二维表格格式 Python

发布于 2024-12-04 18:25:52 字数 849 浏览 0 评论 0原文

我想根据存储在字典中的数据创建二维表格式。

示例：

d = {'ID1':[('Experiment1', 40), (Experiment2, 59), (Experiment3, 65)],    
 'ID2':[('Experiment1', 68), (Experiment2, 21), (Experiment3, 39)],   
 'ID3':[('Experiment1', 57), (Experiment2, 15), (Experiment4, 99)]}

应给出下表：

    Experiment1     Experiment2     Experiment3     Experiment4

ID1 ...40.................................59.... ......................................65...................... …………''
ID2....68.................................21....... ......................................39........................................ ...''
ID3....57.................................15....... ……………………………… ......99

其中 IDx 是行标签，Experimentx 是列名称。如果数据集的列名称没有值，则应添加一个空字符串作为占位符。

有人可以帮助如何做到这一点吗？ python中有没有现成的格式可供我使用？

原文

I want to create a 2D table format from my data stored in a dictionary.

Example:

d = {'ID1':[('Experiment1', 40), (Experiment2, 59), (Experiment3, 65)],    
 'ID2':[('Experiment1', 68), (Experiment2, 21), (Experiment3, 39)],   
 'ID3':[('Experiment1', 57), (Experiment2, 15), (Experiment4, 99)]}

Should give the following table:

    Experiment1     Experiment2     Experiment3     Experiment4

ID1 ...40................................59...............................65................................''
ID2....68...............................21................................39................................''
ID3....57...............................15................................''..................................99

Where IDx are the row labels and Experimentx are the column names.
If a dataset don't have a value to the column name, then an empty string should be added as placeholder.

Can anybody help how to do this? Is there any ready format in python, which I could use?

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青柠芒果 2024-12-11 18:25:52

一个快速而肮脏的实现（修复问题中的拼写错误......）

d = {'ID1':[('Experiment1', 40), ('Experiment2', 59), ('Experiment3', 65)],    
 'ID2':[('Experiment1', 68), ('Experiment2', 21), ('Experiment3', 39)],   
 'ID3':[('Experiment1', 57), ('Experiment2', 15), ('Experiment4', 99)]}

COLUMN_WIDTH = 20
STRING_WHEN_MISSING = '""'
PADDING_STRING = "."


#first determine what are the columns - from the first line
columns = ["id"] #there is always at least the id column
columns = columns + [colName for (colName, colValue) in d.items()[0][1]]

print "".join([ col.ljust(COLUMN_WIDTH) for col in columns])

#browse the lines, order by ID Ascending
for (rowId, rowValues) in sorted(d.items(), key= lambda x: x[0].lower()):
    #print rowId
    #print rowValues
    rowValuesDict = dict(rowValues) #make it a dict with access by key
    rowValuesDict["id"] = rowId
    #print rowValuesDict
    print "".join([ str(rowValuesDict.get(col, STRING_WHEN_MISSING)).ljust(COLUMN_WIDTH, PADDING_STRING) for col in columns])

这会打印出：

id                  Experiment1         Experiment2         Experiment3         
ID1.................40..................59..................65..................
ID2.................68..................21..................39..................
ID3.................57..................15..................""..................

注意：

你的原始字典的格式有点奇怪......我希望有更多这样的东西：

d = [('ID1',{'Experiment1': 40, 'Experiment2':59, 'Experiment3':65}),    
     ('ID2',{'Experiment1': 68, 'Experiment2':21, 'Experiment3':39})]

一些评论：

对于这种类型事情，你想阅读一些Python字符串方法：center(), ljust()和 rjust()在字符串之前和/或之后添加字符以强制其总宽度。

除此之外，这个想法主要是关于循环列表/字典并提取值。

注意 dict 方法 get() 的使用，它允许有一个默认值当给定键不存在值时的值。

A quick and dirty implementation (fixing your typos in the question ...)

d = {'ID1':[('Experiment1', 40), ('Experiment2', 59), ('Experiment3', 65)],    
 'ID2':[('Experiment1', 68), ('Experiment2', 21), ('Experiment3', 39)],   
 'ID3':[('Experiment1', 57), ('Experiment2', 15), ('Experiment4', 99)]}

COLUMN_WIDTH = 20
STRING_WHEN_MISSING = '""'
PADDING_STRING = "."


#first determine what are the columns - from the first line
columns = ["id"] #there is always at least the id column
columns = columns + [colName for (colName, colValue) in d.items()[0][1]]

print "".join([ col.ljust(COLUMN_WIDTH) for col in columns])

#browse the lines, order by ID Ascending
for (rowId, rowValues) in sorted(d.items(), key= lambda x: x[0].lower()):
    #print rowId
    #print rowValues
    rowValuesDict = dict(rowValues) #make it a dict with access by key
    rowValuesDict["id"] = rowId
    #print rowValuesDict
    print "".join([ str(rowValuesDict.get(col, STRING_WHEN_MISSING)).ljust(COLUMN_WIDTH, PADDING_STRING) for col in columns])

This prints out :

id                  Experiment1         Experiment2         Experiment3         
ID1.................40..................59..................65..................
ID2.................68..................21..................39..................
ID3.................57..................15..................""..................

Note :

The format of your orginal dictionary is a bit strange ... I would expect more something like this :

d = [('ID1',{'Experiment1': 40, 'Experiment2':59, 'Experiment3':65}),    
     ('ID2',{'Experiment1': 68, 'Experiment2':21, 'Experiment3':39})]

Some Comments:

For this kind of things, you want to read about some Python string methods : center(), ljust() and rjust() that add characters before and/or after a string to force its total width.

Appart from that, the idea is mostly about looping through lists/dictionaries and extract value.

Note the use of dict method get() that allows to have a default value when no value exists for a given key.

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