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pandas-如何将变量作为列传递到嵌套循环中?

dataframe loops pandas python

Chris Macaluso · 技术社区 · 7 年前

我有一个嵌套循环,我正试图将值从列表传递到中,但它无法识别列表值。如果我替换 value[col] 任何列表值 value['OpNo'] 它起作用了。在列表值或分配代码周围是否有特定的包装器或我需要的东西?

我试过清单 col_list = ["'OpNo'", "'StationNo'", "'Spindle'", "'OpDescription'"] 然后包装 col 类标注 value[[col]]

下面的代码正在获取以下异常: KeyError: 'OpNo'

row_indexer = 0
col_indexer = 1
iloc_indexer = 0
count_row = operationData.shape[0]
col_list = ['OpNo', 'StationNo', 'Spindle', 'OpDescription']
while row_indexer < count_row:    
    value = operationData.iloc[[row_indexer],[iloc_indexer]]
    for col in col_list:
        value = value[col].values[0]
        wb['OneOpSheet'].cell(row = (row_indexer + 12), column = (col_indexer + 1)).value = value
        col_indexer = (col_indexer + 1)
    row_indexer = (row_indexer + 1)
    iloc_indexer = (iloc_indexer + 1)

3 回复 | 直到 7 年前

Mark Moretto 7 年前

我不确定这是否能帮到你,但也许它能让你找到正确的方向。您可以使用pandas.dataframe.itertuples()在数据框中的所有行之间运行,根据需要选择值。

我更进一步,创建了一个快速列标签字典来帮助同步嵌套循环。

我试着在必要的地方发表评论,但如果你不明白,请告诉我!

import pandas as pd
import openpyxl
wb = load_workbook(filename='./generic_workbook_name.xlsx')

# Created smoe data for a dataframe.
fake_data_dict = {
    'OpNo':['1','2','3','4',],
    'StationNo':['11','22','33','44',],
    'Spindle':['S1','S2','S3','S4',],
    'OpDescription':['This','is','a','description',]
    }

# Create the dataframe.
data = pd.DataFrame(fake_data_dict)

我们的数据文件:

  OpNo StationNo Spindle OpDescription
0    1        11      S1          This
1    2        22      S2            is
2    3        33      S3             a
3    4        44      S4   description

脚本的其余部分:

col_list = ['OpNo','StationNo','Spindle','OpDescription']


# Create a column label dictionary; Add 1 to index for Excel cells
col_dict = {i+1:v for i, v in enumerate(col_list)}

# Iterate over each row
for idx, row in enumerate(data.itertuples(), start = 1):
    # For each key in our column dictionary [0, 1, 2, 3]
    for key in col_dict.keys():
        print('Row: {a}\n\tColumn: {b}\n\t\tValue: {c}'.format(a = idx, b = key,
                                                               # Reduce the index by 1; Get column name based on key value.
                                                               c = data.loc[idx - 1, col_dict[key]]))

输出:

Row: 1
    Column: 1
        Value: 1
Row: 1
    Column: 2
        Value: 11
Row: 1
    Column: 3
        Value: S1
Row: 1
    Column: 4
        Value: This
Row: 2
    Column: 1
        Value: 2
Row: 2
    Column: 2
        Value: 22
Row: 2
    Column: 3
        Value: S2
Row: 2
    Column: 4
        Value: is
Row: 3
    Column: 1
        Value: 3
Row: 3
    Column: 2
        Value: 33
Row: 3
    Column: 3
        Value: S3
Row: 3
    Column: 4
        Value: a
Row: 4
    Column: 1
        Value: 4
Row: 4
    Column: 2
        Value: 44
Row: 4
    Column: 3
        Value: S4
Row: 4
    Column: 4
        Value: description

考虑到这一点,这可以简化您的脚本:

for idx, row in enumerate(data.itertuples(), start = 1):
    for key in col_dict.keys():
        wb['OneOpSheet'].cell(row = (idx + 11), column = (key + 1)).value = data.loc[idx - 1, col_dict[key]]

MarMat 7 年前

我想 value = operationData.iloc[[row_indexer],[iloc_indexer]] 返回数据帧。尝试 value = value.loc[col].values[0] .

Chris Macaluso 7 年前

我需要两个命令都在 for 完成后循环并重置两个索引器。下面的嵌套循环完成了我需要的工作:

while row_indexer < count_row:    
    for col in col_list:
        value = operationData.iloc[[row_indexer],[iloc_indexer]]
        value = value[col].values[0]
        wb['OneOpSheet'].cell(row = (row_indexer + 12), column = (col_indexer + 2)).value = value
        col_indexer += 1
        iloc_indexer += 1
    row_indexer += 1
    iloc_indexer = 0
    col_indexer = 0