pandas 用列表中的值替换熊猫数据框中的索引值

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时间:2020-09-14 05:30:06  来源:igfitidea点击:

replace index values in pandas dataframe with values from list

pythonlistpandasloopsdataframe

提问by kiltannen

I have a dataframe and 2 lists.

我有一个数据框和 2 个列表。

the 1st list gives a set of index values from the dataframe I want to replace

第一个列表给出了我想要替换的数据帧中的一组索引值

the 2nd list gives the values I want to use

第二个列表给出了我想要使用的值

I don't want to touch any of the other values

我不想触及任何其他值

Here is the dataframe:

这是数据框:

df =  pd.DataFrame.from_dict({u'Afghanistan': 6532.0,
 u'Albania': 662.0,
 u'Andorra': 2.0,
 u'Angola': 2219.0,
 u'Antigua and Barbuda': 0.0,
 u'Argentina': 6.0,
 u'Armenia': 15.0,
 u'Australia': 108.0,
 u'Azerbaijan': 210.0,
 u'Bahamas': 0.0,
 u'Bahrain': 6.0,
 u'Bangladesh': 5098.0,
 u'Barbados': 0.0,
 u'Belarus': 21.0,
 u'Belize': 0.0,
 u'Benin': 4244.0,
 u'Bhutan': 418.0,
 u'Bolivia (Plurinational State of)': 122.0,
 u'Bosnia and Herzegovina': 43.0,
 u'Botswana': 2672.0,
 u'Brazil': 36.0,
 u'Brunei Darussalam': 42.0,
 u'Bulgaria': 46.0,
 u'Burkina Faso': 6074.0,
 u'Burundi': 18363.0,
 u'Cabo Verde': 2.0,
 u'Cambodia': 12237.0,
 u'Cameroon': 14629.0,
 u'Canada': 206.0,
 u'Central African Republic': 3207.0,
 u'Chad': 3546.0,
 u'Chile': 0.0,
 u'China': 71093.0,
 u'Colombia': 1.0,
 u'Congo': 1678.0,
 u'Cook Islands': 2.0,
 u'Costa Rica': 0.0,
 u'Croatia': 9.0,
 u'Cuba': 0.0,
 u'Cyprus': 0.0,
 u'Czechia': 9.0,
 u"C\xf4te d'Ivtheitroade": 5729.0,
 u'Democratic Republic of the Congo': 8282.0,
 u'Denmark': 14.0,
 u'Djibouti': 183.0,
 u'Dominica': 0.0,
 u'Dominican Republic': 253.0,
 u'Ecuador': 0.0,
 u'Egypt': 2633.0,
 u'El Salvador': 0.0,
 u'Eritrea': 789.0,
 u'Estonia': 9.0,
 u'Ethiopia': 1660.0,
 u'France': 10000.0,
 u'Gabon': 15.0,
 u'Gambia': 336.0,
 u'Georgia': 50.0,
 u'Ghana': 23068.0,
 u'Greece': 56.0,
 u'Grenada': 0.0,
 u'Guatemala': 0.0,
 u'Guinea': 11294.0,
 u'Guyana': 0.0,
 u'Haiti': 992.0,
 u'Honduras': 0.0,
 u'Hungary': 1.0,
 u'Iceland': 0.0,
 u'India': 38835.0,
 u'Indonesia': 3344.0,
 u'Iran (Islamic Republic of)': 11874.0,
 u'Iraq': 726.0,
 u'Israel': 36.0,
 u'Italy': 1457.0,
 u'Jamaica': 0.0,
 u'Japan': 22497.0,
 u'Jordan': 32.0,
 u'Kazakhstan': 245.0,
 u'Kenya': 21002.0,
 u'Kiribati': 0.0,
 u'Kuwait': 6.0,
 u'Kyrgyzstan': 16.0,
 u"Lao People's Democratic Republic": 332.0,
 u'Latvia': 0.0,
 u'Lebanon': 5.0,
 u'Lesotho': 660.0,
 u'Liberia': 5977.0,
 u'Lithuania': 19.0,
 u'Luxembourg': 0.0,
 u'Madagascar': 35256.0,
 u'Malawi': 304.0,
 u'Malaysia': 6187.0,
 u'Maldives': 20.0,
 u'Mali': 1578.0,
 u'Malta': 2.0,
 u'Marshall Islands': 0.0,
 u'Mauritius': 0.0,
 u'Mexico': 30.0,
 u'Micronesia (Federated States of)': 0.0,
 u'Mongolia': 925.0,
 u'Morocco': 7368.0,
 u'Mozambique': 7375.0,
 u'Myanmar': 845.0,
 u'Namibia': 469.0,
 u'Nauru': 0.0,
 u'Nepal': 9397.0,
 u'Netherlands': 1019.0,
 u'New Zealand': 65.0,
 u'Nicaragua': 0.0,
 u'Niger': 21319.0,
 u'Nigeria': 212183.0,
 u'Niue': 0.0,
 u'Norway': 0.0,
 u'Oman': 15.0,
 u'Pakistan': 2064.0,
 u'Palau': 0.0,
 u'Panama': 0.0,
 u'Papua New Guinea': 7135.0,
 u'Paraguay': 0.0,
 u'Peru': 1.0,
 u'Philippines': 7120.0,
 u'Poland': 77.0,
 u'Portugal': 45.0,
 u'Qatar': 46.0,
 u'Republic of Korea': 32647.0,
 u'Republic of Moldova': 687.0,
 u'Romania': 35.0,
 u'Russian Federation': 4800.0,
 u'Rwanda': 2095.0,
 u'Saint Kitts and Nevis': 0.0,
 u'Saint Lucia': 0.0,
 u'Saint Vincent and the Grenadines': 0.0,
 u'San Marino': 1.0,
 u'Sao Tome and Principe': 0.0,
 u'Senegal': 5839.0,
 u'Serbia': 38.0,
 u'Sierra Leone': 3575.0,
 u'Singapore': 141.0,
 u'Slovakia': 0.0,
 u'Somalia': 3965.0,
 u'South Africa': 1459.0,
 u'Spain': 152.0,
 u'Sri Lanka': 16527.0,
 u'Sudan': 2875.0,
 u'Suriname': 0.0,
 u'Swaziland': 10.0,
 u'Sweden': 59.0,
 u'Syrian Arab Republic': 146.0,
 u'Tajikistan': 192.0,
 u'Thailand': 4074.0,
 u'The former Yugoslav republic of Macedonia': 36.0,
 u'Togo': 3578.0,
 u'Tonga': 0.0,
 u'Trinidad and Tobago': 0.0,
 u'Tunisia': 47.0,
 u'Turkey': 16244.0,
 u'Turkmenistan': 113.0,
 u'Uganda': 42554.0,
 u'Ukraine': 817.0,
 u'United Arab Emirates': 69.0,
 u'United Kingdom of Great Britain and Northern Ireland': 104.0,
 u'United Republic of Tanzania': 14649.0,
 u'United States of America': 85.0,
 u'Uruguay': 0.0,
 u'Uzbekistan': 80.0,
 u'Vanuatu': 9.0,
 u'Venezuela (Bolivarian Republic of)': 22.0,
 u'Viet Nam': 16512.0,
 u'Zambia': 30930.0,
 u'Zimbabwe': 1483.0}, orient = 'index')

