How can I compare two data frames in pandas and update values based on keys?

I have two data frames and I want to use pandas syntax or methods to compare them and update values from the larger data frame to the smaller data frame based on similar keys.

import numpy

import pandas as pd



temp = pd.read_csv('.\..\..\test.csv')

temp2 = pd.read_excel('.\..\..\main.xlsx')



lenOfFile = len(temp.iloc[:, 1])

lenOfFile2 = len(temp2.iloc[:, 1])

dict1 = {}

dict2 = {}



for i in range(lenOfFile):

    dict1[temp.iloc[i, 0]] = temp.iloc[i, 1]



for i in range(lenOfFile2):

    dict2[temp2.iloc[i, 0]] = temp2.iloc[i, 1]



for i in dict1:

    if i in dict2:

        dict1[i] = dict2[i]

    else:

        dict1[i] = "Not in dict2"

I want the same behavior as what I wrote.

asked Jan 18 at 23:56

paul

365

add a comment |

I have two data frames and I want to use pandas syntax or methods to compare them and update values from the larger data frame to the smaller data frame based on similar keys.

import numpy

import pandas as pd



temp = pd.read_csv('.\..\..\test.csv')

temp2 = pd.read_excel('.\..\..\main.xlsx')



lenOfFile = len(temp.iloc[:, 1])

lenOfFile2 = len(temp2.iloc[:, 1])

dict1 = {}

dict2 = {}



for i in range(lenOfFile):

    dict1[temp.iloc[i, 0]] = temp.iloc[i, 1]



for i in range(lenOfFile2):

    dict2[temp2.iloc[i, 0]] = temp2.iloc[i, 1]



for i in dict1:

    if i in dict2:

        dict1[i] = dict2[i]

    else:

        dict1[i] = "Not in dict2"

I want the same behavior as what I wrote.

asked Jan 18 at 23:56

paul

365

add a comment |

I have two data frames and I want to use pandas syntax or methods to compare them and update values from the larger data frame to the smaller data frame based on similar keys.

import numpy

import pandas as pd



temp = pd.read_csv('.\..\..\test.csv')

temp2 = pd.read_excel('.\..\..\main.xlsx')



lenOfFile = len(temp.iloc[:, 1])

lenOfFile2 = len(temp2.iloc[:, 1])

dict1 = {}

dict2 = {}



for i in range(lenOfFile):

    dict1[temp.iloc[i, 0]] = temp.iloc[i, 1]



for i in range(lenOfFile2):

    dict2[temp2.iloc[i, 0]] = temp2.iloc[i, 1]



for i in dict1:

    if i in dict2:

        dict1[i] = dict2[i]

    else:

        dict1[i] = "Not in dict2"

I want the same behavior as what I wrote.

asked Jan 18 at 23:56

paul

365

I have two data frames and I want to use pandas syntax or methods to compare them and update values from the larger data frame to the smaller data frame based on similar keys.

import numpy

import pandas as pd



temp = pd.read_csv('.\..\..\test.csv')

temp2 = pd.read_excel('.\..\..\main.xlsx')



lenOfFile = len(temp.iloc[:, 1])

lenOfFile2 = len(temp2.iloc[:, 1])

dict1 = {}

dict2 = {}



for i in range(lenOfFile):

    dict1[temp.iloc[i, 0]] = temp.iloc[i, 1]



for i in range(lenOfFile2):

    dict2[temp2.iloc[i, 0]] = temp2.iloc[i, 1]



for i in dict1:

    if i in dict2:

        dict1[i] = dict2[i]

    else:

        dict1[i] = "Not in dict2"

I want the same behavior as what I wrote.

python pandas dataframe

asked Jan 18 at 23:56

paul

365

asked Jan 18 at 23:56

paul

365

asked Jan 18 at 23:56

paul

365

asked Jan 18 at 23:56

paul

365

asked Jan 18 at 23:56

paul

365

add a comment |

1 Answer
1

active

oldest

votes

You should have put a Minimal, Complete and Verifiable Example. Please, make sure in the future we can run your code just by pasting into our IDE. I spent way too much time on that question haha

import pandas as pd



temp = pd.DataFrame({'A' : [20, 4, 60, 4, 8], 'B' : [2, 4, 5, 6, 7]})

temp2 = pd.DataFrame({'A' : [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 'B' : [1, 2, 3, 10, 5, 6, 70, 8, 9, 10]})

print(temp)

print(temp2)

#     A  B

# 0  20  2

# 1   4  4

# 2  60  5

# 3   4  6

# 4   8  7



#     A   B

# 0   1   1

# 1   2   2

# 2   3   3

# 3   4  10

# 4   5   5

# 5   6   6

# 6   7  70

# 7   8   8

# 8   9   9

# 9  10  10



# Make a mapping of the values of our second mask.

mapping = dict(zip(temp2['A'], temp2['B']))



#   We apply the mapping to each row. If we find the occurence, replace, else, default.

temp['B'] = temp['A'].apply(lambda x:mapping[x] if x in mapping else 'No matching')

print(temp)

#     A            B

# 0  20  No matching

# 1   4           10

# 2  60  No matching

# 3   4           10

# 4   8            8

answered Jan 19 at 2:39

IMCoins

1,531419

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54262874%2fhow-can-i-compare-two-data-frames-in-pandas-and-update-values-based-on-keys%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

