rx   12345678.0        Ugly   NaN
4  5.0  0.0         CA Herbals      12345.0  Appreciate    CA
5  5.0  0.0     Florida Pharma      12345.0        Undo   NaN
6  3.0  NaN              AK RX   12345678.0      Assign    AK
7  1.0  9.0         Ohio Drugs  123456789.0    Unicycle   NaN
8  5.0  0.0              PA Rx    1234567.0      Assign    PA
9  NaN  0.0         USA Pharma          NaN     Unicorn   NaN

In [90]: dfp.C = dfp.C.replace(dfp.state.tolist(),
                               dfp.state.map(statez).tolist(),
                               regex=True)

In [91]: dfp
Out[91]:
     A    B                    C            D           E state
0  NaN  1.0    Pharmacy of IDAHO     123456.0      Assign   NaN
1  NaN  0.0      New York Pharma     123456.0    Unassign    NY
2  3.0  3.0  New Jersey Pharmacy    1234567.0      Assign    NJ
3  4.0  5.0             Idaho Rx   12345678.0        Ugly   NaN
4  5.0  0.0   California Herbals      12345.0  Appreciate    CA
5  5.0  0.0       Florida Pharma      12345.0        Undo   NaN
6  3.0  NaN            Alaska RX   12345678.0      Assign    AK
7  1.0  9.0           Ohio Drugs  123456789.0    Unicycle   NaN
8  5.0  0.0      Pennsylvania Rx    1234567.0      Assign    PA
9  NaN  0.0           USA Pharma          NaN     Unicorn   NaN
我为我的困惑道歉，但是你能解释一下把这个代码放在哪里，或者解释一下它背后的原因吗？当我把它放在循环中时，我得到了疯狂的输出。@MattR我已经添加了额外的代码来帮助您放置它。如果输出不正确，请告诉我您得到了什么。我添加了一些示例代码，以便海报可以使用我的测试数据帧。但这是我当前的输出<代码>爱达荷ID--IDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDXID，如果
len（state）=0
，就会发生这种情况。。。所以
state\u found
可能有一个空字符串
“
”。你能做
类型（状态）
？它不仅仅是州名的小写字符串吗？州是一个字符串
我为我的混淆道歉，但是你能解释一下把这个代码放在哪里，或者解释一下它背后的原因吗？当我把它放在循环中时，我得到了疯狂的输出。@MattR我已经添加了额外的代码来帮助您放置它。如果输出不正确，请告诉我您得到了什么。我添加了一些示例代码，以便海报可以使用我的测试数据帧。但这是我当前的输出<代码>爱达荷ID--IDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDIDXID，如果
len（state）=0
，就会发生这种情况。。。所以
state\u found
可能有一个空字符串
“
”。你能做
类型（状态）
？它不仅仅是州名的一个小写字符串吗？州是一个字符串
我知道这有点违反直觉，但我实际上想从州全名改为缩写版本示例：
oho->OH
@MattR，hmm…，这使得它更具挑战性。让我尝试另一种解决方案…有一些编辑，所以我不确定我之前发布到哪个，但第一部分正是我需要的。然而，我不知道你是怎么做到的！但它的工作非常出色。任何解释都会很棒，但不需要尊重您的时间和帮助@马特，我添加了一些解释-请检查我刚刚回到这篇文章。。。你的大脑完全处于另一个层次。这真是太棒了，我希望有一天能用这种方式思考……我知道这有点违反直觉，但实际上我想从全名改为缩写版本示例：
oho->OH
@MattR，hmm…，这让它更具挑战性。让我尝试另一种解决方案…有一些编辑，所以我不确定我之前发布到哪个，但第一部分正是我需要的。然而，我不知道你是怎么做到的！但它的工作非常出色。任何解释都会很棒，但不需要尊重您的时间和帮助@马特，我添加了一些解释-请检查我刚刚回到这篇文章。。。你的大脑完全处于另一个层次。这真是太棒了，我希望有一天能用这种方式思考。。。