我一直在为该词典编辑信息:
{'Uranus':['2750','3000','2880'],'Mercury':['46','70','57'],'Earth':['147','152 ','150'],'金星':['107','109','108'],'火星':['205','249','228'],'土星':['1350 ','1510','1430'],'木星':['741','817','779'],'海王星':['4450','4550','4500'],'冥王星' :['4440','7380','5910']}
我现在尝试写入文本文件,每一行是一个键,并且每个键和值仅用逗号分隔,因此输出文件类似于此:
Uranus,2750,3000,2880
Mercury,46,70,57
Earth,147,152,150
etc
这是我在下面的内容:
with open(OUTPUT_FILE, "w") as s:
for key in sol:
cap_every_other(key)
combined = [key, sol[key[:]]]
complete = ', '.join(combined)
s.write(complete + '\n')
尽管如此,值列表发送的不是一个字符串的错误标志。如何组合这些键和值以匹配格式?谢谢您的帮助。我肯定已经重写了50次。
[key] + item
或(效率较低)[key, *item]
您可以使用items()
直接通过键和值进行迭代
with open(OUTPUT_FILE, "w") as s:
for key, value in sol.items():
cap_every_other(key) # not sure what this line is supposed to do
complete = ', '.join([key] + value)
s.write(complete + '\n')
.to_csv()
模块中的import pandas as pd
data = your data from question
# create dataframe
df = pd.DataFrame.from_dict(data, orient='index')
# save to csv
df.to_csv('data.csv', sep=',', header=False)
。您可以使用csv.writer
将每一行写为项目列表。csv.writer
output.csv
csv