Here is the 1st list:

这是第一个列表:

list1 = [u'Bolivia (Plurinational State of)', u'Brunei Darussalam', u'Cabo Verde', u'China',
    u'Congo', u'Cook Islands', u'Czechia', u"C\xf4te d'Ivtheitroade", 
    u"Democratic People's Republic of Korea", u'France', u'Iran (Islamic Republic of)', 
    u"Lao People's Democratic Republic", u'Micronesia (Federated States of)', u'Niue', 
    u'Republic of Korea', u'Republic of Moldova', u'Russian Federation', u'Sao Tome and Principe', 
    u'Serbia', u'Somalia', u'Syrian Arab Republic', u'The former Yugoslav republic of Macedonia', 
    u'United Kingdom of Great Britain and Northern Ireland', u'United Republic of Tanzania', 
    u'United States of America', u'Venezuela (Bolivarian Republic of)', u'Viet Nam']

Here is the 2nd list

这是第二个列表

list2 = [u'Bolivia', u'Brunei', u'Cape Verde', u'China[1]', u'Democratic Republic of the Congo', 
    u'Cook Islands (NZ)', u'Czech Republic', u'Ivory Coast', u'North Korea', u'France[2]', 
    u'Iran', u'Laos', u'Federated States of Micronesia', u'Niue (NZ)', u'South Korea', 
    u'Moldova[3]', u'Russia', u'S\xe3o Tom\xe9 and Pr\xedncipe', u'Serbia[5]', 
    u'Somalia[6]', u'Syria', u'Macedonia', u'United Kingdom', u'Tanzania', 
    u'United States', u'Venezuela', u'Vietnam']

This is clearly the sort of thing python excels at - and I suspect a simple for loop will do it but I can't quite wrap my head around the logic (yet)

这显然是 python 擅长的事情 - 我怀疑一个简单的 for 循环可以做到,但我不能完全理解逻辑(还)

Any help gratefully appreciated!

任何帮助,不胜感激!

回答by Scott Boston

Use,

用,

df = df.rename(index=dict(zip(list1,list2)))

回答by Haleemur Ali

zip the two lists to create a dictionary that maps old names to the new names.

压缩这两个列表以创建一个将旧名称映射到新名称的字典。

use the function pandas.DataFrame.rename with with the replacements dictionary and all other default arguments

将函数 pandas.DataFrame.rename 与替换字典和所有其他默认参数一起使用

replacements = {l1:l2 for l1, l2 in zip(list1, list2)}

df2 = df.rename(replacements)