You should have put a Minimal, Complete and Verifiable Example. Please, make sure in the future we can run your code just by pasting into our IDE. I spent way too much time on that question haha

import pandas as pd



temp = pd.DataFrame({'A' : [20, 4, 60, 4, 8], 'B' : [2, 4, 5, 6, 7]})

temp2 = pd.DataFrame({'A' : [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 'B' : [1, 2, 3, 10, 5, 6, 70, 8, 9, 10]})

print(temp)

print(temp2)

#     A  B

# 0  20  2

# 1   4  4

# 2  60  5

# 3   4  6

# 4   8  7



#     A   B

# 0   1   1

# 1   2   2

# 2   3   3

# 3   4  10

# 4   5   5

# 5   6   6

# 6   7  70

# 7   8   8

# 8   9   9

# 9  10  10



# Make a mapping of the values of our second mask.

mapping = dict(zip(temp2['A'], temp2['B']))



#   We apply the mapping to each row. If we find the occurence, replace, else, default.

temp['B'] = temp['A'].apply(lambda x:mapping[x] if x in mapping else 'No matching')

print(temp)

#     A            B

# 0  20  No matching

# 1   4           10

# 2  60  No matching

# 3   4           10

# 4   8            8

answered Jan 19 at 2:39

IMCoins

1,531419

add a comment |

You should have put a Minimal, Complete and Verifiable Example. Please, make sure in the future we can run your code just by pasting into our IDE. I spent way too much time on that question haha

import pandas as pd



temp = pd.DataFrame({'A' : [20, 4, 60, 4, 8], 'B' : [2, 4, 5, 6, 7]})

temp2 = pd.DataFrame({'A' : [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 'B' : [1, 2, 3, 10, 5, 6, 70, 8, 9, 10]})

print(temp)

print(temp2)

#     A  B

# 0  20  2

# 1   4  4

# 2  60  5

# 3   4  6

# 4   8  7



#     A   B

# 0   1   1

# 1   2   2

# 2   3   3

# 3   4  10

# 4   5   5

# 5   6   6

# 6   7  70

# 7   8   8

# 8   9   9

# 9  10  10



# Make a mapping of the values of our second mask.

mapping = dict(zip(temp2['A'], temp2['B']))



#   We apply the mapping to each row. If we find the occurence, replace, else, default.

temp['B'] = temp['A'].apply(lambda x:mapping[x] if x in mapping else 'No matching')

print(temp)

#     A            B

# 0  20  No matching

# 1   4           10

# 2  60  No matching

# 3   4           10

# 4   8            8

answered Jan 19 at 2:39

IMCoins

1,531419

add a comment |

You should have put a Minimal, Complete and Verifiable Example. Please, make sure in the future we can run your code just by pasting into our IDE. I spent way too much time on that question haha

import pandas as pd



temp = pd.DataFrame({'A' : [20, 4, 60, 4, 8], 'B' : [2, 4, 5, 6, 7]})

temp2 = pd.DataFrame({'A' : [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 'B' : [1, 2, 3, 10, 5, 6, 70, 8, 9, 10]})

print(temp)

print(temp2)

#     A  B

# 0  20  2

# 1   4  4

# 2  60  5

# 3   4  6

# 4   8  7



#     A   B

# 0   1   1

# 1   2   2

# 2   3   3

# 3   4  10

# 4   5   5

# 5   6   6

# 6   7  70

# 7   8   8

# 8   9   9

# 9  10  10



# Make a mapping of the values of our second mask.

mapping = dict(zip(temp2['A'], temp2['B']))



#   We apply the mapping to each row. If we find the occurence, replace, else, default.

temp['B'] = temp['A'].apply(lambda x:mapping[x] if x in mapping else 'No matching')

print(temp)

#     A            B

# 0  20  No matching

# 1   4           10

# 2  60  No matching

# 3   4           10

# 4   8            8

answered Jan 19 at 2:39

IMCoins

1,531419

You should have put a Minimal, Complete and Verifiable Example. Please, make sure in the future we can run your code just by pasting into our IDE. I spent way too much time on that question haha

import pandas as pd



temp = pd.DataFrame({'A' : [20, 4, 60, 4, 8], 'B' : [2, 4, 5, 6, 7]})

temp2 = pd.DataFrame({'A' : [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 'B' : [1, 2, 3, 10, 5, 6, 70, 8, 9, 10]})

print(temp)

print(temp2)

#     A  B

# 0  20  2

# 1   4  4

# 2  60  5

# 3   4  6

# 4   8  7



#     A   B

# 0   1   1

# 1   2   2

# 2   3   3

# 3   4  10

# 4   5   5

# 5   6   6

# 6   7  70

# 7   8   8

# 8   9   9

# 9  10  10



# Make a mapping of the values of our second mask.

mapping = dict(zip(temp2['A'], temp2['B']))



#   We apply the mapping to each row. If we find the occurence, replace, else, default.

temp['B'] = temp['A'].apply(lambda x:mapping[x] if x in mapping else 'No matching')

print(temp)

#     A            B

# 0  20  No matching

# 1   4           10

# 2  60  No matching

# 3   4           10

# 4   8            8

answered Jan 19 at 2:39

IMCoins

1,531419

answered Jan 19 at 2:39

IMCoins

1,531419

answered Jan 19 at 2:39

IMCoins

1,531419

answered Jan 19 at 2:39

IMCoins

1,531419

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Brtdku